Joankessler

Chapter 4 Nuclear Physics Multiple Choice Questions Exercise 1 Section (A): Properties Of The Nucleus

Question 1. The mass number of a nucleus is

1. Always less than its atomic number
2. Always more than its atomic number
3. Equal to its atomic number
4. Sometimes more than and sometimes equal to its atomic number

Answer: 4. Sometimes more than and sometimes equal to its atomic number

Question 2. The stable nucleus that has a radius 1/3 that of os189 is –

1. ₃Li⁷
2. ₂He⁴
3. ₅B¹⁰
4. ₆C¹²

Answer: 1. ₃Li⁷

Question 3. For a uranium nucleus, how does its mass vary with volume?

1. m ∝ v
2. M ∝ 1/v
3. M ∝ \(\sqrt{V}\)
4. M ∝ v²

Answer: 1. m ∝ v

Question 4. The graph of ln (R/R₀) versus ln A (r = radius of a nucleus and a = its mass number) is

1. A straight line
2. A parabola
3. An ellipse
4. None of them

Answer: 1. A straight line

Question 5. 1 Amu is equivalent to:

1. 931 Mev
2. 0.51ev
3. 9.31 mev
4. 1.02 mev

Answer: 1. 931 Mev

Question 6. If the mass number for an element is m and the atomic number is z, then several neutrons will be :

1. M – z
2. Z –m
3. M + z
4. Z

Answer: 1. M – z

Question 7. Which of the following particles has a similar mass to an electron?

1. Proton
2. Neutron
3. Positron
4. Neutrino

Answer: 3. Positron

Question 8. Different atoms of the same element which have different masses but have the same chemical properties are called:

1. Isochoric
2. Isotope
3. Isobar
4. Isobaric

Answer: 2. Isotope

Question 9. The Mass-energy equation e = mc² was given by

1. Newton
2. Kepler
3. Einstein
4. Millikan

Answer: 3. Einstein

Question 10. The mass numbers of nuclei a and b are respectively 135 and 5. The ratio of their radii is:

1. 1: 27
2. 27 :1
3. 1 : 3
4. 3: 1

Answer: 2. 27 :1

Question 11. If the nucleus 125 13 52 ai has a nuclear radius of about 3.6 fm, then it would have its radius approximately as :

1. 6.0 FM
2. 9.6 FM
3. 12.0 FM
4. 4.8 FM

Answer: 1. 6.0 FM

Question 12. Two nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be

1. 1 : 3
2. 3: 1
3. 1/3 : 1
4. 1: 1

Answer: 4. 1/3: 1

Question 13. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2: 1. The ratio of their nuclear sizes will be:

1. 21/3 : 1
2. 1: 31/2
3. 31/2 : 1
4. 1: 21/3

Answer: 4. 1: 21/3

Question 14. If the radius of the 125 13 52 al the nucleus is estimated to be 3.6 fermis, then the radius of the nucleus is near:

1. 6 Fermi
2. 8 Fermi
3. 4 Fermi
4. 5 Fermi

Answer: 1. 6 Fermi

Question 15. The uncle of which one of the following pairs of nuclei are isotones:-

1. ₃₄Se⁷⁴, ₃₁Ga⁷¹
2. ₃₈Sr⁸⁴, ₃₈Sr⁸⁶
3. ₄₂Mo⁹²,₄₀Zr⁹²
4. ₂₀Ca⁴⁰,₁₆S³²

Answer: 1. ₃₄Se⁷⁴, ₃₁Ga⁷¹

Question 16. The range of a nuclear force is approximate –

1. 2 × 10^–10 M
2. 1.5 × 10^–20 m
3. 7.2 × 10^–4 m
4. 1.4 × 10^–15 m

Answer: 4. 1.4 × 10^–15 m

Question 17. The order of magnitude of the density of the uranium nucleus is, (m
p = 1.67 × 10–27 kg):

1. 1020 Kg m^–3
2. 1017 Kg m^–3
3. 1014 Kg m^–3
4. 1011 Kg m^–3

Answer: 2.1017 Kg m^–3

Question 18. Which has the highest penetrating power?

1. γ-Rays
2. β-Rays
3. α-Rays
4. Cathode rays

Answer: 1. γ-Rays

Question 19. The penetrating power is minimal for

1. α– Rays
2. β – Rays
3. ϒ – Rays
4. X – rays

Answer: 1. α– Rays

Chapter 4 Nuclear Physics Multiple Choice Questions Section (B): Mass Defect And Binding Energy

Question 1. Two protons are kept at a separation of 50Å. F n is the nuclear force and F e is the electrostatic force between them, then

1. F_n>>F_e
2. F_n=F_e
3. F_n<<F_e
4. F_nF_e

Answer: 3. F_n<<F_e

Question 2. Masses of nucleus, neutron, and protons are M, n m, and m p respectively. If the nucleus has been divided into neutrons and protons, then

1. M=(A-Z)m_n+Zm_p
2. M=ZM_n+(A-Z)m_p
3. M<A(A-X)m_n+Xm_p
4. M>(A-z)m_n+Zm_p

Answer: 2. M=ZM_n+(A-Z)m_p

Question 3. As the mass number A increases, the binding energy per nucleon in a nucleus

1. Increases
2. Decreases
3. Remains The Same
4. Varies In A Way That Depends On The Actual Value Of A.

Answer: 4. Varies In A Way That Depends On The Actual Value Of A.

Question 4. Which of the following is a wrong description of the binding energy of a nucleus?

1. It is the energy required to break a nucleus into its constituent nucleons.
2. It is the energy released when free nucleons combine to form a nucleus
3. It is the sum of the rest of the mass energies of its nucleons minus the rest of the mass energy of the nucleus
4. It is the sum of the kinetic energy of all the nucleons in the nucleus

Answer: 4. It is the sum of the kinetic energy of all the nucleons in the nucleus

Question 5. The energy of the reaction Li7 + p → 2 He4 is (the binding energy per nucleon in Li7 and He4 nuclei are 5.60 and 7.06 MeV respectively.)

1. 17.3 MeV
2. 1.73 MeV
3. 1.46 MeV
4. Depends On Binding Energy Of Proton

Answer: 1. 17.3 MeV

Question 6. Let Fpp, F on, and F nn denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron, and by a neutron on a neutron respectively. When the separation is 1 FM,

1. F_pp > F_pn = F _nn
2. F_pp= F_pn = F_nn
3. F_pp> F_pn > F_nn
4. F_pp< F_pn = F_nn

Answer: 2. F_pp= F_pn = F_nn

Question 7. The binding energies of two nuclei Pⁿ and Q²ⁿ and x and y joules. If 2x > y then the energy released in the reaction Pⁿ + Pⁿ → Q2ⁿ, will be

1. 2x + y
2. 2x – y
3. –(2x – y)
4. x + y

Answer: 3. –(2x – y)

Question 8. ₁H¹ + ₁H¹ + ₁H²→ X + ₁e⁰ + energy .The emitted particle is-

1. Neutron
2. Proton
3. α-particle
4. Neutrino

Answer: 3. α-particle

Question 9. In the following equation, particle X will be ₆C¹¹ → 5B¹¹ + β¹ + X

1. Neutron
2. Antineutrino
3. Neutrino
4. Proton

Answer: 3. Neutrino

Question 10. The mass of a proton is 1.0073 u and that of the neutron is 1.0087 u (u = atomic mass unit). The binding 24He energy of is (Given:- helium nucleus mass 4.0015 u)

1. 0.0305 J
2. 0.0305 erg
3. 28.4 MeV
4. 0.061 U

Answer: 3. 28.4 MeV

Question 11. The mass number of a nucleus is

1. Always Less Than Its Atomic Number
2. Always More Than Its Atomic Number
3. Sometimes Equal To Its Atomic Number
4. Sometimes Less Than And Sometimes More Than Its Atomic Number

Answer: 3. Sometimes Equal To Its Atomic Number

Question 12. For the stability of any nucleus

1. Binding Energy Per Nucleon Will Be More
2. Binding Energy Per Nucleon Will Be Less
3. Number Of Electrons Will Be More
4. None Of The Above

Answer: 1. Binding Energy Per Nucleon Will Be More

Question 13. IF Mo is the mass of an oxygen isotope 8O17, Mp and Mn are the masses of a proton and a neutron, respectively the nuclear binding energy of the isotope is

1. (M0 – 8Mp) C²
2. ( Mo – 8MP – 9Mn) C²
3. Moc²
4. (Mo – 17 Mn) C²

Answer: ( Mo – 8MP – 9Mn) C²

Question 14. If in a nuclear fusion process, the masses of the fusing nuclei are m 1 and m2 and the mass of the resultant nucleus is m 3, then

1. m₃ = |m₁ – m₂|
2. m₃ < (m₁ + m₂)
3. m₃ > (m₁ + m₂)
4. m₃ = m₁ + m₂)

Answer: 2. m₃ < (m₁ + m₂)

Question 15. M p denotes the mass of a proton and M n that of a neutron. A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M (N, Z) of the nucleus is given (c is the velocity of light)

1. M(N, Z) = NM n + ZM p + B/c²
2. M(N, Z) = NM n + ZM p – B/c²
3. M(N, Z) = NM n + ZM p + B/c²
4. M(N, Z) = NM n + ZM p – B/c²

Answer: 2. M(N, Z) = NM n + ZM p – B/c²

Question 16. In the reaction \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{HE}+{ }_0^1 \mathrm{n} .\). If the binding energies of H, H and He are respectively a, b, and c (in MEV), then the energy (in MeV released in this reaction is)

1. a + b + c
2. c + a + b
3. c – (a + b)
4. a + b + c

Answer: 3. c – (a + b)

Question 17. The binding energy of deuteron is 2.2 MeV and that of He is 28 MeV. If two deuterons are fused to 4 form one 2 He then the energy released is:-

1. 25.8 MeV
2. 23.6 MeV
3. 19.2 MeV
4. 30.2 MeV

Answer: 2. 23.6 MeV

Question 18. If the binding energy per nucleon in and nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction \(\mathrm{p}+{ }_3^7 \mathrm{Li} \rightarrow 2{ }_2^4 \mathrm{He}\) energy of proton must be :

1. 39.2 MeV
2. 28.24 MeV
3. 17.28 MeV
4. 1.46 MeV

Answer: 3. 17.28 MeV

Question 19. If Mp is the mass of an oxygen isotope 8O17, M p and MN are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is:

1. (M_ο – 8MP)C²
2. (M_o – 8MP – 9MN)C²
3. M_oC²
4. (M_o – 17M N)C²

Answer: 2. M_oC²

Question 20. Binding energy per nucleon is of the order of –

1. 7.6 eV
2. 7.6 μeV
3. 7.6 MeV
4. 7.6 KeV

Answer: 3. 7.6 MeV

Question 21. A free neutron decays to a proton but a free proton does not decay to a neutron. This is because

1. A neutron Is A Composite Particle Made Of A Proton And An Electron Whereas Proton Is a Fundamental Particle
2. Neutron Is An Uncharged Particle Whereas Proton Is A Charged Particle
3. Neutron Has Larger Rest Mass Than The Proton
4. Weak Forces Can Operate In A Neutron But Not In A Proton.

Answer: 3. Neutron Has Larger Rest Mass Than The Proton

Question 22. M P and MN are masses of proton and neutron, respectively, at rest. If they combine to form a deuterium nucleus. The mass of the nucleus will be:

1. Less Than M_p
2. Less Than (M_p + M_n)
3. Less Than (M_p + 2m_n)
4. Greater Than (M_p + 2m_n)

Answer: 2. Less Than (M_p + M_n)

Question 23. The figure shows a plot of binding energy per nucleon (B.E/A) vs mass number (A) for nuclei. Four nuclei, P, Q, R, and S are indicated on the curve. The process that would release energy is

1. R → 2S
2. P → Q + S
3. P → 2R
4. Q → R + S

Answer: 3. P → 2R

Question 24. A positron of 1MeV collides with an electron of 1 MeV and gets annihilated and the reaction produces two-ray photons. If the effective mass of each photon is 0.0016 amu, then the energy of each ray photon is about-

1. 1.5 MeV
2. 3 MeV
3. 6 MeV
4. 2 MeV

Answer: 1. 1.5 MeV

Question 25. Masses of two isobars \({ }_{29}^{64} \mathrm{Cu} \text { and }{ }_{30}^{64} \mathrm{Zn}\) 63.9298 u and 63.9292 u respectively. It can be concluded from these data that:

1. Both the isobars are stable
2. 64Zn is radioactive, decaying to 64Cu through -decay
3. 64Cu is radioactive, decaying to 64Zn through -decay
4. 64Cu is radioactive, decaying to 64Zn through -decay

Answer: 4. 64Cu is radioactive, decaying to 64Zn through -decay

Question 26. The binding energy per nucleon vs. mass number curve for nuclei is shown in the figure. W, X, Y, and Z are four nuclei indicated on the curve. The process that would release energy is :

1. Y → 2Z
2. W → X + Z
3. W → 2Y
4. X → Y + Z

Answer: 3. W → 2Y

Chapter 4 Nuclear Physics Multiple Choice Questions Section (C): Radioactive Decay And Displacement Law

Question 1. An α-particle is bombarded on 14N. As a result, a 17O nucleus is formed and a particle is emitted. This article is a

1. Neutron
2. Proton
3. Electron
4. Positron

Answer: 2. Proton

Question 2. A free neutron decays into a proton, an electron, and:

1. A neutrino
2. An antineutrino
3. An α-article
4. A α-particle

Answer: 2. An antineutrino

Question 3. The specific activity (per gm) of radium is near –

1. 1 Bq
2. 1 Ci
3. 3.7 × 1010 Ci
4. 1 mCi

Answer: 2. 3.7 × 1010 Ci

Question 4. When a β¯ -particle is emitted from a nucleus, the neutron-proton ratio :

1. Is Decreased
2. Is Increased
3. Remains The Same
4. First Then (2)

Answer: 1. Is Decreased

Question 5. In one α and 2β-emissions:

1. Mass Number Reduces By 2
2. Mass Number Reduces By 6
3. Atomic Number Reduces By 2
4. Atomic Number Remains Unchanged

Answer: 4. Atomic Number Remains Unchanged

Question 6. Which ray contains (+Ve) charge particle :

1. α-rays
2. β-rays
3. γ-rays
4. X-rays

Answer: 1. α-rays

Question 7. Which of the following cannot be bombarded to disintegrate a nucleus –

1. α-ray
2. γ-ray
3. β-ray
4. Laser

Answer: 4. Laser

Question 8. Which of the following is a correct statement?

1. Beta Rays Are the Same As Cathode Rays.
2. Gamma Rays Are High Energy Neutrons.
3. Alpha Particles Are Singly-Ionized Helium Atoms.
4. Protons And Neutrons Have Exactly The Same Mass.

Answer: 1. Beta Rays Are Same As Cathode Rays.

Question 9. A deutron is bombarded on a ₇O¹⁶ nucleus then α-particle is emitted then the product nucleus is :

1. ₇N¹²
2. ₅B¹⁰
3. ₄Be⁹
4. ₇N¹⁴

Answer: 4. ₇N¹⁴

Question 10. An α– particle is bombarded on N¹⁴ As. a result, an O¹⁷ -nucleus is formed and a particle X is emitted. The particle X is

1. Neutron
2. Proton
3. Electron
4. Positron

Answer: 2. Proton

Question 11. In the reaction \({ }_{92} X^{234} \longrightarrow_{87} Y^{222}\) How many α-particles and β-particles are emitted?

1. 3 and 5
2. 5 and 3
3. 3 and 3
4. 3 and 1

Answer: 4. 3 and 1

Question 12. Which of the following radiations has the least wavelength?

1. γ-rays
2. β-rays
3. α-rays
4. X-rays

Answer: 1. γ-rays

Question 13. When U 238 nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is

1. \(\frac{4 u}{238}\)
2. \(-\frac{4 u}{234}\)
3. \(\frac{4 u}{234}\)
4. \(-\frac{4 \mathrm{u}}{238}\)

Answer: 3. \(\frac{4 u}{234}\)

Question 14. A nucleus with Z = 92 emits the following in a sequence : α, α, β–, β–, α, α, α, α; β–, β–, α. The Z of the resulting nucleus is:

1. 76
2. 78
3. 82
4. 74

Answer: 2. 78

Question 15. A nuclear reaction given by \(X^A \rightarrow_{z+1} Y^A+{ }_{-1} \mathrm{e}^0+\bar{v}\)

1. β-decay
2. γ-decay
3. fusion
4. fission

Answer: 1. β-decay

Question 16. In the radioactive decay process, the negatively charged emitted  – particles are

1. The Electrons Present Inside The Nucleus
2. The Electrons Produced Inside As A Result Of The Decay Of Neutrons Inside The Nucleus
3. The Electrons Produced As A Result Of Collisions Between Atoms
4. The Electrons Orbiting Around The Nucleus

Answer: 2. The Electrons Produced Inside As A Result Of The Decay Of Neutrons Inside The Nucleus

Question 17. Ub the disintegration series \(\underset{92}{238} \mathrm{U}{\alpha} X \xrightarrow{\beta^{-}} Y \underset{Z}{A}\) the values of Z and A respectively will be

1. 92,236
2. 88,230
3. 90,234
4. 31,234

Answer: 1. 92,236

Question 18. A nucleus represented by the symbol has:-

1. Z protons and A – Z neutrons
2. Z protons and A neutrons
3. A protons and Z – A neutrons
4. Z protons and A – Z PROTONS

Answer: 3. A protons and Z – A neutrons

Question 19. In the radioactive decay process, the negatively charged emitted -particles are:

1. The electrons present inside the nucleus
2. The electrons produced as a result of the decay of neutrons inside the nucleus
3. The electrons produced as a result of collisions between atoms
4. The electrons orbiting around the nucleus

Answer: 3. The electrons produced as a result of collisions between atoms

Question 20. A nuclear transformation is denoted by \(\mathrm{X}(\mathrm{n}, \alpha) \rightarrow{ }_3^7 \mathrm{Li}\). Which of the following is the nucleus of element X?

1. \({ }_6^{12} \mathrm{C}\)
2. \({ }_5^{10} \mathrm{~B}\)
3. \({ }_5^9 \mathrm{~B}\)
4. \({ }_4^{11} \mathrm{Be}\)

Answer: 2. \({ }_5^{10} \mathrm{~B}\)

Question 21. When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are 4Be⁸, the emitted particles will be

1. Neutrons
2. Alpha Particles
3. Beta Particles
4. Gamma Photons

Answer: 4. Gamma Photons

Question 22. The ‘rad’ is the correct unit used to report the measurement of

1. The Rate Of Decay Of Radioactive Source
2. The Ability Of A Beam Of Gamma Ray Photons To Produce Ions In A Target
3. The Energy Delivered By Radiation To A Target.
4. The Biological Effect Of Radiation

Answer: 4. The Biological Effect Of Radiation

Question 23. In gamma-ray emission from a nucleus:

1. Both The Neutron Number And The Proton Number Change
2. There Is No Change In The Proton Number And The Neutron Number
3. Only The Neutron Number Changes
4. Only The Proton Number Changes

Answer: 2. There Is No Change In The Proton Number And The Neutron Number

Question 24. Bombardment of a neutron \({ }_0 \mathrm{n}^1+{ }_5 \mathrm{~B}^{10} \rightarrow{ }_2 \mathrm{He}^4+\mathrm{x}\) on boron, forms a nucleus x with emission of  particle Nuclear x is –

1. ₆C¹²
2. ₃Li⁶
3. ₃Li⁷
4. ₄Be⁹

Answer: 3. ₃Li⁷

Question 25. 22Ne nucleus, after absorbing energy, decays into two -particles and an unknown nucleus. The unknown nucleus is:

1. Nitrogen
2. Carbon&14
3. Boron&12
4. Oxygen

Answer: 2. Carbon&14

Question 26. Consider a sample of a pure beta-active material

1. All the beta particles emitted have the same energy
2. The beta particles originally exist inside the nucleus and are ejected at the time of beta decay
3. The antineutrino emitted in a beta decay has zero rest mass and hence zero momentum.
4. The active nucleus changes to one of its isobars after the beta decay

Answer: 4. The active nucleus changes to one of its isobars after the beta decay

Question 27. \(\mathrm{X}+\mathrm{n} \rightarrow \alpha+{ }_3 \mathrm{Li}^7\) then X will be :-

1. \({ }_5^{10} \mathrm{~B}\)
2. \({ }_5^9 \mathrm{~B}\)
3. \({ }_4^{11} \mathrm{~B}\)
4. \({ }_2^4 \mathrm{He}\)

Answer: 1. \({ }_5^{10} \mathrm{~B}\)

Question 28. M n and M p represent the mass of neutron and proton respectively An element having mass M has N neutron and Z-protons, then the correct relation will be:-

1. M < {N.mn + Z.Mp}
2. M > {N.mn + Z.M p}
3. M = {N.mn + Z.M p}
4. M = N {.mn + Mp}

Answer: 1. M < {N.mn + Z.Mp}

Question 29. In the nucleus of an atom, neutrons are in excess, then emitted particles are:

1. Neutron
2. Electron
3. Proton
4. Positron

Answer: 2. Electron

Question 30. A nuclei X with mass number A and charge number Z disintegrates into one α-particle and one β-particle. The resulting R has atomic mass and atomic number, equal to:

1. (A – Z) and (Z – 1)
2. (A – Z) and (Z – 2)
3. (A – and (A – 2)
4. (A – and (Z – 1)

Answer: 4. (A – and (Z – 1)

Question 31. In gamma-ray emission from a nucleus

1. Both The Neutron Number And The Proton Number Change
2. There Is No Change In The Proton Number And The Neutron Number
3. Only The Neutron Number Changes
4. Only The Proton Number Changes

Answer: 1. Both The Neutron Number And The Proton Number Change

Question 32. Which one of the following is a possible nuclear reaction?

1. \({ }_5^{10} \mathrm{~B}+{ }_2^4 \mathrm{He} \longrightarrow{ }_7^{13} \mathrm{~N}+{ }_1^1 \mathrm{H}\)
2. \({ }_{11}^{23} \mathrm{Na}+{ }_1^1 \mathrm{H} \longrightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_2^4 \mathrm{He}\)
3. \({ }_{93}^{239} \mathrm{~Np} \longrightarrow{ }_{94}^{239} \mathrm{pu}+\mathrm{B}^{-}+\overline{\mathrm{v}}\)
4. \({ }_7^{11} \mathrm{~N}+{ }_1^1 \mathrm{H} \longrightarrow{ }_6^{12} \mathrm{C}+\mathrm{B}^{-}+\overline{\mathrm{v}}\)

Answer: 3. \({ }_{93}^{239} \mathrm{~Np} \longrightarrow{ }_{94}^{239} \mathrm{pu}+\mathrm{B}^{-}+\overline{\mathrm{v}}\)

Question 33. In an α-decay, the Kinetic energy of the Q particle is 48 MeV and the Q-value of the reaction is 50 MeV. The mass number of the mother nucleus is:- (Assume that the daughter nucleus is in the ground state)

1. 96
2. 100
3. 104
4. None of these

Answer: 2. 100

Question 34. Protons and singly ionized atoms of U²³⁵ & U²³⁸ are passed in turn (which means one after the other and not at the same time) through a velocity selector and then enter a uniform magnetic field. The protons describe semicircles of radius 10 mm. The separation between the ions of U235 and U238 after describing the semicircle is given by

1. 60mm
2. 30mm
3. 2350mm
4. 2380mm

Answer: 1. 60mm

Question 35. Which of the following processes represents a gamma decay?

1. \({ }^A X_Z+\gamma \longrightarrow{ }^A X_{Z-1}+a+b\)
2. \({ }^A X_Z+{ }^1 n_0 \longrightarrow A-3 X_{Z-2}+c\)
3. \({ }^A X_Z \longrightarrow{ }^A X_Z+f\)
4. \({ }^A X_Z \longrightarrow{ }^A X_Z+f\)

Answer: 3. \({ }^A X_Z \longrightarrow{ }^A X_Z+f\)

Question 36. A nucleus with mass number 220 initially at rest emits an α-particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the α-particle

1. 4.4 MeV
2. 5.4 MeV
3. 5.6 MeV
4. 6.5 MeV

Answer: 2. 5.4 MeV

Chapter 4 Nuclear Physics Multiple Choice Questions Section (D): Statistical Law Of Radioactive Decay

Question 1. In one average-life

1. Half The Active Nuclei Decay
2. Less Than Half The Active Nuclei Decay
3. More Than Half The Active Nuclei Decay
4. All The Nuclei Decay

Answer: 3. More Than Half The Active Nuclei Decay

Question 2. A freshly prepared radioactive source of half-life 2h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is –

1. 6 h
2. 12 h
3. 24 h
4. 128 h

Answer: 2. 12 h

Question 3. 10 grams of 57Co kept in an open container decays β–a particle with a half-life of 270 days. The weight of the material inside the container after 540 days will be very nearly –

1. 10 g
2. 7.5 g
3. 5 g
4. 2.5 g

Answer: 2. 7.5 g

Question 4. After a time equal to four half-lives, the amount of radioactive material remaining undecayed is –

1. 6.25 %
2. 12.50 %
3. 25.0 %
4. 50.0 %

Answer: 1. 6.25 %

Question 5. The decay constant of the parent nuclide in the Uranium series is. Then the decay constant of the stable end product of the series will be –

1. λ/238
2. λ/206
3. λ/208
4. zero

Answer: 4. zero

Question 6. The half-life of thorium (Th 23is 1.4 × 1010 years. Then the fraction of thorium atoms decaying per year is very nearly –

1. 1 × 10–11μ
2. 4.95 × 10–11
3. 0.69 × 10–11
4. 7.14 × 10–11

Answer: 1. 1 × 10–11

Question 7. The half-life of 215At is 100 μs. The time taken for the radioactivity of a sample of 215At to decay to 1/16th of its initial value is :

1. 400 μs
2. 6.3 μs
3. 40 μs
4. 300 μs

Answer: 4. 300 μs

Question 8. Two identical samples (same material and same amount) P and Q of a radioactive substance having mean life T are observed to have activities A P & AQ respectively at the time of observation. If P is older than Q, then the difference in their ages is:

1. \(\mathrm{T} \ell \mathrm{n}\left(\frac{\mathrm{A}_{\mathrm{P}}}{\mathrm{A}_{\mathrm{Q}}}\right)\)
2. \(T \ell n\left(\frac{A_Q}{A_P}\right)\)
3. \(\frac{1}{T} \ln \left(\frac{A_P}{A_Q}\right)\)
4. \(T\left(\frac{A_P}{A_Q}\right)\)

Answer: 1. \(\mathrm{T} \ell \mathrm{n}\left(\frac{\mathrm{A}_{\mathrm{P}}}{\mathrm{A}_{\mathrm{Q}}}\right)\)

Question 9. Two radioactive sources A and B initially contain an equal number of radioactive atoms. Source A has a half-life of 1 hour and source B has a half-life of 2 hours. At the end of 2 hours, the ratio of the rate of disintegration of A to that of B is:

1. 1: 2
2. 2: 1
3. 1: 1
4. 1: 4

Answer: 4. 1: 4

Question 10. If 10% of a radioactive material decays in 5 days then the amount of original material left after 15 days is about –

1. 65%
2. 73%
3. 70%
4. 63%

Answer: 2. 73%

Question 11. The half-life of radioactive Polonium (Po) is 138.6 days. For ten lakh Polonium atoms, the number of disintegrations in 24 hours is –

1. 2000
2. 3000
3. 4000
4. 5000

Answer: 4. 5000

Question 12. Half-lives of two radioactive substances A and B are respectively 20 min and 40 min. Initially, the samples of A and B have an equal number of nuclei. After 80 min the ratio of the remaining number of A and B nuclei is:

1. 1: 16
2. 4: 1
3. 1: 4
4. 1: 1

Answer: 1. 1: 16

Question 13. A nucleus n Xm emits one and two particles. The resulting nucleus is :

1. Nx^m–4
2. N –2 Y^{m – 4}
3. N – 4 Z ^{m – 4}
4. None Of These

Answer: 3. N – 4 Z ^{m – 4}

Question 14. The half-life of a radioactive element is 12.5 Hours and its quantity is 256 gm. After how much time its \quantity will remain 1 gm:-

1. 50 Hrs
2. 100 Hrs
3. 150 Hrs
4. 200 Hrs

Answer: 2. 100 Hrs

Question 15. If the half-life of a substance is 38 days and its quantity is 10.38 g. The quantity remaining after 19 days will be:

1. 0.151g
2. 7.0 g
3. 0.51g
4. 0.16 g

Answer: 1. 0.151g

Question 16. Remaining quantity (in %) of the radioactive element after 5 half-lives:

1. 4.125%
2. 3.125%
3. 31.1%
4. 42.125%

Answer: 2. 3.125%

Question 17. When neutrons are bvaomardexd on ⎯⎯→ 5B10 then +on1 X + 2He4, X is :

1. 3Li7
2. 3Li6
3. 5Be8
4. 2Li7

Answer: 1. 3Li7

Question 18. A sample of radioactive element containing 4 × 1016 active nuclei. The half-life of the element is 10 days, then the number of decayed nuclei after 30 days:-

1. 0.5 × 1016
2. 2 × 1016
3. 3.5 × 1016
4. 1 × 1016

Answer: 3. 3.5 × 1016

Question 19. A sample of radioactive element has a mass of 10 gm at an instant t = 0. The approximate mass of this element in the sample after two mean lives is:_

1. 1.35 gm
2. 2.50 gm
3. 3.70 gm
4. 6.30 gm

Answer: 1. 1.35 gm

Question 20. The decay constant of a radioactive substance is. The life and mean life of substance are respectively given by

1. \(\frac{1}{\lambda} \text { and } \frac{\log _e 2}{\lambda}\)
2. \(\frac{\log _e 2}{\lambda} \text { and } \frac{1}{\lambda}\)
3. \(\frac{\lambda}{\log _e 2} \text { and } \frac{1}{\lambda}\)
4. \(\frac{\lambda}{\log _e 2} \text { and } 2 \lambda\)

Answer: 2. \(\frac{\log _e 2}{\lambda} \text { and } \frac{1}{\lambda}\)

Question 21. The half-life of a certain radioactive substance is 12 days. The time taken for 8th of the sample to decay is

1. 36 days
2. 12 days
3. 4 days
4. 24 days

Answer: 1. 36 days

Question 22. If N 0 is the original mass of the substance of half-life period tl/2 = 5 years, then the amount of substance left after 15 years is:

1. N 0 / 8
2. N 0 / 16
3. N 0 / 2
4. N 0 / 4

Answer: 1. N 0 / 8

Question 23. The atomic bomb was first made by

1. Otto Hahn
2. Fermi
3. Oppenheimer
4. Taylor

Answer: 1. Otto has

Question 24. A radioactive element has a half-life of 3.6 days. At what time will it be left 1/32nd undecayed?

1. 4 days
2. 12 days
3. 18 days
4. 24 days

Answer: 3. 18 days

Question 25. If a sample of 16 g radioactive substance disintegrates to 1g in 120 days, then what will be the half-life of the sample?

1. 15 days
2. 7.5 days
3. 30 days
4. 60 days

Answer: 3. 30 days

Question 26. n α-particles per second are being emitted by N atoms of a radioactive element. The half-life of element will be

1. \(\left(\frac{\mathrm{n}}{\mathrm{N}}\right) \mathrm{s}\)
2. \(\left(\frac{N}{n}\right) s\)
3. \(\frac{0.693 \mathrm{~N}}{n} s\)
4. \(\frac{0.693 n}{N} s\)

Answer: 3. \(\frac{0.693 \mathrm{~N}}{n} s\)

Question 27. In a sample of radioactive material, what percentage of the initial number of active nuclei will decay during one mean life?

1. 37%
2. 50%
3. 63%
4. 69.3%

Answer: 3. 63%

Question 28. If the half-life of any sample of a radioactive substance is 4 days, then the fraction of the sample will remain undecayed after 2 days and will be

1. \(\sqrt{2}\)
2. \(\frac{1}{\sqrt{2}}\)
3. \(\frac{\sqrt{2}-1}{\sqrt{2}}\)
4. \(\frac{1}{2}\)

Answer: 2. \(\frac{1}{\sqrt{2}}\)

Question 29. A nucleus with Z = 92 emits the following in a sequence: α β–, β –, α α α β-; β–, β–, α, α+, β+, α: The Z of the resulting nucleus is

1. 76
2. 80
3. 82
4. 74

Answer: 2. 80

Question 30. If a radioactive substance decays \(\frac{1}{16} \text { th }\) of its original amount in 2 h, then the half-life of that substance is

1. 15 min
2. 30 min
3. 45 min
4. None Of These

Answer: 2. 30 min

Question 31. If N 0 is the original mass of the substance of half-life period T1/2 = 5 ye, then the amount of substance left after 15 yr is

1. N 0 /8
2. N 0/16
3. N 0/2
4. N 0/4

Answer: 1. N 0 /8

Question 32. The half-life of radium is about 1600 years. Of 100g of radium existing now, 25g will remain undocked after:-

1. 6400 years
2. 2400 years
3. 3200 years
4. 4800 years

Answer: 3. 3200 years

Question 33. In a radioactive material the activity at time t1 is R1 and at a later time t2, it is R2. If the decay constant of the material is , then

1. R₁ = R₂ e–λ(t₁–t₂)
2. R₁ = R₂ eλ(t₁–t₂)
3. R₁ = R₂ (t₁ / t₂)
4. R₁ = R₂

Answer: 1.R₁ = R₂ e–λ(t₁–t₂)

Question 34. The atomic bomb was first made by

1. Otto Hahn
2. Fermi
3. Oppenheimer
4. Taylor

Answer: 1. Otto hahn

Question 35. Two radioactive materials X₁ and X₂ have decay constants 5λ and λ  respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X₁ to that of X₂ will be e after a time

1. \(\frac{1}{2} \lambda\)
2. \(\frac{1}{4 \lambda}\)
3. \(\frac{e}{\lambda}\)

Answer: 3. \(\frac{e}{\lambda}\)

Question 36. Starting with a sample of pure 66Cu, 7/8 of it decays into Zn in 15 minutes. The corresponding half-life is:

1. 10 minute
2. 15 minute
3. 5 minute
4. 7 minute

Answer: 3. 5 minute

Question 37. The activity of the Po sample is 5 millicuries. Half-life of Po is 138 days, what amount of Po was initially taken? (Avogadro’s no. = 6.02×1026 Per k mole)

1. 3.18 × 1015 atoms
2. 3.18 × 1013 atoms
3. 3.18 × 1016 atoms
4. 3.18 × 1014 atoms

Answer: 1. 3.18 × 1015 atoms

Question 38. The half-life period of a radioactive element X is the same as the mean-life time of another radioactive element Y. Initially both of them have the same number of atoms. Then

1. X and Y have the same decay rate initially
2. X and Y decay at the same rate always
3. Y will decay at a faster rate than X
4. X will decay at a faster rate than Y

Answer: 3. Y will decay at a faster rate than X

Question 39. A radioactive element ThA (₈₄Po²¹⁶) can undergo α and βis a type of disintegration with half-lives, T1 and T 2 respectively. Then the half-life of ThA is

1. T1 + T2
2. T1 T2
3. T1 – T2
4. \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\mathrm{~T}_1+\mathrm{T}_2}\)

Answer: 4. \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\mathrm{~T}_1+\mathrm{T}_2}\)

Question 40. The mean lives of radioactive substances are 1620 years and 405 years for -emission and -emission respectively. Then the time in which three-fourths of a sample will decay is –

1. 224 years
2. 324 years
3. 449 years
4. 810 years

Answer: 3. 449 years

Question 41. Two radioactive materials X 1 and X2 have decay constants 10 and  respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X 1 to that of X2 will be 1/e after a time.

1. 1/(10)
2. 1/(11)
3. 11/(10)
4. 1/(9)

Answer: 4. 1/(10)

Question 42. A sample of radioactive material has mass m, decay constant, and molecular weight M. Avogadro constant = N The initial activity of the sample is

1. \(\frac{\lambda \mathrm{m}}{\mathrm{M}}\)
2. \(\frac{\lambda \mathrm{mN}_{\mathrm{A}}}{\mathrm{M}}\)
3. \(m N_A e^\lambda\)

Answer: 3. \(m N_A e^\lambda\)

Question 43. The activity of a radioactive element is 103 dis/sec. Its half-life is 1 sec. After 3 sec. its activity will be :

1. 1000 dis/sec
2. 250 dis/sec
3. 125 dis/sec
4. none of these

Answer: 3. 125 dis/sec

Question 44. A 280-day-old sample of a radioactive substance has the activity of 6000 DPS. In the next 140 days, activity falls to 3000 dps. The initial activity of the sample would have been

1. 9000
2. 24000
3. 12,000
4. 18,000

Answer: 2. 24000

Question 45. A radioactive sample at any instant has a disintegration rate of 5000 disintegrations per minute. After
5 min, the rate is 1250 disintegrations per min. then the decay constant (per – minute) is

1. 0.4 In 2
2. 0.2 In2
3. 0.1 In 2
4. 0.8 In2

Answer: 4. 0.8 In2

Question 46. A radioactive sample consists of two distinct species having an equal number of atoms initially. The mean lifetime of one species is and that of the other is 5. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figures best represents the form of this plot?

Question 47. A 280-day-old sample of a radioactive substance has an activity of 6000 DPS. In the next 140 days, activity falls to 3000 dps. The initial activity of the sample would have been

1. 9000
2. 24000
3. 12,000
4. 18,000

Answer: 1. 9000

Question 48. A radioactive sample at any instant has a disintegration rate of 5000 disintegrations per minute. After 5 min, the rate is 1250 disintegrations per minute. then the decay constant (per – minute) is

1. 0.4 In 2
2. 0.2 In2
3. 0.1 In 2
4. 0.8 In2

Answer: 1. 0.4 In 2

Question 49. Radioactivity is –

1. Irreversible Process
2. Spontaneous Disintegration Process
3. Not Effected By Temperature Or Pressure
4. Process Obeying All Of The Above

Answer: 4. Process Obeying All Of The Above

Chapter 4 Nuclear Physics Multiple Choice Questions Section (E): Nuclear Fission And Fusion

Question 1. If the mass of the fissionable material is less than the critical mass, then

1. Fission And Chain Reactions Both Are Impossible
2. Fission Is Possible But Chain Reaction Is Impossible
3. Fission Is Impossible But Chain Reaction Is Possible
4. Fission And Chain Reaction Are Possible.

Answer: 2. Fission Is Possible But Chain Reaction Is Impossible

Question 2. Which of the following materials is used for controlling the fission

1. Heavy Water
2. Graphite
3. Cadmium
4. Beryllium Oxide

Answer: 3. Cadmium

Question 3. The atomic reactor is based on

1. Controlled Chain Reaction
2. Uncontrolled Chain Reaction
3. Nuclear Fission
4. Nuclear Fusion

Answer: 1. Controlled Chain Reaction

Question 4. Thermal neutron means

1. Neutron Being Heated
2. The Energy Of These Neutrons Is Equal To The Energy Of Neutrons in a heated atom
3. These neutrons have the Energy Of A Neutron In A Nucleus At a normal temperature
4. Such Neutrons Gather Energy Released In The Fission Process

Answer: 3. These neutrons have the Energy Of A Neutron In A Nucleus At a normal temperature

Question 5. \({ }_{92} U^{235}\) nucleus ab sorbs a slow neutron and undergoes fission into 54X139 and 38Sr94 nuclei.The other particles produced in this fission process are

1. 1β and 1α
2. 2β and 1 neutron
3. 2 Neutrons
4. 3 Neutrons

Answer: 4. 3 Neutrons

Question 6. Two lithium 6Li nuclei in a lithium vapor at room temperature do not combine to form a carbon 12C nucleus because

1. A Lithium Nucleus Is More Tightly Bound Than A Carbon Nucleus
2. The Carbon Nucleus Is An Unstable Particle
3. It Is Not Energetically Favourable
4. Coulomb Repulsion Does Not Allow The Nuclei To Come Very Close

Answer: 4. Coulomb Repulsion Does Not Allow The Nuclei To Come Very Close

Question 7. Choose the true statement.

1. The energy released per unit mass is more in fission than in fusion
2. The energy released per atom is more in fusion than in fission.
3. The energy released per unit mass is more in fusion and that per atom is more in fission.
4. Both fission and fusion produce the same amount of energy per atom as well as per unit mass.

Answer: The energy released per unit mass is more in fusion and that per atom is more in fission

Question 8. A fusion reaction is possible at high temperatures because –

1. Atoms Are Ionised At High Temperature
2. Molecules Break-Up At High Temperature
3. Nuclei Break-Up At High Temperature
4. Kinetic Energy Is High Enough To Overcome Repulsion Between Nuclei.

Answer: 4. Kinetic Energy Is High Enough To Overcome Repulsion Between Nuclei.

Question 9. In a uranium reactor whose thermal power is P = 100 MW, if the average number of neutrons liberated in each nuclear splitting is 2.5. Each splitting is assumed to release an energy E = 200 MeV. The number of neutrons generated per unit of time is –

1. 4 × 1018 s–1
2. 8 × 1023 s–1
3. 8 × 1019 s–1
4. 16 × 1018 s–1

Answer: 4. 16 × 1018 s–1

Question 10. Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice given below.

1. Fusion Of Two Nuclei With Mass Numbers Lying In The Range Of 1 < A < 50 Will Release Energy
2. Fusion Of Two Nuclei With Mass Numbers Lying In The Range Of 51 < A < 100 Will Release Energy
3. Fission Of A Nucleus Lying In The Mass Range Of 100 < A < 200 Will Release Energy When Broken Into Two Equal Fragments
4. Both and (2)

Answer: 3. Fission Of A Nucleus Lying In The Mass Range Of 100 < A < 200 Will Release Energy When Broken Into Two Equal Fragments

Question 11. A fission reaction is given by \({ }_{92}^{236} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+\mathrm{x}+\mathrm{y}\) where x and y are two particle considering \({ }_{92}^{236} U\) to be at rest, the kinetic energies of the products are denoted by KXe, KSr, Kx (2MeV) and ky (2mev), respectively. let the binding energies per nucleon of ,\({ }_{92}^{236} \mathrm{U},{ }_{54}^{140} \mathrm{Xe}\) and \({ }_{38}^{94} \mathrm{Sr}\) MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s)
is(are)

1. \(x=n, y=n, K_{s r}=129 \mathrm{MeV}, K_{\mathrm{xe}}=86 \mathrm{MeV}\)
2. \(x=p, y=e-, K_{s t}=129 \mathrm{MeV}, \mathrm{K}_{\mathrm{xe}}=86 \mathrm{MeV}\)
3. \(x=p, y=n, K_{s r}=129 \mathrm{MeV}, K_{x_e}=86 \mathrm{MeV}\)
4. \(x=n, y=n, K_{s r}=86 \mathrm{MeV}, K_{x e}=129 \mathrm{MeV}\)

Answer: 1. \(x=n, y=n, K_{s r}=129 \mathrm{MeV}, K_{\mathrm{xe}}=86 \mathrm{MeV}\)

Question 12. In a fission reaction \({ }_{92}^{236} \mathrm{U} \longrightarrow{ }^{117} \mathrm{X}+{ }^{117} \mathrm{Y}+n+n\) the average binding energy per nucleon of X and Y is 8.5 MeV whereas that of 236U is 7.6 MeV. The
total energy liberated will be about \({ }_{92}^{236} \mathrm{U} \longrightarrow{ }^{117} \mathrm{X}+{ }^{117} \mathrm{Y}+n+n\)

1. 200ev
2. 2 me v
3. 200 me v
4. 20000 meV

Answer: 3. 200 me v

Question 13. Energy is released in nuclear fission due to

1. Few mass is converted into energy
2. The total binding energy of fragments is more than the B. E. of the parental element
3. The total B.E. of fragments is less than the B.E. of parental element
4. The total B.E. of fragments is equal to the B.E. of the parental elements is

Answer: 2. Total B.E. of fragments is less than the B.E. of parental element

Question 14. Boron rods in nuclear reactors are used as a:

1. Moderator
2. Control Rods
3. Coolant
4. Protective Shield

Answer: 2. Control Rods

Question 15. 200 Me V energy is obtained by fission of 1 nuclei of 92U 235, to obtain 1 kW energy number of fission per second will be:

1. 3.215 × 1013
2. 3.215 × 1014
3. 3.215 × 1015
4. 3.215 × 1016

Answer: 1. 3.215 × 1013

Question 16. The best moderator for neutrons is –

1. Beryllium Oxide
2. Pure Water
3. Heavy Water
4. Graphite

Answer: 3. Heavy Water

Question 17. Which of the following are suitable for the fusion process:-

1. Light nuclei
2. heavy nuclei
3. Element must be lying in the middle of the periodic table
4. Middle elements, which are lying on the binding energy curve

Answer: 1. Light nuclei

Question 18. Solar energy is mainly caused due to:-

1. Burning of hydrogen in the oxygen
2. Fusion of uranium present in the sun
3. Fusion of protons during the synthesis of heavier elements
4. Gravitational contraction

Answer: 3. Fusion of protons during synthesis of heavier elements

Question 19. Which one of the following acts as a neutron absorber in a nuclear reactor?

1. Cd-rod
2. Heavy water (D2O)
3. Graphite
4. Distilled water (H2O)

Answer: 1. Cd-rod

Question 20. The functions of mediators in nuclear reactors are:

1. Decrease The Speed Of Neutrons
2. Increase The Speed Of Neutrons
3. Decrease The Speed Of Electrons
4. Decrease The Speed Of Electrons

Answer: 1. Decrease The Speed Of Neutrons

Question 21. A chain reaction in the fission of uranium is possible, because:

1. Two Intermediate Sized Nuclear Fragments Are Formed
2. Three Neutrons Are Given Out In Each Fission
3. Fragments In Fission Are Radioactive
4. Large Amount Of Energy Is Released

Answer: 2. Three Neutrons Are Given Out In Each Fission

Question 22. Nuclear fusion is common to the pair:

1. Thermonuclear Rector, Uranium-Based Nuclear Reactor
2. Energy Production In Sun, Uranium-Based Nuclear Reactor
3. Energy Production Of Heavy Nuclei Hydrogen Bomb
4. Disintegration Of Heavy Nuclei Hydrogen Bomb

Answer: 3. Energy Production Of Heavy Nuclei Hydrogen Bomb

Question 23. IN any fission process the ratio \(\frac{\text { mass of fission products }}{\text { maas of parent nucleus }}\)

1. Greater than 1
2. Depends on the mass of the parent nucleus
3. Less than 1
4. Less than 1

Answer: 3. Less than 1

Question 24. Fission of nuclei is possible because the binding energy in nucleons in them-

1. Decreases with mass number at low mass numbers
2. Increases with mass number at low mass numbers
3. Decreases with mass number at high mass numbers
4. Increases with mass number at high mass numbers

Answer: 3. Decreases with mass number at high mass numbers

Question 25. In the nuclear fusion reaction, \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \longrightarrow{ }_2^4 \mathrm{He}+\mathrm{n}\) given that the repulsive potential energy between the two nuclei is ~ 7.7 × 10 –14 J, the temperature at which the gases must be heated to initiate the reaction is nearly (Boltzmann’s constant k = 1.38 × 10 –23 J/K):

1. 107K
2. 105K
3. 103K
4. 109K

Answer: 4. 109K

Question 26. The operation of a nuclear reactor is said to be critical if the multiplication factor (k) has a value

1. 1
2. 1.5
3. 2.1
4. 2.5

Answer: 1. 1

Question 27. This question contains Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.  Statement-1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. and Statement 2: For heavy nuclei, the binding energy per nucleon increases with increasing Z while for light nuclei it decreases with increasing Z.

1. Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
2. Statment-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
3. Statement-1 is true, Statement-2 is false
4. Statement-1 is false, Statement-2 is true

Answer: 3. Statement-1 is true, Statement-2 is false

Chapter 4 Nuclear Physics Multiple Choice Questions Exercise 2

Question 1. The dynamic mass of an electron having rest mass m0 and moving with speed 0.8 c is

1. 0.6 m₀
2. 0.8 m₀
3. 3 m₀
4. 1.25 m₀

Answer: 3. 3 m₀

Question 2. An alpha nucleus of energy 2 mv2 bombards a heavy closet approach for the alpha nuclieus will be proportional to

1. v²
2. 1/M
3. 1/ v⁴
4. 1 / ze

Answer: 2. 1/M

Question 3. An α-particle of energy 5 MeV is scattered through 180º by a fixed uranium nucleus. The distance of the closest approach is of the order of:

1. 1 Å
2. 10–10 cm
3. 10-12 cm
4. 10–15 cm

Answer: 3. 10-12 cm

Question 4. Helium nuclei combine to form an oxygen nucleus. The energy released in the reaction is if mO = 15.9994 amu and mHe = 4.0026 amu

1. 10.24 MeV
2. 0 MeV
3. 5.24 MeV
4. 4 MeV

Answer: 1. 10.24 MeV

Question 5. A nucleus z X has mass represented by M(A, Z). If M p and M n denote the mass of proton and neutron respectively and BE the binding energy (in MeV), then:

1. \(B E=\left[M(A, Z)-Z M_p-(A-Z) M_n\right] c^2\)
2. \(B E=\left[Z M_p+(A-Z) M_n-M(A, Z)\right] c^2\)
3. \(B E=\left[Z M_p+A M_n-M(A, Z)\right] c^2\)
4. \(B E=M(A, Z)-Z M_p-(A-Z) M_n\)

Answer: 2. \(B E=\left[Z M_p+(A-Z) M_n-M(A, Z)\right] c^2\)

Question 6. If M (A, Z), M p and M n A denote the masses of the nucleus Z X, proton, and neutron respectively in units of u (1 u = 931.5 MeV/cand BE represents its binding energy in MeV, then

1. M (A, Z) = ZM p + (A – Z) Mn – BE/c²
2. M (A, Z) = ZM p + (A – Z) Mn – BE
3. M (A, Z) = ZM p + (A – Z) Mn – BE
4. M (A, Z) = ZM p + (A – Z) Mn + BE/c2

Answer: 1. M (A, Z) = ZM p + (A – Z) Mn – BE/c²

Question 7. The binding energy per nucleon of the deuteron and helium nuclei is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is:

1. 13.9 MeV
2. 26.9 MeV
3. 23.6 MeV
4. 19.2 MeV

Answer: 3. 23.6 MeV

Question 8. ₆C¹¹ undergoes a decay by emitted β+ then writes its complete equation. Given the mass value of m(5C1= 11.011434 u, m(5B1= 11.009305 u. m e = 0.000548 u and 1 u = 931.5 MeV/c² Calculate the Q-value of reaction.

1. 0.962 Mev
2. 09.62Mev
3. 096.2MeV
4. 962.0MeV

Answer: 1. 0.962 Mev

Question 9. The atomic weight of boron is 10.81 and it has two isotopes 5B10 and 5B11 in nature would be:

1. 19: 81
2. 10: 11
3. 15: 16
4. 81: 19

Answer: 1. 19: 81

Question 10. A nucleus of mass number 332 after many disintegrations α and β radiations, decays into another nucleus whose mass number is 220 and atomic number is 86. The numbers of α and β radiations will be:

1. 4, 0
2. 3, 6
3. 3, 2
4. 2, 1

Answer: 3. 3, 2

Question 11. A radioactive sample at any instant has a disintegration rate of 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is :

1. 0.4 ln 2
2. 0.2 ln 2
3. 0.1 ln 2
4. 0.8 ln 2

Answer: 1. 0.4 ln 2

Question 12. At time t = 0, some radioactive gas is injected into a sealed vessel. At time T, some more of the same gas is injected into the same vessel. Which one of the following graphs best represents the variation of the logarithm of activity A of the gas with time t?

Answer: 2.

Question 13. N atoms of a radioactive element emit n alpha particles per second at an instant. Then the half-life of the element is

1. \(\frac{\mathrm{n}}{\mathrm{N}} \mathrm{sec} \text {. }\)
2. \(1.44 \frac{\mathrm{n}}{\mathrm{N}} \mathrm{sec} .\)
3. \(0.69 \frac{\mathrm{n}}{\mathrm{N}} \text { sec. }\)
4. \(0.69 \frac{N}{n} \text { sec. }\)

Answer: 4. \(0.69 \frac{N}{n} \text { sec. }\)

Question 14. Two isotopes P and Q of atomic weight 10 and 20, respectively are mixed in equal amounts by weight. After 20 days their weight ratio is found to be 1: 4. Isotope P has a half-life of 10 days. The half-life of isotope Q is

1. Zero
2. 5 days
3. 20 days
4. Infinite

Answer: 4. Infinite

Question 15. The half-life of a radioactive substance is 4 days. Its 100 g is kept for 16 days. After this period, the amount of substance that remained was:

1. 25 g
2. 15 g
3. 10 g
4. 6.25 g

Answer: 4. 6.25 g

Question 16. Two radioactive substances A and B have decay constants 5 and respectively. At = 0 they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be \(\left(\frac{1}{e}\right)^2\) after a time interval:

1. \(\frac{1}{4 \lambda}\)
2. 4
3. 2
4. \(\frac{1}{2 \lambda}\)

Answer: 4. \(\frac{1}{2 \lambda}\)

Question 17. The half-life period of a radio-active element X is the same as the mean lifetime of another radio-active element Y. Initially they have the same number of atoms. Then:

1. X will decay faster than Y
2. Y will decay faster than X
3. X and Y have the same decay rate initially
4. X and Y decay at the same rate always

Answer: 2. Y will decay faster than X

Question 18. How much uranium is required per day in a nuclear reactor of power capacity of 1 MW

1. 15 mg
2. 1.05 gm
3. 105 gm
4. 10.5 kg

Answer: 2. 1.05 gm

Question 19. Complete the equation for the following fission process \({ }_{92} \mathrm{U}^{235}+{ }_0 \mathrm{n}^1 \longrightarrow{ }_{38} \mathrm{Sr}^{90}+\ldots \ldots.\)

1. \({ }_{54} X^{143}+3{ }_0 n\)
2. \({ }_{54} X e^{145}\)
3. \({ }_{57} \mathrm{Xe}^{142}\)
4. \({ }_{54} \mathrm{Xe}^{142}+{ }_0 \mathrm{n}^1\)

Answer: 1. \({ }_{54} X^{143}+3{ }_0 n\)

Chapter 4 Nuclear Physics Multiple Choice Questions Part – 1: Neet / Aipmt Question (Previous Years)

Question 1. In the nuclear decay given below \({ }_Z^A X \longrightarrow{ }_{Z+1}^A Y \longrightarrow{ }_{Z-1}^{A-4} B^{\circ} \longrightarrow{ }_{Z-1}^{A-4} B \text {, }\)  the particles emitted in the sequence are

1. β,α, ϒ
2. ϒ,β,α
3. β,γ,α
4. α,β,γ

Answer: 1. β,α, ϒ

Question 2. The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an

1. Isobar Of Parent
2. Isomar Of Parent
3. Isotone Of Parent
4. Isotope Of Parent

Answer: 4. Isotope Of Parent

Question 3. A radioactive nucleus X converts into to stable nucleus Y. Half-life of X is 50 yr. Calculate the age of the radioactive sample when the radio of X and Y is 1:15.

1. 200 yr
2. 350 yr
3. 150 yr
4. 250 yr

Answer: 1. 200 yr

Question 4. The mass of a 3Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding 73Li energy per nucleon of a nucleus is nearly

1. 46 MeV
2. 5.6 MeV
3. 3.9 MeV
4. 23 MeV

Answer: 2. 5.6 MeV

Question 5. The activity of a radioactive sample is measured as N₀ counts per minute at t = 0 and N₀/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

1. \(\log _e \frac{2}{5}[latex]
2. [latex]\frac{5}{\log _e 2}\)
3. 5 log₁₀2
4. 5 log_e2

Answer: 4. 5 log₁₀2

Question 6. 2 An alpha nucleus of energy 2 bombards a heavy nuclear target of charge Ze. Then the distance of the closest approach for the alpha nucleus will be proportional to

1. \(\frac{1}{\mathrm{Ze}}\)
2. u²
3. \(\frac{1}{\mathrm{~m}}\)
4. \(\frac{1}{u^4}\)

Answer: 3. \(\frac{1}{\mathrm{~m}}\)

Question 7. The decay constant of a radioisotope is λ. If A₁ and A2 are its activities at times t₁ and t₂ respectively, the number of nuclei that have decayed during the time (t₂ – t

1. A₁t₁ – A₂t₂
2. A₁ – A₂
3. (A₁ – A₂)/λ
4. λ(A₁ – A₂)

Answer: 3. (A₁ – A₂)/λ

Question 8. The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

1. 23.6 MeV
2. 2.2 MeV
3. 28.0 MeV
4. 30.2 MeV

Answer: 1. 23.6 MeV

Question 9. Two radioactive nuclei P and Q, in a given sample decay into a stable nucleolus R. At time t = 0, many P species are 4 N₀ and that of Q are N₀. The half-life of P (for conversion to R) is 1 minute whereas that of Q is 2 minutes. Initially, there are no nuclei of R present in the sample. When the number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be –

1. 9N₀
2. 3N₀
3. 2 5N₀
4. 22N₀

Answer: 2. 3N₀

Question 10. The half life of a radioactive isotope ‘X’ is 50 years. It decays to another element ‘Y’ which is stable. The two elements ‘X’ and ‘Y’ were found to be in the ratio of 1: 15 in a sample of a given rock. The age of the rock was estimated to be:

1. 150 years
2. 200 years
3. 250 years
4. 100 years

Answer: 2. 200 years

Question 11. The power obtained in a reactor using U 235 disintegration is 1000 kW. The mass decay of U 235 per hour is:

1. 10 microgram
2. 20 microgram
3. 40 microgram
4. 1 microgram

Answer: 3. 40 microgram

Question 12. A radioactive nucleus of mass M emits a photon of frequency  and the nucleus recoils. The recoil energy will be:

1. Mc² – hν
2. h²v² / 2Mc²
3. zero
4. hν

Answer: 2. h²v² / 2Mc²

Question 13. A nucleus \({ }_n^m x\) emits one –particle and two – particles. The resulting nucleus is :

1. \({ }_{n-4}^{m-6} Z\)
2. \(\mathrm{m}_{\mathrm{m}}-6\)
3. \({ }_n^{m-4} X\)
4. \({ }_{n-2}^{m-4} Y\)

Answer: 3. \({ }_n^{m-4} X\)

Question 14. Fusion reaction takes place at high temperatures because:

1. Nuclei Break Up At High Temperature
2. Atoms Get Ionised At High Temperature
3. Kinetic Energy Is High Enough To Overcome The Coulomb Repulsion Between Nuclei
4. Molecules Break Up At High Temperature

Answer: 3. Kinetic Energy Is High Enough To Overcome The Coulomb Repulsion Between Nuclei

Question 15. If the nuclear radius of 27 Al is 3.6 Fermi, the approximate nuclear radius of 64 Cu in Fermi is:

1. 2.4
2. 1.2
3. 4.8
4. 3.6

Answer: 3. 4.8

Question 16. A mixture consists of two radioactive materials A₁ and A₂ with half-lives of 20s and 10 s respectively. Initially, the mixture has 40 g of A1 and 160 g of A₂. The amount of the two in the mixture will become equal after:

1. 60 s
2. 80 s
3. 20 s
4. 40 s

Answer: 4. 40 s

Question 17. The half life of a radioactive nucleus is 50 days. The time interval (t₂ – between the time t₂ when 3 of 1 it has decayed and the time t1 when 3 of it had decayed is:

1. 30 days
2. 50 days
3. 60 days
4. 15 days

Answer: 2. 50 days

Question 18. A certain mass of Hydrogen is changed to Helium by the process of fusion. The Mass defect in the fusion reaction is 0.02866 u. The energy liberated per u is : (given 1u = 931 MeV)

1. 26.7 MeV
2. 6.675 MeV
3. 13.35MeV
4. 2.67 MeV

Answer: 2. 6.675 MeV

Question 19. The half life of a radioactive isotope ‘X’ is 20 years. It decays to another element ‘Y’ which is stable. The two elements ‘X’ and ‘Y’ were found to be in the ratio 1: 7 in a sample of a given rock. The age of the rock is estimated to be:

1. 60 years
2. 80 years
3. 100 years
4. 40 years

Answer: 1. 60 years

Question 20. The Binding energy per nucleon of\({ }_3^7 \mathrm{Li} \text { and }{ }_2^4 \mathrm{He}\)  and nucleon are 5.60 MeV and 7.06 MeV, respectively. In 7 1 4 4 the nuclear reaction \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_2^4 \mathrm{He}+\mathrm{Q}\), the value of energy Q released is

1. 19.6MeV
2. –2.4 MeV
3. 8.4 MeV
4. 17.3 MeV

Answer: 4. 17.3 MeV

Question 21. A radioisotope ‘X’ with a half-life of 1.4 × 109 years decays to ‘Y’ which is stable. A sample of the rock from a cave was found to contain ‘X’ and ‘Y’ in the ratio 1: 7. The age of the rock is

1. 1.96 × 109 years
2. 3.92 × 109 years
3. 4.20 × 109 years
4. 8.40 × 109 years

Answer: 3. 4.20 × 109 years

Question 22. If radius of the \({ }_{12}^{27} \mathrm{Al}\) Al Te nucleus is taken to be \(\mathrm{R}_{\mathrm{Al}^{\prime}}\) the the radius of \({ }_{53}^{125} \mathrm{Te}\) nucleus is nearly:

1. \(\frac{5}{3} R_{A l}\)
2. \(\frac{3}{5} R_{A I}\)
3. \(\left(\frac{13}{53}\right)^{1 / 3} \mathrm{R}_{\mathrm{Al}}\)
4. \(\left(\frac{53}{13}\right)^{1 / 3} R_{\mathrm{Al}}\)

Answer: 1. \(\frac{5}{3} R_{A l}\)

Question 23. The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is :

1. 60
2. 15
3. 30
4. 45

Answer: 1. 60

Question 24. Radioactive material ‘A’ has a decay constant of ‘8 ’ and material ‘B’ has a decay constant of ‘λ’. Initially, they have same number of nuclei. After what time, the ratio of many nuclei of material ‘B’ to that ‘A’ will be \(\frac{1}{\mathrm{e}}\)?

1. \(\frac{1}{\lambda}\)
2. \(\frac{1}{7 \lambda}\)
3. \(\frac{1}{8 \lambda}\)
4. \(\frac{1}{9 \lambda}\)

Answer: 2. \(\frac{1}{7 \lambda}\)

Question 25. For a radioactive material, the half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is:

1. 20
2. 15
3. 30
4. 10

Answer: 1. 20

Question 26. The rate of radioactive disintegration at an instant for a radioactive sample of half life 2.2 × 109 s is 1010 s–1. The number of radioactive atoms in that sample at that instant is

1. 3.17×1020
2. 3.17×1017
3. 3.17×1018
4. 3.17×1019

Answer: 4. 3.17×1019

Question 27. The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively:

1. –3.4 eV, –3.4 eV
2. –3.4 eV, –6.8 eV
3. 3.4 eV, –6.8 eV
4. 3.4 eV, 3.4 eV

Answer: 3. 3.4 eV, –6.8 eV

Question 28. The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 Å and its− ground state energy equals –13.6 eV. If the electron in the hydrogen atom is replaced by muon [charge same as electron and mass 207 me], the first Bohr radius and ground state energy will be:

1. 0.53 × 10–13 m, –3.6 eV
2. 25.6 × 10–13 m, –2.8 eV
3. 2.56 × 10–13 m, –2.8 eV
4. 2.56 × 10–13 m, –13.6 eV

Answer: 3. 2.56 × 10–13 m, –2.8 eV

Question 29. The total energy of an electron in the nth stationary orbit of the hydrogen atom can be obtained by

1. \(E_n=\frac{13.6}{n^2} e V\)
2. \(E_n=-\frac{13.6}{n^2} e V\)
3. \(\mathrm{E}_{\mathrm{n}}=-\frac{1.36}{\mathrm{n}^2} \mathrm{eV}\)
4. En=-13.6xn2ev

Answer: 2. \(E_n=-\frac{13.6}{n^2} e V\)

Question 30. For which one of the following. Bohr model is not valid?

1. Singly Ionized Neon Atom (Ne+)
2. Hydrogen Atom
3. Singly Ionized Helium Atom (He+)
4. Deuteron Atom

Answer: 1. Singly Ionized Neon Atom (Ne+)

Question 31. What happens to the mass number and atomic number of an element when it emits -radiation?

1. The mass number decreases by four and the atomic number decreases by two.
2. Mass number and atomic number remain unchanged.
3. The mass number remains unchanged while the atomic number decreases by one.
4. Mass number increases by four and atomic number increases by two.

Answer: 2. Mass number and atomic number remain unchanged.

Question 32. The half life of a radioactive sample undergoing -decay is 1.4 × 1017s. If the number of nuclei in the sample is 2.0 × 1021, the activity of the sample is nearly:

1. 104 Bq
2. 105 Bq
3. 106 Bq
4. 103 B

Answer: 1. 104 Bq

Question 33. When a uranium isotope \({ }_{92}^{235} U\) is bombarded with a neutron it generate \({ }_{36}^{89} \mathrm{Kr}\) three neutrons and

1. \({ }_{36}^{103} \mathrm{Kr}[latex]
2. [latex]{ }_{56}^{144} B a\)
3. \({ }_{40}^{91} \mathrm{Zr}\)
4. \({ }_{36}^{101} \mathrm{Kr}\)

Answer: 2. \({ }_{56}^{144} B a\)

Question 34. The energy equivalent of 0.5 g of a substance is

1. 0.5x1013J
2. 4.5x1016J
3. 4.5x1013J
4. 1.5x1013J

Answer: 3. 4.5x1013J

Question 35. A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragmentsis 8.5 MeV. The total gain in the Binding Energy in the process is

1. 9.4MeV
2. 804Mev
3. 216MeV
4. 0.9MeV

Answer: 3. 216MeV

Question 36. A 9. Radioactive nucleus \({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}\) undergoes spontaneous decay in the sequence \({ }_{\mathrm{z}}^{\mathrm{A}} \mathrm{X} \rightarrow_{\mathrm{z}-1} \mathrm{~B} \rightarrow_{\mathrm{Z}-3} \mathrm{C} \rightarrow_{\mathrm{z}-2} \mathrm{D}\) where Z is the atomic number of element X. The possible decay particles
in the sequence are:

1. \(\alpha, \beta^{+}, \beta^{-}\)
2. \(\beta^{+}, \alpha, \beta^{-}\)
3. \(\beta^{-}, \alpha, \beta^{+}\)
4. \(\alpha, \beta^{-}, \beta^{+}\)

Answer: 2. \(\beta^{+}, \alpha, \beta^{-}\)

Question 37. The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be

1. \(\frac{1}{2 \sqrt{2}}\)
2. \(\frac{2}{3}\)
3. \(\frac{2}{3 \sqrt{2}}\)
4. 1/2

Answer: 1. \(\frac{1}{2 \sqrt{2}}\)

Chapter 4 Nuclear Physics Multiple Choice Questions Part – 2: Jee (Main) / Aieee Problems (Previous Years)

Question 1. The above is a plot of binding energy per nucleon Eb, against the nuclear mass M; A, B, C, D, and E, correspond to different nuclei. Consider four reactions:

1. A + B → C + e
2. C → A + B +e
3. D + E → F + e and
4. F → D + E + e,

Answer: 4. F → D + E + e,

Question 2. The half life of a radioactive substance is 20 minutes. The approximate time interval (t2 – between the time t2 when \(\frac{2}{3}\) of it has decayed and time t1 when \(\frac{1}{3}\) of it had decayed is:

1. 7 min
2. 14 min
3. 20 min
4. 28 min

Answer: 3. 20 min

Question 3. Statement – 1: A nucleus having energy E1 decays by b– emission to a daughter nucleus having energy E 2, but the b– rays are emitted with a continuous energy spectrum having endpoint energy E1 – E2. Statement – 2: To conserve energy and momentum in B-decay at least three particles must take part in the transformation.

1. Statement 1 is correct but statement 2 is not correct.
2. Statement-1 and statement-2 both are correct and statement-2 is the correct explanation of statement-1.
3. Statement-1 is correct, statement-2 is correct and statement-2 is not the correct explanation of statement-1.
4. Statement-1 is incorrect, and statement-2 is correct.

Answer: 2. Statement-1 and statement-2 both are correct and statement-2 is the correct explanation of statement-1.

Question 4. Assume that a neutron breaks into a proton and an electron. The energy released during this process is: (mass of neutron = 1.6725 × 10–27 kg, Mass of proton = 1.6725 × 10–27 kg, mass of electron = 9 × 10–31 kg)

1. 0.73 MeV
2. 7.10 MeV
3. 6.30 MeV
4. 5.4 MeV

Answer: 1. 0.73 MeV

Question 5. In a hydrogen-like atom, electrons make a transition from an energy level with quantum number n to another with a quantum number (n–1). If n>>1, the frequency of radiation emitted is proportional to :

1. \(\frac{1}{n}\)
2. \(\frac{1}{n^2}\)
3. \(\frac{1}{n 3 / 2}\)

Answer: 4. \(\frac{1}{n 3 / 2}\)

Question 6. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively, initially the samples have equal numbers of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be:

1. 4:1
2. 1:4
3. 5:4
4. 1:16

Answer: 3. 5:4

Question 7. A radioactive nucleus A with a half-life T, decays into a nucleus B. At t = 0, there is no nucleus B. At some time t, the ratio of the number of B to that of A is 0.3. Then, t is given by

1. \(t=\frac{T}{\log (1.3)}\)
2. \(\mathrm{t}=\frac{\mathrm{T}}{2} \frac{\log 2}{\log 1.3}\)
3. \(\mathrm{t}=\mathrm{T} \frac{\log 1.3}{\log 2}\)
4. t=T log (1.3)

Answer: 3. \(\mathrm{t}=\mathrm{T} \frac{\log 1.3}{\log 2}\)

Question 8. A sample of radioactive material A, which has an activity of 10 mCi(1 Ci = 3.7 × 10 10 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively:

1. 10 days and 40 days
2. 20 days and 5 days
3. 5 days and 10 days
4. 20 days and 10 days

Answer: 1. 10 days and 40 days

Question 9. At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio \(\frac{R_B}{R_A}\) of Their Activities after time itself decays with time \(t \text { as } e^{-3 t}\) If the half-life of A is n2, the half-life of B is :

1. \(\frac{\ell n 2}{4}\)
2. 2ln2
3. 4ln2
4. \(\frac{\ell \mathrm{n} 2}{2}\)

Answer: 2. 2ln2

Question 10. Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0, it was 1600 counts per second and at t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:

1. 150
2. 400
3. 360
4. 2000

Answer: 2. 400

Question 11. Consider the nuclear fission \(\mathrm{Ne}^{20} \rightarrow 2 \mathrm{He}^4+\mathrm{C}^{12}\)  Given that the binding energy/nucleon of Ne²⁰, He⁴, and C¹² are, respectively, 8.03 MeV, 7.07 MeV, and 7.86 MeV, identify the correct statement:

1. Energy Of 11.9 Mev Has To Be Supplied
2. 8.3 Mev Energy Will Be Released
3. Energy Of 12.4 Mev Will Be Supplied
4. Energy Of 3.6 Mev Will Be Released

Answer: 3. Energy Of 12.4 Mev Will Be Supplied

Question 12. In a radioactive decay chain, the initial nucleus is \({ }_{90}^{232} \mathrm{Th}.\) At the end there are 6 β-particles and 4 βparticles that are emitted. If the end nucleus is

1. A=202;X=8
2. A=208;Z=80
3. A=200;Z=81
4. A=208; Z=82

Answer: 4. A=208; Z=82

Chapter 4 Nuclear Physics Multiple Choice Questions Self Practice Paper

Question 1. Binding Energy per nucleon of a fixed nucleus XA is 6 MeV. It absorbs a neutron moving with KE = 2 MeV, and converts into Y at the ground state, emitting a photon of energy 1 MeV. The Binding Energy per nucleon of Y (in MeV) is

1. \(\frac{(6 A+1)}{(A+1)}\)
2. \(\frac{(6 A-1)}{(A+1)}\)
3. 7
4. \(\frac{7}{6}\)

Answer: 2. \(\frac{(6 A-1)}{(A+1)}\)

Question 2. The half life of a radioactive substance ‘A’ is 4 days. The probability that a nucleus will decay in two-half

1. \(\frac{1}{4}\)
2. \(\frac{3}{4}\)
3. \(\frac{1}{2}\)
4. 1

Answer: 3. \(\frac{1}{4}\)

Question 3. To determine the half life of a radioactive element, a student plots a graph of \(\ln \left|\frac{d N(t)}{d t}\right|\) versus t. Here \(\frac{\mathrm{dN}(\mathrm{t})}{\mathrm{dt}}\) is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is:

1. 8
2. 6
3. 7
4. 9

Answer: 1. 8

Question 4. The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 10–25 kg. The mass (in mg) of the radioactive sample is

1. 8
2. 3
3. 5
4. 1

Answer: 4. 1

Question 5. An accident in a nuclear laboratory resulted in the deposition of a certain amount of radioactive material with a half-life of 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for the safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?

1. 64
2. 90
3. 108
4. 12

Answer: 3. 108

Question 6. The energy spectrum of b -particles (number N(E) as a function of -energy E) emitted from a radioactive source is:

Answer: 4.

Question 7. The ‘rad’ is the correct unit used to report the measurement of

1. The Rate Of Decay Of Radioactive Source
2. The Ability Of A Beam Of Gamma Ray Photons To Produce Ions In A Target
3. The Energy Delivered By Radiation To A Target.
4. The Biological Effect Of Radiation

Answer: 4. The Biological Effect Of Radiation

Chapter 3 Modern Physics Multiple Choice Questions Exercise -1 Section (A): photoelectric effect

Question 1. A metal surface is illuminated by a light of a given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would be:

1. Unchanged
2. 1/16Th of the original value
3. Twice the original value
4. Four times the original value

Answer: 1. Unchanged

Question 2. Mark the correct statement: in photoelectric effect –

1. Electrons are emitted from a metal surface when light falls on it.
2. The kinetic energy of photoelectrons is more for light of longer wavelength in comparison to that due to shorter wavelength.
3. Both of the above
4. None of the above

Answer: 4. None of the above

Question 3. If the threshold wavelength of light for the photoelectric effect from the sodium surface is 6800 aº then, the work function of sodium is

1. 1.8 ev
2. 2.9 ev
3. 1.1 ev
4. 4.7 ev

Answer: 1. 1.8 ev

Question 4. When the distance of a point light source from a photocell is r1, the photoelectric current is i1, if the distance becomes r₂, then the current is i₂, and the ratio (i₁:i₂) is equal to

1. R²₂ : r²₁
2. R² : r₁
3. R₁²: r²₂
4. R₁ : r₂

Answer: 1. R²₂ : r²₁

Question 5. The maximum energy of the electrons released in photocell is independent of

1. Frequency of incident light.
2. Intensity of incident light.
3. Nature of cathode surface.
4. None of these.

Answer: 2. Intensity of incident light.

Question 6. In the photoelectric effect, we assume the photon energy is proportional to its frequency and is completely absorbed by the electrons in the metal. Then the photoelectric current

1. Decreases when the frequency of the incident photon increases.
2. Increases when the frequency of the incident photon increases.
3. Does not depend on the photon frequency but only on the intensity of the incident beam.
4. Depends both on the intensity and frequency of the incident beam.

Answer: 3. Does not depend on the photon frequency but only on the intensity of the incident beam.

Question 7. When stopping potential is applied in an experiment on the photoelectric effect, no photocurrent is observed. This means that

1. The emission of photoelectrons is stopped
2. The photoelectrons are emitted but are reabsorbed by the emitter metal
3. The photoelectrons accumulated near the collector plate
4. The photoelectrons are dispersed from the sides of the apparatus.

Answer: 2. The photoelectrons are emitted but are reabsorbed by the emitter metal

Question 8. If the frequency of light in a photoelectric experiment is doubled then the stopping potential will

1. Be doubled
2. Be halved
3. Become more than double
4. Become less than double

Answer: 3. Become more than double

Question 9. The energy of a photon of frequency ν is e = hν and the momentum of a photon of wavelength λ is p = h/λ. From this statement, one may conclude that the wave velocity of light is equal to E 3 × 108 ms–1

1. 3×10-8 ms-1
2. \(\frac{E}{p}\)
3. Ep
4. \(\left(\frac{E}{p}\right)^2\)

Answer: 2. \(\frac{E}{p}\)

Question 10. Which one of the following graphs in the figure shows the variation of photoelectric current (I) with voltage between the electrodes in a photoelectric cell?

Answer: 1.

Question 11. The collector plate in an experiment on the photoelectric effect is kept vertically above the emitter plate. The light source is put on and a saturation photocurrent is recorded. An electric field is switched on which has a vertically downward direction.

1. The photocurrent will increase
2. The kinetic energy of the electrons will increase
3. The stopping potential will decrease
4. The threshold wavelength will increase

Answer: 2. The kinetic energy of the electrons will increase

Question 12. The frequency and intensity of a light source are both doubled. Consider the following statements.

The saturation photocurrent remains almost the same.

The maximum kinetic energy of the photoelectrons is doubled.

1. Both 1 and 2 are true
2. Is true but 2 is false
3. 1 is false but 2 is true
4. both 1 and 2 are false

Answer: 2. Is true but 2 is false

Question 13. A point source of light is used in a photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential.

1. Will Increase
2. Will Decrease
3. Will Remain Constant
4. Will Either Increase Or Decrease

Answer: 3. Will Remain Constant

Question 14. A point source causes a photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal?

Answer:

Question 15. The photoelectrons emitted from a metal surface:

1. Are all at rest
2. Have the same kinetic energy
3. Have the same momentum
4. Have speeds varying from zero up to a certain maximum value

Answer: 4. Have speeds varying from zero up to a certain maximum value

Question 16. The stopping potential as a function of the frequency of incident radiation is plotted for two different photoelectric surfaces A and B. The graphs show the work function of A is

1. Greater than that of B
2. Smaller than that of B
3. Same as that of B
4. No comparison can be made from the given graphs.

Answer: 2. Smaller than that of B

Question 17. In an electron gun electrons are accelerated through a potential difference V. If e = charge of electron and m = mass of electron then maximum electron velocity will be

1. 2ev/m
2. \(\sqrt{2 \mathrm{eV} / \mathrm{m}}\)
3. \(\sqrt{2 \mathrm{~m} / \mathrm{eV}}\)

Answer: 2. \(\sqrt{2 \mathrm{eV} / \mathrm{m}}\)

Question 18. Light of wavelength 5000 Å falls on a sensitive plate with a photoelectric work function of 1.9 eV. The kinetic energy of the photoelectron emitted will be:

1. 0.58ev
2. 2.48ev
3. 1.24ev
4. 0.58ev1.16ev

Answer: 1. 0.58ev

Question 19. In the photo-emissive cell, with an exciting wavelength, the fastest electron has speed v. If the exciting wavelength is changed to 3 v1/4, the speed of the fastest emitted electron will be:

1. v (3/4)1/2
2. v (4/3)1/2
3. Less than v (4/3)1/2
4. Greater than v (4/3)

Answer: 4. Greater than v (4/3)

Question 20. When the intensity of incident light increases:

1. Photo – Current Increases
2. Photo – Current Decreases
3. Kinetic Energy Of Emitted Photoelectrons Increases
4. Kinetic Energy Of Emitted Photoelectrons Decreases

Answer: 1. Photo – Current Increases

Question 21. If the wavelength of the photo is 6000 Å, then its energy will be :

1. 0.66 eV
2. 1.66 eV
3. 2.66 eV
4. 3.5 eV

Answer: 3. 2.66 eV

Question 22. Work function a metal is 5.26 × 10–18 then its threshold wavelength will be:

1. 736.7 Å
2. 760.7 Å
3. 301 Å
4. 344.4 Å

Answer: 4. 344.4 Å

Question 23. A radio station transmits waves of wavelength 300 m. Radiation capacity of the transmitter is 10 KW. Find out the number of photons that are emitted per unit of time:

1. 1.5 × 1035
2. 1.5 × 1031
3. 1.5 × 1029
4. 1.5 × 103

Answer: 2. 1.5 × 1031

Question 24. Work function a metal is 5.26 × 10–18 then its threshold wavelength will be:

1. 736.7 Å
2. 760.7 Å
3. 301 Å
4. 344.4 Å

Answer: 2. 760.7 Å

Question 25. A radio station transmits waves of wavelength 300 m. The Radiation capacity of the transmitter is 10 KW. Find out the number of photons that are emitted per unit of time:

1. 1.5 × 1035
2. 1.5 × 1031
3. 1.5 × 1029
4. 1.5 × 103

Answer: 1. 1.5 × 1035

Question 26. The accelerating voltage of an electron gun is 50,000 volts. De-Broglie wavelength of the electron will be

1. 0.55 Å
2. 0.055 Å
3. 0.077 Å
4. 0.095 Å

Answer: 2. 0.055 Å

Question 27. photo-cell is illuminated by a source of light, which is placed at a distance d from the cell, If the sustained becomes d/2, then the number of electrons emitted per second will be:-

1. Remain same
2. Four times
3. Two times
4. One-fourth

Answer: 2. Four times

Question 28. Relation between wavelength of photon and electron of same energy is :

1. \(\lambda_{\text {ph }}>\lambda_e\)
2. \(\lambda_{\mathrm{ph}}<\lambda_{\mathrm{e}}\)
3. \(\lambda_{\mathrm{ph}}<\lambda_{\mathrm{e}}\)
4. \(\frac{\lambda_e}{\lambda_{\mathrm{ph}}}=\text { constant }\)

Answer: 1. \(\lambda_{\text {ph }}>\lambda_e\)

Question 29. The wavelength associated with an electron accelerated through a potential difference of 100 V is nearly

1. 100 Å
2. 123 Å
3. 1.23 Å
4. 0.123 Å

Answer: 4. 0.123 Å

Question 30. The work function of a photometal is 6.63 eV. The threshold wavelength is

1. 3920 Å
2. 1866 Å
3. 186.6 Å
4. 18666 Å

Answer: 2. 1866 Å

Question 31. The speed of an electron having a wavelength of 10–10 m is :

1. 4.24 × 106 m/s
2. 5.25 × 106 m/s
3. 6.25 × 106 m/s
4. 7.25 × 106 m/s

Answer: 4. 7.25 × 106 m/s

Question 32. Sodium and copper have work functions of 2.3 eV and 4.5 eV respectively. Then the ratio of threshold wavelengths is nearest to:

1. 1: 2
2. 4: 1
3. 2: 1
4. 1: 4

Answer: 3. 2: 1

Question 33. The de-Broglie wavelength :

1. Is proportional to the mass
2. Is proportional to the impulse
3. Inversely proportional to the impulse
4. Does not depend on impulse

Answer: 3. Inversely proportional to impulse

Question 34. The minimum wavelength of a photon is 5000 Å, its energy will be :

1. 2.5 eV
2. 50 eV
3. 5.48 eV
4. 7.48 eV

Answer: 1. 2.5 eV

Question 35. The wavelength associated with an electron accelerated through a potential difference of 100 V is of the order of:

1. 1.2Å
2. 10.5 Å
3. 100 Å
4. 1000 Å

Answer: 1. 1.2Å

Question 36. The slope of a graph drawn between threshold frequency and stopping potential is :

1. e
2. h
3. h/e
4. he

Answer: 3. h/e

Question 37. The photoelectric work function of a metal is 3.3 eV. The threshold frequency for this metal is approximately:

1. 3.3 × 1013 Hz
2. 8.0 × 1014 Hz
3. 1.65 × 1015 Hz
4. 9.9 × 1015 Hz

Answer: 2. 8.0 × 1014 Hz

Question 38. A particle of mass 11 × 10–12 kg is moving with a velocity of 6 × 10–7 m/s. Its de–Broglie wavelength is nearly:

1. 10–20 m
2. 10–16 m
3. 10–12 m
4. 10–8 m

Answer: 2. 10–16 m

Question 39. According to Einstein’s photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

Answer: 3.

Question 40. A photosensitive metallic surface has a work function, h v0. If photons of energy 2h v0 fall on this surface, the electrons come, out with a maximum velocity of 4 × 106 m/s. when the photon energy is increased to 5hv 0, then the maximum velocity of photoelectrons will be

1. 2 × 107 m/s
2. 2 × 107 m/s
3. 8 × 105 m/s
4. 8 × 106 m/s

Answer: 4. 8 × 106 m/s

Question 41. When photons of energy hv fall on an aluminum plate (of work function E 0), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

1. K + E0
2. 2K
3. K
4. K + hv

Answer: 4. K + hv

Question 42. The momentum of a photon of energy 1 MeV is kgm/s, will be:-

1. 0.33 × 106
2. 7 × 10–24
3. 10–22
4. 5 × 10–22

Answer: 4. 5 × 10–22

Question 43. A 5 W source emits monochromatic light of wavelength 5000 Å. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface, When the source is moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of :

1. 4
2. 8
3. 16
4. 2

Answer: 1.4

Question 44. The work function of the surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the.

1. Ultraviolet region
2. Visible region
3. Infrared region
4. X-ray region

Answer: 1. Ultraviolet region

Question 45. Radiation of energy E falls normally on a perfecting reflecting surface. The momentum transferred to the surface is:

1. E/c
2. 2E/c
3. Ec
4. E/c2

Answer: 2. 2E/c

Question 46. According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal Vs the frequency, of the incident radiation gives a straight line whose slope:

1. Depends on the nature of the metal used
2. Depends on the intensity of the radiation
3. Depends both on the intensity of the radiation and the metal used
4. Is the same for all metals and independent of the intensity of the radiation

Answer: 4. Is the same for all metals and independent of the intensity of the radiation

Question 47. The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately:

1. 540 nm
2. 400 nm
3. 310 nm
4. 220 nm

Answer: 3. 310 nm

Question 48. If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor:

1. 1/2
2. 2
3. \(\frac{1}{\sqrt{2}}\)
4. \(\sqrt{2}\)

Answer: 3. \(\frac{1}{\sqrt{2}}\)

Question 49. The time that a photoelectron comes out after the photon strikes is approximately

1. 10–1 s
2. 10–4 s
3. 10–10 s
4. 10–16 s

Answer: 3. 10–10 s

Question 50. A photon of frequency v has a momentum associated with it. If c is the velocity of light, the momentum is:

1. v/c
2. svc
3. h v/c2
4. h v/c

Answer: 4. h v/c

Question 51. The threshold frequency for a certain metal is 0. When light of frequency  = 20 is incident on it, the maximum velocity of photoelectrons is 4 x 106 m/s. If the frequency of incident radiation is increased to 5 o, then the maximum velocity of photo-electrons in m/s will be

1. (4/5) × 106
2. 2 × 106
3. 8 × 106
4. 2 × 107

Answer: 3. 8 × 106

Question 52. If the energy of a photon corresponding to a wavelength of 6000 Aº is 3.32 × 10 –19 joule, the photon energy (in joule) for a wavelength of 4000 Aº will be

1. 1.11 × 10–19
2. 2.22 × 10–19
3. 4.44 × 10–19
4. 4.98 × 10–19

Answer: 3. 4.44 × 10–19

Question 53. Light of frequency 1.5 times the threshold frequency is incident on photo-sensitive material. If the frequency is halved and intensity is doubled, the photo-current becomes

1. Quadrupled
2. Doubled
3. Halved
4. Zero

Answer: 4. Zero

Question 54. Which of the following figure, represents the variation of the particle momentum and associated de-Broglie wavelength

Answer: 4.

Question 55. The linear momenta of a proton and an electron are equal. Relative to an electron

1. The kinetic energy of the proton is more.
2. De-Broglie wavelength of the proton is more.
3. De-Broglie wavelength of the proton is less.
4. De-Broglie wavelength of proton and electron are equal.

Answer: 4. De-Broglie wavelength of proton and electron are equal.

Question 56. The maximum velocity of an electron emitted by light of wavelength incident on the surface of a metal of work function  is

1. \(\left[\frac{2(\mathrm{hc}+\lambda \phi)}{\mathrm{m} \lambda}\right]^{1 / 2}\)
2. \(\left[\frac{2(\mathrm{hc}+\lambda \phi)}{\mathrm{m} \lambda}\right]^{1 / 2}\)
3. \(\frac{2(\mathrm{hc}-\lambda \phi)}{\mathrm{m} \lambda}\)
4. \(\frac{2(\mathrm{hc}+\lambda \phi)}{\mathrm{m} \lambda}\)

Answer: 2. \(\left[\frac{2(\mathrm{hc}+\lambda \phi)}{\mathrm{m} \lambda}\right]^{1 / 2}\)

Question 57. The graph is plotted between the maximum kinetic energy of the electron with the frequency of incident photons in the Photoelectric effect. The slope of the curve will be

1. Charge of electron
2. The work function of metal
3. Planck’s constant
4. The ratio of the Planck constant and the charge of the electron

Answer: 3. Planck’s constant

Question 58. Light of frequency v is incident of photon v0. Then work function of the device will be

1. hv
2. hv0
3. h[v-v0]
4. h[v0-v]

Answer: 2. hv0

Question 59. Choose the correct equation

1. \(\frac{\mathrm{h} \lambda}{\mathrm{c}}=\mathrm{E}\)
2. \(\mathrm{h} \lambda=\frac{\mathrm{E}}{\mathrm{c}}\)
3. \(\frac{\mathrm{hc}}{\mathrm{E}}=\lambda\)
4. None of these

Answer: 3. \(\frac{\mathrm{hc}}{\mathrm{E}}=\lambda\)

Question 60. The energy of an electron with a de-Broglie wavelength of 10–10 meters, in [ev] is

1. 13.6
2. 12.27
3. 1.27
4. 150.6

Answer: 4. 150.6

Question 61. If particles are moving with the same velocity, then the maximum de-Broglie wavelength is for

1. Proton
2. α-particle
3. Neutron
4. β-particle

Answer: 4. β-particle

Question 62. The photoelectric effect can be explained by assuming that light

1. Is a form of transverse waves
2. Is a form of longitudinal waves
3. Can be polarised
4. Consists of quanta

Answer: 4. Consists of quanta

Question 63. A proton and photon both have the same energy of E = 100 K eV. The de Broglie wavelength of proton and photon be λ1 and λ2 then λ1/λ2 is proportional to –

1. E–1/2
2. E1/2
3. E–1
4. E

Answer: 2. E1/2

Question 64. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is:

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Question 65. For the photo-electric effect with incident photon wavelength λ, the stopping potential is V0. Identify the correct variation(s) of V0 with λ

Answer: 1.

Question 66. The work functions for metals A, B, and C are respectively 1.92 eV, 2.0 eV, and 5eV According to Einstein’s equation, the metals that will emit photoelectrons for radiation of wavelength 4100 Å is /are:-

1. None
2. An only
3. A and B only
4. All the three metals

Answer: 3. A and B only

Question 67. When a monochromatic source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are respectively 0.6 V and 18 mA. If the same source is placed 0.6 m away from the cell, then:

1. The Stopping Potential Will Be 0.2 V
2. The Stopping Potential Will Be 1.8 V
3. The Saturation Current Will Be 6.0 Ma
4. The Saturation Current Will Be 2.0 Ma

Answer: 4. The Saturation Current Will Be 2.0 Ma

Question 68. A cesium photocell, with a steady potential difference of 60 volts across it, is illuminated by a small bright light placed 50 cm away. When the same light is placed one meter away, the photoelectrons emerge from the photocell: (assume that the potential difference applied is sufficient to produce saturation current)

1. Each Carry One-Quarter Of Their Previous Energy
2. Each Carry One-Quarter Of Their Previous Momentum
3. Are Half As Numerous
4. Are One Quarter As Numerous

Answer: 4. Are One Quarter As Numerous

Question 69. The work function for aluminium surface is 4.2 eV and that for sodium surface is 2.0 ev. The two metals were illuminated with appropriate radiations to cause photoemission. Then :

1. Both aluminum and sodium will have the same threshold frequency
2. The threshold frequency of aluminum will be more than that of sodium
3. The threshold frequency of aluminium will be less than that of sodium
4. The threshold wavelength of aluminum will be more than that of sodium

Answer: 2. The threshold frequency of aluminum will be more than that of sodium

Question 70. A photoelectric cell is illuminated by a point source of light 1 mm away. When the source is shifted to 2m then

1. Each Emitted Electron Carries One-Quarter Of The Initial Energy
2. Number Of Electrons Emitted Is Half The Initial Number
3. Each Emitted Electron Carries Half The Initial Energy
4. Number Of Electrons Emitted Is A Quarter Of The Initial Number

Answer: 2. Number Of Electrons Emitted Is Half The Initial Number

Question 71. Light of wavelength 4000 Å is incident on a metal plate whose work function is 2eV. What is the maximum kinetic energy of the emitted photoelectron?

1. 0.5 eV
2. 1.1 eV
3. 2.0 eV
4. 1.5 eV

Answer: 2. 1.1 eV

Question 72. The maximum wavelength of radiation that can produce the photoelectric effect in a certain metal is 200 nm. The maximum kinetic energy acquired by an electron due to radiation of wavelength 100 nm will be

1. 12.4 eV
2. 6.2 eV
3. Manganin
4. Aluminium

Answer: 2. 6.2 eV

Question 73. A photoelectric cell is illuminated by a point source of light 1 m away. When the source is shifted to 2 m then-

1. Each emitted electron carries one-quarter of the initial energy.
2. Number of electrons emitted is half the initial number.
3. Each emitted electron carries half the initial energy.
4. Number of electrons emitted is a quarter of the initial number.

Answer: 4. Number of electrons emitted is a quarter of the initial number.

Question 74. A photon of light enters a block of glass after traveling through a vacuum. The energy of the photon on entering the glass block

1. Increases because its associated wavelength decreases
2. Decreases because the speed of the radiation decreases
3. Stays the same because the speed of the radiation and the associated wavelength do not change
4. Stays the same because the frequency of the radiation does not change

Answer: 4. Stays the same because the frequency of the radiation does not change

Question 75. Two separate monochromatic light beams A and B of the same intensity (energy per unit area per unit time) are falling normally on a unit area of a metallic surface. Their wavelength is respectively. Assuming that all the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam A to that from B is

1. \(\left(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}\right)\)
2. \(\left(\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}\right)\)
3. \(\left(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}\right)^2\)
4. \(\left(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}\right)^2\)

Answer: 1. \(\left(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}\right)\)

Question 76. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. The power of the pulse is 30mV and the speed of light is 3 × 10 ms–1. The final momentum of the object is:

1. 0.3 × 10–17 kg ms–1
2. 1.0 × 10–17 kg ms–1
3. 3.0 × 10–17 kg ms–1
4. 9.0 × 10–17 kg ms–1

Answer: 2. 1.0 × 10–17 kg ms–1

Question 77. When photons of energy hv fall on a photo-sensitive metallic surface (work function hv0) electrons are emitted from the metallic surface. This is known as the photoelectric effect. The electrons coming out of the surface have a K.E. It is possible to say that

1. All ejected electrons have the same K.E. equal to hv – hv0
2. The ejected electrons have a distribution of K.E., the most energetic ones having equal to hv – hv 0
3. The most energetic ejected electrons have K.E. equal to hv.
4. The K.E. of the ejected electrons is hv0

Answer: 2. The ejected electrons have a distribution of K.E., the most energetic ones having equal to hv – hv 0

Question 78. A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is 1 placed 2 m away, the number of electrons emitted by the photocathode would:

1. Decrease By A Factor Of 4
2. Increase By A Factor Of 4
3. Decrease By A Factor Of 2
4. Increase By A Factor Of 2

Answer: 1. Decrease By A Factor Of 4

Chapter 3 Modern Physics Multiple Choice Questions Section (B): De–Broglie Wave (Matterwaves)

Question 1. The ratio of de Broglie wavelengths of a proton and an alpha particle of the same energy is.

1. 1
2. 2
3. 4
4. 0.25

Answer: 2. 2

Question 2. The ratio of de Broglie wavelengths of a proton and an alpha particle moving with the same velocity is

1. 1
2. 2
3. 4
4. 0.25

Answer: 3. 4

Question 3. The ratio of de Broglie wavelengths of a proton and a neutron moving with the same velocity is nearly

1. 1
2. √2
3. 1/√2
4. None of the above

Answer: 1. 1

Question 4. Two particles have identical charges. If they are accelerated through identical potential differences, then the ratio of their de Broglie wavelength would be

1. \(\lambda_1: \lambda_2=1: 1\)
2. \(\lambda_1: \lambda_2=m_2: m_1\)
3. \(\lambda_1: \lambda_2=\sqrt{m_2}: \sqrt{m_1}\)
4. \(\lambda_1: \lambda_2=\sqrt{m_1}: \sqrt{m_2}\)

Answer: 3. \(\lambda_1: \lambda_2=\sqrt{m_2}: \sqrt{m_1}\)

Question 5. If the velocity of a moving particle is reduced to half, then the percentage change in its wavelength will be

1. 100% decrease
2. 100% increase
3. 50% decrease
4. 50% increase

Answer: 2. 100% increase

Question 6. The momentum of the r-ray photon of energy 3 keV in kg-m/s will be

1. 1.6 × 10–19
2. 1.6 × 10–21
3. 1.6 × 10–24
4. 1.6 × 10–27

Answer: 3. 1.6 × 10–24

Question 7. Which one of the following statements is NOT true for de Broglie waves?

1. All atomic particles in motion have waves of a definite wavelength associated with them
2. The higher the momentum, the longer the wavelength
3. The faster the particle, the shorter the wavelength
4. For the same velocity, a heavier particle has a shorter wavelength

Answer: 2. The higher the momentum, the longer the wavelength

Question 8. In a TV tube, the electrons are accelerated by a potential difference of 10 kV. Then, their de Broglie wavelength is nearly

1. 1.2 Å
2. 0.12 Å
3. 12 Å
4. 0.01 Å

Answer: 2. 0.12 Å

Question 9. The de Broglie waves are associated with moving particles. This article may be

1. Electrons
2. He+, li2+ ions
3. Cricket ball
4. All of the above

Answer: 4. All of the above

Question 10. What voltage must be applied to an electron microscope to produce electrons λ = 1.0 Å

1. 190 volt
2. 180 volt
3. 160 volt
4. 150 volt

Answer: 4. 150 volt

Question 11. An α-particle moves along a circular path of radius 0.83 cm in a magnetic field of 0.25 Wb/m 2. The de Broglie wavelength associated with it will be

1. 10 Å
2. 1 Å
3. 0.1 Å
4. 0.01 Å

Answer: 4. 0.01 Å

Question 12. The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is approximately – (Planck’s constant, h = 6.63 × 10–34 Js)

1. 10–33 metre
2. 10–31 metre
3. 10–16 metre
4. 10–25 metre

Answer: 1. 10–33 metre

Question 13. The de Broglie wavelength of an electron moving with a velocity of 1.5 × 108 ms–1 is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the energy of a photon is:

1. 2
2. 4
3. 1/2
4. 1/4

Answer: 4. 1/4

Question 14. Let p and E denote the linear momentum and the energy of a photon. For another photon of a smaller wavelength (in the same medium)

1. Both P And E Increase
2. P Increases And E Decreases
3. P Decreases And E Increases
4. Both P And E Decreases

Answer: 1. Both P And E Increase

Chapter 3 Modern Physics Multiple Choice Questions Section(C): Bohr’S Atomic Model Of H-Atom And H-Like Species (Properties)

Question 1. The Lyman series of hydrogen spectrum lies in the region

1. Infrared
2. Visible
3. Ultraviolet
4. Of x – rays

Answer: 3. Ultraviolet

Question 2. Which one of the series of hydrogen spectrum is in the visible region

1. Lyman series
2. Balmer series
3. Paschen series
4. Bracket series

Answer: 2. Balmer series

Question 3. The Rutherford α-particle experiment shows that most of the α–particles pass through almost unscattered while some are scattered through large angles. What information does it give about the structure of the atom:

1. Atom is hollow
2. The whole mass of the atom is concentrated in a small centre called the nucleus
3. The nucleus is positively charged
4. All the above

Answer: 1. Atom is hollow

Question 4. The energy required to knock out the electron in the third orbit of a hydrogen atom is equal to

1. 13.6 eV
2. \(+\frac{13.6}{9} \mathrm{eV}\)
3. \(+\frac{13.6}{9} \mathrm{eV}\)
4. \(+\frac{13.6}{9} \mathrm{eV}\)

Answer: 3. \(+\frac{13.6}{9} \mathrm{eV}\)

Question 5. The ionization potential for the second He electron is

1. 13.6 eV
2. 27.2 eV
3. 54.4 eV
4. 100 eV

Answer: 3. 54.4 eV

Question 6. An electron makes a transition from orbit n 4 to orbit n = 2 of a hydrogen atom. The wave number of the emitted radiation (R = Rydberg’s constant) will be

1. \(\frac{16}{3 R}\)
2. \(\frac{2 R}{16}\)
3. \(\frac{3 R}{16}\)
4. \(\frac{4 \mathrm{R}}{16}\)

Answer: 3. \(\frac{3 R}{16}\)

Question 7. If a 0 is the Bohr radius, the radius of the n = 2 electronic orbit in triply ionized beryllium is –

1. 4a0
2. a0
3. a0/4
4. a0/16

Answer: 2. a0

Question 8. Which energy state of doubly ionized lithium (Li++) has the same energy as that of the ground state of hydrogen? Given Z for lithium = 3 :

1. n = 1
2. n = 2
3. n = 3
4. n = 4

Answer: 3. n = 3

Question 9. If an orbital electron of the hydrogen atom jumps from the ground state to a higher energy state, its orbital speed reduces to half its initial value. If the radius of the electron orbit in the ground state is r, then the radius of the new orbit would be

1. 2r
2. 4r
3. 8r
4. 16r

Answer: 2. 4r

Question 10. The relation between λ₁: the wavelength of the series limit of the Lyman series, λ₂: the wavelength of the series limit of the Balmer series & λ₃: the wavelength of the first line of the Lyman series is:

1. λ₁ = λ₂ + λ₃
2. λ₃ = λ₁ + λ₂
3. λ₂ = λ₃ −λ₁
4. none of these

Answer: 4. none of these

Question 11. let v1 be the frequency of the series limited of the Lyman series, v2 be the frequency of the first line of the Lyman series, and v3 be the frequency of the series limited of the Balmer series.

1. ν₁ = ν₂ + ν₃
2. ν₂ = ν₁ + ν₃
3. \(v_3=\frac{1}{2}\left(v_1+v_3\right)\)
4. ν₁ = ν₂ = ν₃

Answer: 1. ν₁ = ν₂ + ν₃

Question 12. The innermost orbit of the hydrogen atom has a diameter of 1.06 Å. What is the diameter of the tenth orbit?

1. 5.3 Å
2. 10.6 Å
3. 53 Å
4. 106 Å

Answer: 4. 106 Å

Question 13. The energy difference between the first two levels of hydrogen atoms is 10.2 eV. What is the corresponding energy difference for a singly ionized helium atom?

1. 10.2 eV
2. 20.4 eV
3. 40.8 eV
4. 81.6 eV

Answer: 3. 40.8 eV

Question 14. An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (In eV) required to remove both the electrons from a neutral helium atom is:

1. 38.2
2. 49.2
3. 51.8
4. 79.0

Answer: 4. 79.0

Question 15. In Bohr’s model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electron. If a0 is the radius of the ground state orbit, m is the mass and e is the charge on the electron, 0 is the vacuum permittivity, the speed of the electron is:

1. Zero
2. \(\frac{\mathrm{e}}{\sqrt{\varepsilon_0 \mathrm{a}_0 \mathrm{~m}}}\)
3. \(\frac{\mathrm{e}}{\sqrt4\pi{\varepsilon_0 \mathrm{a}_0 \mathrm{~m}}}\)

Answer: 3. \(\frac{\mathrm{e}}{\sqrt4\pi{\varepsilon_0 \mathrm{a}_0 \mathrm{~m}}}\)

Question 16. The energy of an electron in the excited state of H-tom is –1.5 eV, then according to Bohr’s model, its angular momentum will be:

1. 3.15 × 10–34 J-sec
2. 2.15 × 10–34 J-sec
3. 5.01 × 10–30 J-sec
4. 3.15 × 10–33 J-sec

Answer: 1. 3.15 × 10–34 J-sec

Question 17. The wavelength of radiation emitted is λ0 when an electron jumps from the third to the second orbit of the hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be

1. \(\frac{16}{25} \lambda_0\)
2. \(\frac{20}{27} \lambda_0\)
3. \(\frac{27}{20} \lambda_0\)
4. \(\frac{25}{16} \lambda_0\)

Answer: 2. \(\frac{20}{27} \lambda_0\)

Question 18. In which of the following systems will be radius of the first orbit (n =1) be the minimum

1. Doubly Ionized Lithium
2. Singly Ionized Helium
3. Deuterium Atom
4. Hydrogen Atom

Answer: 3. Deuterium Atom

Question 19. The hydrogen atom is excited through a monochromatic radiation of wavelength 975 Å. In the emission spectrum, the number of possible lines are:

1. 2
2. 4
3. 5
4. 6

Answer: 4. 6

Question 20. According to Bohr’s model of the hydrogen atom, the relation between principal quantum number n and radius of stable orbit:

1. \(r \propto \frac{1}{n}\)
2. \(r \propto n\)
3. \(r \propto \frac{1}{n_2}\)
4. \(r \propto \frac{1}{n_2}\)

Answer: 4. \(r \propto \frac{1}{n_2}\)

Question 21. The minimum orbital angular momentum of the electron in a hydrogen atom is

1. h
2. h/2
3. h/2π
4. h/π

Answer: 3. h/2π

Question 22. The energy of a hydrogen-like atom in its ground state is – 54.4 eV. It may be

1. Hydrogen
2. Deuterium
3. Helium
4. Lithium

Answer: 3. Helium

Question 23. The wavelength of light emitted due to the transition of an electron from the second orbit to the first orbit in a hydrogen atom is

1. 6563 Å
2. 4102 Å
3. 4861 Å
4. 1215 Å

Answer: 4. 1215 Å

Question 24. The ratio of the specific charge of an α-particle to that of a proton is

1. 2: 1
2. 1: 1
3. 1: 2
4. 1: 3

Answer: 3. 1: 2

Question 25. If 13.6 eV energy is required to lionize the hydrogen atom, then the energy required to remove an electron from n = 2 is:

1. 10.2 eV
2. 0 eV
3. 3.4 eV
4. 6.8 eV

Answer: 3. 3.4 eV

Question 26. In the Bohr series of lines of the hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits is an atom of hydrogen?

1. 3 → 2
2. 5 → 2
3. 4 → 1
4. 2 → 5

Answer: 2. 5 →2

Question 27. When an electron in a hydrogen atom makes a transition from the first Bohr orbit to the second Bohr orbit, how much energy it absorbs?

1. 3.4 eV
2. 10.2 eV
3. 13.6 eV
4. 1.51 eV

Answer: 2. 10.2 eV

Question 28. The radius of the first Bohr orbit is 0.5 Ã, then the radius of the fourth Bohr orbit will be :

1. 0.03 Å
2. 0.12 Å
3. 2.0 Å
4. 8.0 Å

Answer: 4. 8.0 Å

Question 29. The ionization energy of 10 times ionized sodium atom is

1. \(\frac{13.6}{11} \mathrm{eV}\)
2. \(\frac{13.6}{(11)^2} \mathrm{eV}\)
3. 13.6x(11)2ev
4. 13.6ev

Answer: 3. 13.6x(11)2ev

Question 30. An electron with a kinetic energy of 5 eV is incident on an H-atom in its ground state. The collision

1. Must Be Elastic
2. May Be Partially Elastic
3. May Be Completely Elastic
4. May Be Completely Inelastic

Answer: 1. Must Be Elastic

Question 31. An electron makes a transition from orbit n = 4 to orbit n = 2 of a hydrogen atom. The wave number of the emitted radiation (R = Rydberg’s constant) will be

1. 16/3R
2. 2R/16
3. 3R/16
4. 4R/16

Answer: 3. 3R/16

Question 32. The transition from state n = 4 to b = 3 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation may be obtained in the transition

1. 2 → 1
2. 3 → 2
3. 4 → 2
4. 5 → 4

Answer: 4. 5 → 4

Question 33. Energy E of a hydrogen atom with principal quantum number n is given by E = The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 states of hydrogen is passionate:

1. 0.85 eV
2. 3.4 eV
3. 1.9 eV
4. 1.5 eV

Answer: 3. 1.9 eV

Question 34. The total energy of an electron in the first excited state of a hydrogen atom is about – 3.4 V. Its kinetic energy in this state is –

1. –6.8 eV
2. 3.4 eV
3. 6.8 eV
4. –3.4 eV

Answer: 2. 3.4 eV

Question 35. The ionization potential of a hydrogen atom is 13.6eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be:-

1. Two
2. Three
3. Four
4. One

Answer: 2. Three

Question 36. In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of

1. Excitation of electrons in the atoms
2. Collision between the atoms of the gas
3. Collisions between the charged particles emitted from the cathode and the atoms of the gas
4. Collision between different electrons of the atoms of the gas

Answer: 3. Collisions between the charged particles emitted from the cathode and the atoms of the gas

Question 37. The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

1. 3
2. 4
3. 1
4. 2

Answer: 1.

Question 38. An alpha nucleus of energy 2 mv2 bombards a heavy nuclear target of charge Ze. Then the distance of the closest approach for the alpha nucleus will be proportional to the following:

1. \(\frac{1}{\mathrm{Ze}}\)
2. v²
3. 1/m
4. 1/v4

Answer: 3. 1/m

Question 39. The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in

1. X-ray region
2. Ultra-violet region
3. Infra-red region
4. Visible region

Answer: 2. Ultra-violet region

Question 40. Which of the following transitions in hydrogen atoms emit photons of the highest frequency?

1. n = 2 to n = 6
2. n = 6 to n = 2
3. n = 2 to n = 1
4. n = 1 to n = 2

Answer: 3. n = 2 to n = 1

Question 41. Suppose an electron is attracted towards the origin by a force r where ‘k’ is a constant and ‘r’ is the distance of the electron from the origin. By applying the Bohr model to this system, the radius of the nth orbital of the electron is found to be ‘r n’ and the kinetic energy of the electron to be ‘Tn’. Then which of the following is true?

1. Tn independent of n, rn independent of n, Tn1
2. \(T_n \propto \frac{1}{n}, r_n \propto n\)
3. \(T_n \propto \frac{1}{n} n_1, r_n \propto n^2\)
4. \(T_n \propto \frac{1}{n}, r_n \propto n^2\)

Answer: 1. Tn independent of n, rn independent of n, Tn1

Question 42. The ratio of the kinetic energy of the n = 2 electron for the H atom to that of the He + ion is:

1. 1/4
2. 1/2
3. 1
4. 2

Answer: 1. 1/4

Question 43. Energy levels A, B and C of a certain atom correspond to increasing values of energy i.e., E A < EB < EC. If 1, λ2 and λ3 are wave lengths of radiations corresponding to transitions C to B, to A and C to A respectively, which of the following relations is correct –

1. \(\lambda_3=\lambda_1+\lambda_3\)
2. \(\lambda_3=\lambda_2+\lambda_3=0\)
3. \(\lambda_3^2=\lambda_1^2+\lambda_2^2\)
4. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

Answer: 4. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

Question 44. If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li2+ is :

1. 30.6 eV
2. 13.6 eV
3. 13.6 eV
4. 122.4 eV

Answer: 1. 30.6 eV

Question 45. A hydrogen atom (ionisation potential 13.6 eV) transitions from the third excited state to the first excited state. The energy of the photon emitted in the process is

1. 1.89 eV
2. 2.55 eV
3. 12.09 eV
4. 1275 eV

Answer: 2. 2.55 eV

Question 46. Energy levels A, B and C of a certain atom correspond to increasing energy values, i.e. EA < EB < EC.

If λ1, λ2 and λ3 are the wavelengths of radiations corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct?

1. \(\lambda_3=\lambda_1+\lambda_2\)
2. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)
3. \(\lambda_1+\lambda_2+\lambda_3=0\)
4. \(\lambda_3^2=\lambda_1^2+\lambda_2^2\)

Answer: 2. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

Question 47. In a mixture of H – He+ gas (He+ is singly ionized He atom), H atoms and He+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He + ions (by collisions). Assume that the Bohr model of the atom is exactly valid. The quantum number n of the state finally populated in He+ ions is:

1. 2
2. 3
3. 4
4. 5

Answer: 3. 4

Question 48. The wavelength of the first spectral line in the Balmer series of hydrogen atoms is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly ionized helium atoms is:

1. 1215 Å
2. 1640 Å
3. 2430 Å
4. 4687 Å

Answer: 1. 1215 Å

Question 49. Which of the following statements is wrong

1. Infrared photon has more energy than photons of visible light.
2. Photographic plates are sensitive to ultraviolet rays.
3. Photographic plates can be made sensitive to infrared rays.
4. Infrared rays are invisible but can cast shadows like visible light rays.

Answer: 1. Infrared photon has more energy than photons of visible light.

Chapter 3 Modern Physics Multiple Choice Questions Section (D): Electronic Transition In The H/H-Like Atom

Question 1. Three photons coming from excited atomic-hydrogen samples are picked up. Their energies are 12.1eV, 10.2eV and 1.9eV. These photons must come from

1. A single atom
2. Two atoms
3. Three atom
4. Either two atoms or three atoms

Answer: 4. Either two atoms or three atoms

Question 2. In a hypothetical atom, if the transition from n = 4 to n = 3 produces visible light then the possible transition to obtain infrared radiation is:

1. n = 5 to n = 3
2. n = 4 to n = 2
3. n = 3 to n = 1
4. None Of These

Answer: 4. None Of These

Question 3. The ionization energy of the hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy 12.1 eV. How many spectral lines will be emitted by the hydrogen atoms?

1. One
2. Two
3. Three
4. Four

Answer: 3. Three

Question 4. The wavelength of the first line in the Balmer series in the hydrogen spectrum is λ. What is the wavelength of the second line :

1. \(\frac{20 \lambda}{27}\)
2. \(\frac{3 \lambda}{16}\)
3. \(\frac{5 \lambda}{36}\)
4. \(\frac{3 \lambda}{4}\)

Answer: 1. \(\frac{20 \lambda}{27}\)

Chapter 3 Modern Physics Multiple Choice Questions Section (E): X–Rays

Question 1. Why do we not use X-rays in the RADAR

1. They can damage the target
2. They are absorbed by the air
3. Their speed is low
4. They are not reflected by the target

Answer: 4. They are not reflected by the target

Question 2. Production of continuous X-rays is caused by

1. Transition of electrons from higher levels to lower levels in target atoms.
2. Retardiation of the incident electron when it enters the target atom.
3. Transition of electrons from lower levels to higher levels in target atoms.
4. Neutralizing the incident electron.

Answer: 2. Retardiation of the incident electron when it enters the target atom.

Question 3. The graph between the square root of the frequency of a specific line of characteristic spectrum of Xrays and the atomic number of the target will be

Question 4. The minimum wavelength min in the continuous spectrum of X-rays is

1. Proportional to the potential difference V between the cathode and anode.
2. Inversely proportional to the potential difference V between the cathode and anode.
3. Proportional to the square root of the potential difference V between the cathode and the anode.
4. Inversely proportional to the square root of the potential difference V between the cathode and the anode.

Answer: 2. Inversely proportional to the potential difference V between the cathode and anode.

Question 5. For the structural analysis of crystals, X-rays are used because

1. X-rays have wavelengths of the order of the inter-atomic spacing.
2. X-rays are highly penetrating radiations.
3. The wavelength of X-rays is of the order of nuclear size.
4. X-rays are coherent radiations.

Answer: 1. X-rays have wavelengths of the order of the inter-atomic spacing.

Question 6. A direct X-ray photograph of the intestines is not generally taken by radiologists because

1. Intestines would burst on exposure to X-rays.
2. The X-rays would not pass through the intestines.
3. The X-rays will pass through the intestines without causing a good shadow for any useful diagnosis.
4. A very small exposure to X-rays causes cancer in the intestines.

Answer: 3. The X-rays will pass through the intestines without causing a good shadow for any useful diagnosis.

Question 7. The characteristic X-ray radiation is emitted when

The bombarding electrons knock out electrons from the inner shell of the target atoms and one of the outer electrons falls into this vacancy.

1. The valance electrons are removed from the target atoms as a result of the collision.
2. The source of electrons emits a mono-energetic beam.
3. The electrons are accelerated to a fixed energy.

Answer: 1. The bombarding electrons knock out electrons from the inner shell of the target atoms and one of the outer electrons falls into this vacancy.

Question 8. X-rays are produced

1. During electric discharge at low pressure.
2. During nuclear explosions.
3. When cathode rays are reflected from the target.
4. When electrons from a higher energy state come back to a lower energy state.

Answer: 4. When electrons from a higher energy state come back to a lower energy state.

Question 9. If the current in the circuit for heating the filament is increased, the cutoff wavelength

1. Will increase
2. Will decrease
3. Will remain unchanged
4. Will change

Answer: 3. Will remain unchanged

Question 10. The characteristic X-ray spectrum is emitted due to the transition of

1. Valence electrons of the atom
2. Inner electrons of the atom
3. Nucleus of the atom
4. Both, the inner electrons and the nucleus of the atom

Answer: 2. Inner electrons of the atom

Question 11. The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is:

1. 4125 Å
2. 3000 Å
3. 6000 Å
4. 2062.5 Å

Answer: 2. 3000 Å

Question 12. If λmin is the minimum wavelength produced in an X-ray tube and kα is the wavelength of the line. As the operating tube voltage is increased.

1. (λk – λmin) increases
2. (λk – λmin) decreases
3. λkx increases
4. λka decreases

Answer: 1. (λk – λmin) increases

Question 13. X-rays obtained by Coolidge tube:

1. Are mono-chromatic
2. Have all wavelengths are below a maximum wavelength.
3. Have all wavelengths above a minimum wavelength.
4. Have all wavelengths be between a maximum and a minimum wavelength.

Answer: 3. Have all wavelengths are above a minimum wavelength.

Question 14. Penetration power of X-rays depends on 

1. Current flowing in filament
2. Nature of target
3. Applied potential difference
4. All of the above

Answer: 3. Applied potential difference

Question 15. The wavelength of an x-ray photon is 0.01 Å, and its momentum in Kg m/sec is

1. 6.6 x 10–22
2. 6.6 x 10–20
3. 6.6 x 10–46
4. 6.6 x 10–27

Answer: 1. 6.6 x 10–22

Question 16. For hard X-rays.

1. The wavelength is higher
2. The intensity is higher
3. The frequency is higher
4. The photon energy is lower

Answer: 3. The frequency is higher

Question 17. If X-rays are passed through a strong magnetic field, then X-rays

1. Will deviate maximum
2. Will deviate minimum
3. Pass undeviated
4. None of these

Answer: 3. Pass undeviated

Question 18. The minimum wavelength of X-rays produced in a Coolidge tube operated at a potential difference of 40 k V is

1. 0.31 Å
2. 3.1 Å
3. 31 Å
4. 311 Å

Answer: 1. 0.31 Å

Question 19. An X-ray photon has a wavelength of 0.01Å. Its momentum (in kg ms–1) is :

1. 6.66 × 10–22
2. 3.3 × 10–32
3. 6.6 × 10–22
4. 0

Answer: 3. 6.6 × 10–22

Question 20. The minimum wavelength of X-rays emitted by an X-ray tube is 0.4125 Å. The accelerating voltage is :

1. 30 kV
2. 50 kV
3. 80 kV
4. 60 kV

Answer: 1. 30 kV

Question 21. If the frequency of Kα, X-ray of the element of atomic number 31 is f, then the frequency of Kα, X-ray for atomic number 51 is

1. 25/9 f
2. 16/25 f
3. 9/25 f
4. zero

Answer: 1. 25/9 f

Question 22. The intensity of gamma radiation from a given source is. On passing through 36 mm of lead, it is reduced to 1/8. The thickness of lead, which will reduce the intensity to 1/2 will be:

1. 6 mm
2. 9 mm
3. 18 mm
4. 12 mm

Answer: 4. 12 mm

Question 23. Which of the following X-rays has maximum energy, if they are produced by the collision of electrons of energy 40 keV with the same target

1. 300 Å
2. 10 Å
3. 4 Å
4. 0.31 Å

Answer: 4. 0.31 Å

Question 24. Both X-rays and γ-rays are electromagnetic waves, which of the following statements is true for them

1. The energy of X-rays is more than that of γ-rays.
2. The wavelength of X-rays in general, is larger than that of γ-rays.
3. The frequency of X-rays is greater than that of γ-rays.
4. The velocity of X-rays is greater than that of γ-rays.

Answer: 2. Wavelength of X-rays in general, is larger than that of γ-rays.

Question 25. According to Moseley’s law, the ratio of the slopes of the graph between Z for Kβ and Kα is :

1. \(\sqrt{\frac{32}{27}}\)
2. \(\sqrt{\frac{27}{32}}\)
3. \(\sqrt{\frac{33}{22}}\)
4. \(\sqrt{\frac{22}{33}}\)

Answer: 1. \(\sqrt{\frac{32}{27}}\)

Question 26. If the frequency of Kα X-ray emitted from the element with atomic number 31 is f, then the frequency of Kα X-ray emitted from the element with atomic number 51 would be (assume that screening constant for Kα is 1):

1. \(\frac{5}{3} f\)
2. \(\frac{51}{31} f\)
3. \(\frac{9}{25} \mathrm{f}\)
4. \(\frac{25}{9} f\)

Answer: 4. \(\frac{25}{9} f\)

Question 27. Which one of the following statements is WRONG in the context of X-rays generated from an X-ray tube?

1. The wavelength of characteristic X-rays decreases when the atomic number of the target increases
2. The cut-off wavelength of the continuous X-rays depends on the atomic number of the target
3. The intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube
4. The cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube

Answer: 2. Cut-off wavelength of the continuous X-rays depends on the atomic number of the target

Question 28. 50% of the X-rays coming from a Coolidge tube can pass through a 0.1 mm thick aluminum foil. The potential difference between the target and the filament is increased. The thickness of aluminum foil, which will allow 50% of the X-ray to pass through, will be –

1. zero
2. < 0.1 mm
3. 0.1 mm
4. > 0.1 mm

Answer: 4. > 0.1 mm

Question 29. When ultraviolet rays incident on metal plate the photoelectric effect does not occur, it occurs by incidence of:-

1. Infrared rays
2. X – rays
3. Radio wave
4. Lightwave

Answer: 2. X – rays

Question 30. An X-ray photon of wavelength λ and frequency ν collides with an initially stationary electron (but free to move) and bounces off. If λ’ and ν’ are respectively the wavelength and frequency of the scattered photon, then:

1. λ’=λ;ν’=v
2. λ'<λ;ν’>v
3. λ’>λ;ν’>v
4. λ’>λ;ν'<v

Answer: 4. λ’>λ;ν'<v

Chapter 3 Modern Physics Multiple Choice Questions Exercise -2

Question 1. An image of the sun is formed by a lens of focal length 30 cm on the metal surface of a photo-electric cell and it produces a current I. The lens forming the image is then replaced by another lens of the same diameter but of focal length of 15 cm. The photoelectric current in this case will be : (In both cases the plate is kept at the focal plane and normal to the axis lens).

1. I/2
2. 2I
3. I/2
4. 4I

Answer: 3. I/2

Question 2. Two identical, photocathodes receive light of frequencies f1 and f2. If the velocities of the photoelectrons (of mass m) coming out are respectively ν₁ and ν₂, then:

1. \(v_1^2-v_2^2=\frac{2 h}{m}\left(f_1-f_2\right)\)
2. \(v_1-v_2=\left[\frac{2 h}{m}\left(f_1+f_2\right)\right]^{1 / 2}\)
3. \(v_1^2-v_2^2=\frac{2 h}{m}\left(f_1+f_2\right)\)
4. \(v_1-v_2=\left[\frac{2 h}{m}\left(f_1-f_2\right)\right]^{1 / 2}\)

Answer: 1. \(v_1^2-v_2^2=\frac{2 h}{m}\left(f_1-f_2\right)\)

Question 3. Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2 × 10 –3 W. The number of photons emitted, on the average, by the source per second is:

1. 5 × 1015
2. 5 × 1016
3. 5 × 1017
4. 5 × 1014

Answer: 1. 5 × 1015

Question 4. A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 × 10 6 ms–1. The velocity of the particle is

1. 2.7 × 10–18 ms–1
2. 9 × 10–2 ms–1
3. 3 × 10–31 ms–1
4. 2.7 × 10–21 ms–1
5. (Mass of electron = 9.1 × 10–31 kg)

Answer: 1. 2.7 × 10–18 ms–1

Question 5. The anode voltage of a photocell is kept fixed. The wavelength of the light falling on the cathode is gradually changed. The plate current  of the photocell varies as follows:

Answer: 3.

Question 6. The graph is showing the photocurrent with the applied voltage of a photoelectric effect experiment. Then

1. A and B will have the same intensity and B & C have the same frequency
2. B and C have the same intensity and A & B have the same frequency
3. A and B will have the same frequency and B & C have the same intensity
4. A and C will have the same intensity and B & C have the same frequency

Answer: 1. A and B will have the same intensity and B & C have the same frequency

Question 7. If λ=10–10 m changes to λ’= 0.5 × 10–10 m, find the energy difference (λE) given to the particle :

1. \(\Delta \mathrm{E} \text { is equal to }\left(\frac{1}{4}\right)^{\mathrm{th}} \text { of initial energy }\)
2. \(\Delta E \text { is equal to }\left(\frac{1}{2}\right)^{\text {th }} \text { of initial energy }\)
3. ΔE is equal to twice of initial energy
4. ΔE is equal to the initial energy

Answer: 4. ΔE is equal to the initial energy

Question 8. The wavelength involved in the spectrum of deuterium is slightly different from that of the hydrogen spectrum, because:

1. Size Of The Two Nuclei Are Different
2. Nuclear Forces Are Different In The Two Cases
3. Masses Of The Two Nuclei Are Different
4. Attraction Between The Electron And The Nucleus Is Different In The Two Cases

Answer: 3. Masses Of The Two Nuclei Are Different

Question 9. The total energy of an electron in the ground state of a hydrogen atom is –13.6 eV. The kinetic energy of an electron in the first excited state is:

1. 3.4 eV
2. 6.8 eV
3. 13.6 eV
4. 1.7 eV

Answer: 1. 3.4 eV

Question 10. In the Davisson-Germer experiment when an electron strikes the Ni-crystal which of the following is produced-

1. X-rays
2. ϒ-rays
3. Electron
4. photon

Answer: 3. Electron

Question 11. The wavelengths of Kα x-rays of two metals ‘A’ and ‘B’ are \(\frac{4}{1875 R} \text { and } \frac{1}{675 R}\) respectively, where ‘R’ is rydberg constant. The number of elements lying between ‘A’ and ‘B’ according to their atomic numbers is

1. 3
2. 6
3. 5
4. 4

Answer: 4. 4

Question 12. If Bohr’s theory applies to 100Fm257, then the radius of this atom in Bohr’s unit is:

1. 4
2. 1/4
3. 100
4. 200

Answer: 2. 1/4

Question 13. An electron in an excited state of Li 2+ ions has an angular momentum of 3h/2π. The de-Broglie wavelength of the electron in this state is pa0 (where a0 is the Bohr radius). The value of p is

1. 2
2. 4
3. 6
4. 10

Answer: 1. 2

Question 14. If m is the mass of the electron, ν its velocity, r is the radius of a stationary circular orbit around a nucleus with 1 charge Ze, then from Bohr’s first postulate, the kinetic energy \(\mathrm{K}=\frac{1}{2} m v^2\) of the electron in C.G.S. System is equal to:

1. \(\frac{1}{2} \frac{\mathrm{Ze}^2}{\mathrm{r}}\)
2. \(\frac{1}{2} \frac{\mathrm{Ze}^2}{\mathrm{r}^2}\)
3. \(\frac{1}{2} \frac{\mathrm{Ze}}{\mathrm{r}}\)
4. \(\frac{1}{2} \frac{Z e}{r^2}\)

Answer: 1. \(\frac{1}{2} \frac{\mathrm{Ze}^2}{\mathrm{r}}\)

Question 15. The attractive potential between electron and nucleus is given by v = v0 \(v=v_0 \ell n \frac{r}{r_0}, v_0 \text { and } r_0\) are constants and ‘ r ‘ is the radius. The radius ‘ r ‘ of the nth Bohr’s orbit depends upon principal quantum number ‘n’ as:

1. r∝n2²
2. r∝n
3. \(r \propto \frac{1}{n}\)
4. \(r \propto \frac{1}{n^2}\)

Answer: 2. \(r \propto \frac{1}{n}\)

Question 16. 50% of the x-ray coming from a Coolidge tube can pass through 0.1 mm thick aluminium foil. If the potential difference between the target and the filament is increased, the fraction of the X-ray passing through the same foil will be

1. 0%
2. <50%
3. >50%
4. 50%

Answer: 3. >50%

Question 17. Figure shows the intensity-wavelength relations of X-rays coming from two different Coolidge tubes. The solid curve represents the relation for tube A in which the potential difference between the target and the filament is VA and the atomic number of the target material is ZA. These quantities are VB and Z B for the other tube. Then,

1. V_A > VB Z_A > Z_B
2. V_A > VB, Z_A > Z_B
3. V_A < VB Z_A > Z_B
4. V_A < VB Z_A < Z_B

Answer: 2. V_A > VB Z_A > Z_B

Question 18. Wavelengths of the Kx lines of the two elements are 250 and 179 pm respectively. The number of elements between these elements in the sequence will be

1. Zero
2. 3
3. 2
4. 1

Answer: 2. 2

Question 19. Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

1. \(\lambda_0=\frac{2 \mathrm{mc} \lambda^2}{\mathrm{~h}}\)
2. \(\lambda_0=\frac{2 h}{m c}\)
3. \(\lambda_0=\frac{2 \mathrm{~m}^2 \mathrm{c}^2 \lambda^3}{\mathrm{~h}^2}\)
4. \(\lambda_0=\lambda\)

Answer: 1. \(\lambda_0=\frac{2 \mathrm{mc} \lambda^2}{\mathrm{~h}}\)

Question 20. Which of the following wavelengths is not possible for an X-ray tube which is operated at 40 kV?

1. 0.25 Å
2. 0.5 Å
3. 0.52 Å
4. 1A

Answer: 1. 0.25 Å

Question 21. For soft X-rays, the attenuation constant for aluminum is 1.73 per cm. Then the percentage of X-rays that will pass through an aluminium sheet of thickness 1.156 cm will be

1. 13.5%
2. 6.8%
3. 20.4%
4. 27.0%

Answer: 1. 13.5%

Question 22. An x-ray tube, when operated at 50 kV tube voltage, records an anode current of 20 mA. If the efficiency of the tube for the production of X-rays is 1% then the heat produced per second in calories is nearly

1. 249
2. 238
3. 10
4. 2.38

Answer: 4. 2.38

Question 23. Which of the following are the characteristics required for the target to produce X-rays

1. Atomic number Low High Low High
2. Melting point High High Low Low

Answer: 2. Melting point High High Low Low

Question 24. In a discharge tube when a 200-volt potential difference is applied 6.25 1018 electrons move from cathode to anode and 3.125  1018 singly charged positive ions move from anode to cathode in one second. Then the power of the tube is:

1. 100 watt
2. 200 watt
3. 300 watt
4. 400 watt

Answer: 3. 300 watt

Question 25. The wavelength of Kα X-ray of an element having atomic number z = 11 isλ. The wavelength of Kα

1. 11
2. 44
3. 6
4. 4

Answer: 3. 6

Question 26. The angle volute of the photocell is kept fixed. The wavelength of the light falling on the cathode is gradually changed. The maximum kinetic energy (K.E.) of the photoelectrons emitted varies with as

Answer: 4.

Question 27. A photon of wavelength (less than threshold wavelength0) is incident on a metal surface of work function W 0. The de Broglie wavelength of the ejected electron of mass ‘m’ is

1. \(h\left[2 m\left(\frac{h c}{\lambda}-W_0\right)\right]\)
2. \(\frac{h}{2 m\left(\frac{h c}{\lambda}-W_0\right)}\)
3. \(\frac{h}{\sqrt{2 m\left(\frac{h c}{\lambda}-W_0\right)}}\)
4. \(\frac{1}{h \sqrt{2 m\left(\frac{h c}{\lambda}-W_0\right)}}\)

Answer: 3. \(\frac{h}{\sqrt{2 m\left(\frac{h c}{\lambda}-W_0\right)}}\)

Chapter 3 Modern Physics Multiple Choice Questions Exercise 3 Part -1: Neet / Aipmt Question (Previous Years)

Question 1. Monochromatic light of wavelength 667 nm is produced by a helium-neon laser. The power emitted is 9 mW. The number of photons arriving per second on average at a target irradiated by this beam is

1. 9 × 1017
2. 3 × 1016
3. 9 × 1015
4. 9 × 1019

Answer: 2. 3 × 1016

Question 2. The figure shows a plot of photocurrent versus anode potential for a photo-sensitive surface for three different radiations. Which one of the following is a correct statement?

1. Curves a and b represent incident radiations of different frequencies and different intensities
2. Curves a and b represent incident radiations of the same frequencies but of different intensities
3. Curves b and c represent incident radiations of different frequencies and different intensities
4. Curves b and c represent incident radiations of the same frequencies having the same intensity

Answer: 2. Curves a and b represent incident radiations of the same frequencies but of different intensities

Question 3. The number of photoelectrons emitted for light of a frequency ν is proportional to (higher than the threshold frequency ν0)

1. ν – ν0
2. Threshold Frequency (ν0)
3. Intensity Of Light
4. Frequency Of Light (ν)

Answer: 4. Frequency Of Light (ν)

Question 4. The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between

1. n = 3 to n = 2 states
2. n = 3 to n = 1 states
3. n = 2 to n = 1 states
4. n = 4 to n = 3 states

Answer: 4. n = 4 to n = 3 states

Question 5. The energy of a hydrogen atom in the ground state is –13.6 eV. The energy of a He+ ion in the first excited state will be

1. – 13.6 eV
2. – 27.2 eV
3. – 54.4 eV
4. – 6.8 eV

Answer: 1. – 13.6 eV

Question 6. A source S 1 is producing 1015 photons per second of wavelength 5000Å. Another source S 2 is producing 1.02 × 1015 photons per second of wavelength 5100 Å. Then (power of S2)/(power of S1) is equal to

1. 1.00
2. 1.02
3. 1.04
4. 0.98

Answer: 1. 1.00

Question 7. The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be

1. 2.4 V
2. –1.2 V
3. – 2.4 V
4. 1.2 V

Answer: 2. –1.2 V

Question 8. When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy is respectively

1. N and 2T
2. 2N and T
3. 2N and 2T
4. N and T

Answer: 2. 2N and T

Question 9. The electron in the hydrogen atom jumps from its excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, 13.62 eV the stopping potential is estimated to be (the energy of the electron in n th state \(E_n=-\frac{13.6}{n^2} e V\)

1. 5.1 V
2. 12.1 V
3. 17.2 V
4. 7V

Answer: 4. 7v

Question 10. The threshold frequency for a photosensitive metal is 3.3 × 10 14 Hz. If the light of frequency 8.2 × 1014 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is near:

1. 2V
2. 3V
3. 5V
4. 1V

Answer: 1. 2V

Question 11. An electron in the hydrogen atom jumps from the excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having a work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, the value of n is:

1. 3
2. 4
3. 5
4. 2

Answer: 2. 4

Question 12. Out of the following which one is not a possible energy for a photon to be emitted by a hydrogen atom according to Bohr’s atomic model?

1. 1.9 eV
2. 11.1 eV
3. 13.6 eV
4. 0.65 eV

Answer: 2. 11.1 eV

Question 13. Photoelectric emission occurs only when the incident light has more than a certain minimum:

1. Power
2. Wavelength
3. Intensity
4. Frequency

Answer: 4. Frequency

Question 14. The wavelength of the first line of the Lyman series for hydrogen atoms is equal to that of the second line of the Balmer series for a hydrogen-like ion. The atomic number Z of a hydrogen-like ion is:

1. 3
2. 4
3. 1
4. 2

Answer: 4. 2

Question 15. In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by:

1. Increasing The Potential Difference Between The Anode And Filament
2. Increasing The Filament Current
3. Decreasing The Filament Current
4. Decreasing The Potential Difference Between The Anode And Filament

Answer: 1. Increasing The Potential Difference Between The Anode And Filament

Question 16. The decreasing order of wavelength of infrared, microwave, ultraviolet, and gamma rays is:

1. Microwave, Infrared, Ultraviolet, Gamma Rays
2. Gamma Rays, Ultraviolet, Infrared, Microwaves
3. Microwaves, Gamma Rays, Infrared, Ultraviolet
4. Infrared, Microwave, Ultraviolet, Gamma Rays

Answer: 1. Gamma Rays, Ultraviolet, Infrared, Microwaves

Question 17. Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. The ratio of maximum speeds emitted electrons will be:

1. 1: 4
2. 1: 2
3. 1: 1
4. 1: 5

Answer: 2. 1: 2

Question 18. Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100kV then the de–Broglie wavelength associated with the electrons would:

1. Increases By 2 Times
2. Decrease By 2 Times
3. Decrease By 4 Times
4. Increases By 4 Times

Answer: 2. Decrease By 2 Times

Question 19. In the photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is:

1. 1.8 V
2. 1.2 V
3. 0.5 V
4. 2.3 V

Answer: 3. 0.5 V

Question 20. The electron in a hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited to the first excited state. The ratio of the wavelength 1: 2 emitted in the two cases is

1. 7/5
2. 27/20
3. 27/5
4. 20/7

Answer: 4. 20/7

Question 21. A 200 W sodium street lamp emits yellow light of wavelength 0.6 m. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is.

1. 1.5 × 1020
2. 6 × 1018
3. 62 × 1020
4. 3×1019

Answer: 1. 1.5 × 1020

Question 22. An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be

1. \(\frac{24 \mathrm{hR}}{25 \mathrm{~m}}\)
2. \(\frac{25 \mathrm{hR}}{24 \mathrm{~m}}\)
3. \(\frac{25 \mathrm{~m}}{24 \mathrm{hR}}\)
4. \(\frac{25 \mathrm{~m}}{24 \mathrm{hR}}\)

Answer: 1. \(\frac{24 \mathrm{hR}}{25 \mathrm{~m}}\)

Question 23. Monochromatic radiation emitted when an electron on a hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the materials is :

1. 4 × 1015 Hz
2. 5 × 1015 Hz
3. 1.6 × 1015 Hz
4. 2.5 × 1015 Hz

Answer: 3. 1.6 × 1015 Hz

Question 24. An α-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. The de Broglie wavelength associated with the particle will be:

1. 1 Å
2. 0.1 Å
3. 10 Å
4. 0.01 Å

Answer: 4. 0.01 Å

Question 25. If the momentum of the electron is changed by P, then the de Broglie wavelength associated with it changes by 0.5%. The initial momentum of the electron will be :

1. 200p
2. 400p
3. \(\frac{P}{200}\)
4. 100p

Answer: 1. 200p

Question 26. Two radiations of photon energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is:

1. 1: 4
2. 1: 2
3. 1: 1
4. 1: 5

Answer: 2. 1: 2

Question 27. The transition from the state n = 3 to n = 1 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:

1. 2 → 1
2. 3 → 2
3. 4 → 2
4. 4 → 3

Answer: 4. 4 → 3

Question 28. For photoelectric emission from certain metals, the cutoff frequency is . If radiation of frequency 2 impinges on the metal plate the maximum possible velocity of the emitted electron will be (m is the electron mass):

1. \(\sqrt{\mathrm{h} v / \mathrm{m}}\)
2. \(\sqrt{\mathrm{h} v / \mathrm{m}}\)
3. \(2 \sqrt{h v / m}\)
4. \(2 \sqrt{h v / m}\)

Answer: 2. \(\sqrt{\mathrm{h} v / \mathrm{m}}\)

Question 29. The wavelength λe of an electron and λP of a photon of the same energy E are related by:

1. \(\lambda_{\mathrm{p}} \propto \lambda_{\mathrm{e}}\)
2. \(\lambda_{\mathrm{P}} \propto \sqrt{\lambda_e}\)
3. \(2 \sqrt{h v / m}\)
4. \(\sqrt{\mathrm{h} /(2 \mathrm{~m})}\)

Answer: 4. \(\sqrt{\mathrm{h} /(2 \mathrm{~m})}\)

Question 30. The ratio of longest wavelengths corresponding to the Lyman and Blamer series in the hydrogen spectrum is:

1. \(\frac{3}{23}\)
2. \(\frac{7}{29}\)
3. \(\frac{9}{31}\)
4. \(\frac{9}{31}\)

Answer: 4. \(\frac{9}{31}\)

Question 31. When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increases from emitted 0.5 eV to 0.8eV. The work function of the metal is:

1. 0.65 eV
2. 1.0 eV
3. 1.3 eV
4. 1.5 eV

Answer: 2. 1.0 eV

Question 32. A hydrogen atom in the ground state is excited by a monochromatic radiation of λ = 975 Å. The number of spectral lines in the resulting spectrum emitted will be:

1. 3
2. 2
3. 6
4. 10

Answer: 3. 6

Question 33. Which of the following figures represents the variation of particle momentum and the associated de Broglie wavelength?

Answer: 1.

Question 34. A certain metallic surface is illuminated with monochromatic light of wavelength. The stopping potential for photo-electric current for this light is 3V0. If the same surface is illuminated with light of wavelength 2 the stopping potential is V0, and the threshold wavelength for this surface for photo-electric effect is :

1. 4
2. \(\frac{\lambda}{4}\)
3. \(\frac{\lambda}{6}\)
4. 6

Answer: 1. 4

Question 35. A radiation of energy ‘E’ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light):

1. \(\frac{2 E}{C}\)
2. \(\frac{2 E}{C^2}\)
3. \(\frac{2 E}{C^2}\)
4. \(\frac{2 E}{C^2}\)

Answer: 1. \(\frac{2 E}{C}\)

Question 36. When a metallic surface is illuminated with radiation of wavelength E the stopping potential is V. If the V same surface is illuminated with radiation of wavelength 2λ, the stopping potential is 4. The threshold wavelength for the metallic surface is:

1. 3
2. 4
3. 5
4. \(\frac{5}{2} \lambda\)

Answer: 1. 3

Question 37. Given the value of the Rydberg constant is 107 m–1, the wave number of the last line of the Balmer series in the hydrogen spectrum will be:

1. 2.5×107 m–1
2. 0.025 ×104 m–1
3. 0.5 ×107 m–1
4. 0.25×107 m–1

Answer: 4. 0.25×107 m–1

Question 38. An electron of mass m and a photon have the same energy E. The ratio of de-Broglie wavelengths associated with them is:

1. \(\frac{1}{c}\left(\frac{2 m}{E}\right)^{\frac{1}{2}}\)
2. \(\frac{1}{c}\left(\frac{E}{2 m}\right)^{\frac{1}{2}}\)
3. \(\left(\frac{E}{2 m}\right)^{\frac{1}{2}}\)
4. \(\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2}\)

Answer: 2. \(\frac{1}{c}\left(\frac{E}{2 m}\right)^{\frac{1}{2}}\)

Question 39. Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ0) of the emitted X-ray is:

1. \(\lambda_0=\lambda\)
2. \(\lambda_0=\frac{2 m c \lambda^2}{h}\)
3. \(\lambda_0=\frac{2 h}{m c}\)
4. \(\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2}\)

Answer: 2. \(\lambda_0=\frac{2 m c \lambda^2}{h}\)

Question 40. Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is:

1. –3 V
2. +3 V
3. +4 V
4. –1 V

Answer: 1. –3 V

Question 41. If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be:

1. \(\frac{20}{13} \lambda\)
2. \(\frac{16}{25} \lambda\)
3. \(\frac{9}{16} \lambda\)
4. \(\frac{20}{7} \lambda\)

Answer: 4. \(\frac{20}{7} \lambda\)

Question 42. The photoelectric threshold wavelength of silver is 3250 × 10 –10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10–10 m is : (Given h = 4.14 × 10–15 eVs and c = 3 × 108 ms–1)

1. 6 × 105 ms–1
2. 0.6 × 106 ms–1
3. 61 × 103 ms–1
4. 0.3 × 106 ms–1 43.

Answer: 1. 6 × 105 ms–1

Question 43. The ratio of wavelengths of the last line of the Balmer series and the last line of Lyman series is

1. 2
2. 1
3. 4
4. 0.5

Answer: 3. 4

Question 44. The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is:

1. \(\frac{h}{\sqrt{\mathrm{mkT}}}\)
2. \(\frac{h}{\sqrt{3 \mathrm{mkT}}}\)
3. \(\frac{2 \mathrm{~h}}{\sqrt{3 \mathrm{mkT}}}\)
4. \(\frac{2 h}{\sqrt{\mathrm{mkT}}}\)

Answer: 2. \(\frac{h}{\sqrt{3 \mathrm{mkT}}}\)

Question 45. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is:

1. 1: 1
2. 1: – 2
3. 2: –1
4. 1: –1

Answer: 4. 1: –1

Question 46. When the light of frequency 2v0 (where v0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v1. When the frequency of the incident radiation is increased to 5v0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is:

1. 1 : 2
2. 2: 1
3. 4: 1
4. 1: 4

Answer: 1. 1: 2

Question 47. An electron of mass m with an initial velocity = V0 (V0 > 0) enters an electric field = – E0 (E0 = constant > 0) at t = 0. If 0 is its de-Broglie wavelength at time t is:

1. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m V_0} t\right)}\)
2. \(\lambda_0\left(1+\frac{\mathrm{eE}_0}{\mathrm{mV}_0} \mathrm{t}\right)\)

Answer: 1. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m V_0} t\right)}\)

Question 48. The total energy of an electron in an atom in an orbit is – 3.4 eV. Its kinetic and potential energies are, respectively:

1. 3.4 eV, 3.4 eV
2. –3.4 eV, –3.4 eV
3. –3.4 eV, –6.8 eV
4. 3.4 eV, –6.8 eV

Answer: 4. 3.4 eV, –6.8 eV

Question 49. ∝-particle consists of:

1. 2 protons only
2. 2 protons and 2 neutrons only
3. 2 electrons, 2 protons and 2 neutrons
4. 2 electrons and 4 protons only

Answer: 2. 2 protons and 2 neutrons only

Question 50. An electron is accelerated through a potential difference of 10,000V. Its de Broglie wavelength is, (nearly) : (me = 9×10–31 kg)

1. 12.2 nm
2. 12.2 × 10–13 m
3. 12.2 × 10–12 m
4. 12.2 × 10–14 m

Answer: 3. 12.2 × 10–12 m

Question 51. The radius of the first permitted Bohr orbit, for the electron, in a hydrogen atom equals 0.51Å and its ground state energy equals –13.6 eV. If the electron in the hydrogen atom is replaced by a muon (u–) [charge same as electron and mass 207 me], the first Bohr radius and ground state energy will be

1. 0.53 × 10–13 m, – 3.6 eV
2. 25.6 × 10–13 m, – 2.8 eV
3. 2.56 × 10–13 m, – 2.8 keV
4. 2.56 × 10–13 m, – 13.6 eV

Answer: 3. 2.56 × 10–13 m, – 2.8 keV

Question 52. The work function of a photosensitive material is 4.0 eV. The longest wavelength of light that can cause photon emission from the substance is (approximately)

1. 3100 nm
2. 966 nm
3. 31 nm
4. 310 nm

Answer: 4. 310 nm

Question 53. A proton and an α-particle are accelerated from rest to the same energy. The de Broglie wavelength and λp λα are in the ratio:

1. 2 :1
2. 2 : 1
3. 1 : 1
4. 4 : 1

Answer: 1. 102 × 10–3 nm

Question 54. The de Broglie wavelength of an electron moving with kinetic energy of 144 eV is nearly

1. 102 × 10–3 nm
2. 102 × 10–4 nm
3. 102 × 10–5 nm
4. 102 × 10–2 nm

Answer: 1. 102 × 10–3 nm

Question 55. The wave nature of electrons was experimentally verified by,

1. de Broglie
2. Hertz
3. Einstein
4. Davisson and Germer

Answer: 4. Davisson and Germer

Question 56. An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength 1.227 10 −2 mm of the electron is the potential difference is

1. 104v
2. 10v
3. 102v
4. 103v

Answer: 1. 104v

Question 57. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and the intensity is doubled?

1. Zero
2. Doubled
3. Four times
4. One-fourth

Answer: 1. Zero

Question 58. Light with an average flux of 20 W/cm2 falls on a non-reflecting surface at normal incidence having a surface area of 20 cm2. The energy received by the surface during 1 minute

1. 48x103J
2. 10x103J
3. 12x103J
4. 24x103J

Answer: 4. 24x103J

Question 59. An electromagnetic wave of wavelength ‘ λ’ is incident on a photosensitive surface of negligible work λd function. If m’ mass is of photoelectron emitted from the surface has de-Broglie wavelength ‘ λd’ then

• \(\lambda_{\mathrm{d}}=\left(\frac{2 \mathrm{mc}}{\mathrm{h}}\right) \lambda^2\)
• \(\lambda=\left(\frac{2 \mathrm{mc}}{\mathrm{h}}\right) \lambda_{\mathrm{d}}^2\)
• \(\lambda=\left(\frac{2 \mathrm{~h}}{\mathrm{mc}}\right) \lambda_{\mathrm{d}}^2\)
• \(\lambda=\left(\frac{2 \mathrm{~h}}{\mathrm{mc}}\right) \lambda_{\mathrm{d}}^2\)

Answer: 2. \(\lambda=\left(\frac{2 \mathrm{mc}}{\mathrm{h}}\right) \lambda_{\mathrm{d}}^2\)

Question 60. The number of photons per second on average emitted by the source of monochromatic light of h Js 3.3 10-3 = h6.6 10−34 −3 wavelength 600 nm, when it delivers the power of watt be

1. 1017
2. 1016
3. 1015
4. 1018

Answer: 2. 1016

Chapter 3 Modern Physics Multiple Choice Questions Part – 2: Jee (Main) / Aieee Problems (Previous Years)

Question 1. The transition from the state n =4 to n =3 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:

1. 3 → 2
2. 4 → 2
3. 5 → 4
4. 2 → 1

Answer: 3. 5 → 4

Question 2. The surface of a metal is illuminated with a light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is : (hc = 1240 eV.nm)

1. 1.41 eV
2. 1.51 eV
3. 1.68 eV
4. 3.09 eV

Answer: 1. 1.41 eV

Question 3. Statement-1: When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is replaced by X-rays, both V 0 and Kmax increase.

Statement 2: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

1. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
2. Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1
3. Statement-1 is false, Statement-2 is true.
4. Statement-1 is true, Statement-2 is false.

Answer: 4. Statement-1 is true, Statement-2 is false.

Question 4. If a source of power 4 kW produces 1020 photons/second, the radiation belongs to a part of the spectrum called:

1. X-rays
2. ultraviolet rays
3. microwaves
4. γ-rays

Answer: 1. X-rays

Question 5. The energy required for the electron excitation in Li ++ from the first to the third Bohr orbit is:

1. 12.1 eV
2. 36.3 eV
3. 108.8 eV
4. 122.4 eV

Answer: 3. 108.8 eV

Question 6. This question has Statement –1 and Statement –2. Of the four choices given after the statements, choose the one that best describes the two statements: Statement –1: A metallic surface is irradiated by a monochromatic light of frequency  (the threshold frequency). The maximum kinetic energy and the stopping potential are Kmax and V0 respectively. If the frequency incident on the surface is doubled, both the K max and V 0 are also doubled.

Statement –2:

1. The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
2. Statement –1 is true, statement –2 is false.
3. Statement –1 is true, Statement –2 is true, Statement –2 is the correct explanation of
   Statement –1
4. Statement –1 is true, Statement –2 is true, Statement –2 is not the correct explanation of Statement–1
5. Statement–1 is false, Statement –2 is true

Answer: 4. Statement–1 is false, Statement –2 is true

Question 7. After absorbing a slowly moving neutron of Mass m N (momentum ≈ 0) a nucleus of mass M breaks into two nuclei of masses m 1 and 5m1 (6 m1 = M + mN ) respectively. If the de Broglie wavelength of the nucleus with mass m 1 is λ, the de Broglie wavelength of the nucleus will be:

1. 5λ
2. λ/5
3. λ
4. 25λ

Answer: 3.

Question 8. The hydrogen atom is excited from ground state to another state with a principal quantum number equal to

1. 2
2. 3
3. 5
4. 6

Answer: 4. 6

Question 4. Then the number of spectral lines in the emission spectra will be:

1. 2
2. 3
3. 5
4. 6

Answer: 4. 6

Question 9. The anode voltage of a photocell is kept fixed. The wavelength of the light falling on the cathode is gradually changed. The plate current of the photocell varies as follows: 

Answer: 4.

Question 10. The radiation corresponding to the 3 → 2 transition of the hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10 –4 T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to:

1. 1.8 eV
2. 1.1 eV
3. 0.8 eV
4. 1.6 eV

Answer: 2. 1.1 eV

Question 11. Hydrogen (1H¹), Deuterium (1H³), singly ionized Helium (2He4)+, and doubly ionized lithium (3Li6)++ all have one electron around the nucleus. Consider an elelctron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are λ1, λ2, λ3, and λ4 respectively then approximately which one of the following is correct?

1. 4λ1 = 2λ2 = 2λ3 = λ4
2. λ1 = 2λ2 = 2λ3 = λ4
3. λ1 =λ2 = 4λ3 = λ4
4. λ1 = 2λ2 = 3λ3 = 4λ4

Answer: 3. λ1 =λ2 = 4λ3 = λ4

Question 12. As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion

1. Its Kinetic Energy Increases But Potential Energy And Total Energy Decrease
2. Kinetic Energy, Potential Energy, And Total Energy Decrease
3. Kinetic Energy Decreases, Potential Energy Increases But Total Energy Remains the Same
4. Kinetic Energy And Total Energy Decrease But Potential Energy Increases

Answer: 1. Its Kinetic Energy Increases But Potential Energy And Total Energy Decrease

Question 13. Match List-1 Fundamental Experiment) with List-2 (its conclusion) and select the correct option from the choices given below the list :

1. (A) – (1) (B) – (4) (C) – (3)
2. (A) – (2) (B)-(4) (C) – (3)
3. (A) – (2) (B) (1) (C) -(3)
4. (A) – (4) (B) – (3) (C) – (2)

Answer: 3. (A) – (4) (B) – (3) (C) – (2)

Question 14. Radiation of wavelength λis incident on a photocell. The fastest emitted electron has speed υ. If the 3 wavelength is changed to \(\frac{3 \lambda}{4}\), the speed of the fastest emitted electron will be:

1. \(<v\left(\frac{4}{3}\right)^{\frac{1}{2}}\)
2. \(=v\left(\frac{4}{3}\right)^{\frac{1}{2}}\)
3. \(=v\left(\frac{4}{3}\right)^{\frac{1}{2}}\)
4. \(>v\left(\frac{4}{3}\right)^{\frac{1}{2}}\)

Answer: 4. \(>v\left(\frac{4}{3}\right)^{\frac{1}{2}}\)

Question 15. An electron beam is acceleration by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λmin is the smallest possible wavelength of Xray in the spectrum, the variation of log λmin with log V is correctly represented in:

Answer: 2.

Question 16. A particle A of mass m and initial velocity υ collides with a particle B of mass 2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is:

1. \(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{2}\)
2. \(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{3}\)
3. \(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=2\)
4. \(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{2}{3}\)

Answer: 3. \(\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=2\)

Question 18. If the series limit frequency of the Lyman series is vL, then the series limit frequency of the Pfund series is:

1. vL/16
2. vL/25
3. 25vL
4. 16vL

Answer: 2. vL/25

Question 19. An electron from various excited states of a hydrogen atom emits radiation to come to the ground state. Let λn, and λg be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let λn be the wavelength of the emitted photon in transition from the nth state to the ground state. For large n, (A, B are constants)

1. \(\hat{n}_{\mathrm{n}}^2 \approx \mathrm{A}+\mathrm{B} \lambda_{\mathrm{n}}^2\)
2. \(\Lambda_n^2 \approx \lambda\)
3. \(\lambda_n \approx A+\frac{B}{\lambda_n^2}\)
4. \(\wedge_n \approx A+B \lambda_n\)

Answer: 3. \(\lambda_n \approx A+\frac{B}{\lambda_n^2}\)

Question 20. The surface of certain metal is first illuminated with light of wavelength λ1 = 350 nm and then, by the light of wavelength λ2 = 540 nm. It is found that the maximum speed of the photoelectrons in the two cases differs by a factor of 2. The work function of the metal (in eV) is close to:

1. 2.5
2. 5.6
3. 1.4
4. 1.8

Answer: 4. 1.8

Question 21. The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin (6.28 × 107)ct]. If this light falls on a sliver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photoelectrons? (c = 3 × 108 ms–1, h = 6.6 × 10–34 J-s)

1. 6.82 eV
2. 12.5 eV
3. 7.72 eV
4. 8.52 eV

Answer: 3. 7.72 eV

Question 22. In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. The resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to:

1. 500 keV
2. 25 keV
3. 1 keV
4. 100 keV

Answer: 2. 25 keV

Question 23. A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980Å. The radius of the atom in the excited state, in terms of Bohr radius a0, will be : (hc = 12500 eV- Å)

1. 16 a0
2. 25 a0
3. 9a0
4. 4a0

Answer: 1. 16 a0

Question 24. If the de-Broglie wavelength of an electron is equal to 10–3 times the wavelength of a photon of frequency 6 × 1014 Hz, then the speed of the electron is equal to : (Speed of light = 3 × 108 m/s, Planck’s constant = 6.63 × 10–34J. s Mass of electron = 9.1 × 10-31 kg)

1. 1.7 × 106 m/s
2. 1.45 × 106 m/s
3. 1.8 × 106 m/s
4. 1.1 × 106 m/s

Answer: 2. 1.45 × 106 m/s

Question 25. In a hydrogen-like atom, when an electron jumps from the M – M-shell to the L – L-shell, the wavelength of emitted radiation is. If an electron jumps from the N-shell to the L-shell, the wavelength of emitted radiation will be:

1. \(\frac{16}{25} \lambda\)
2. \(\frac{25}{16} \lambda\)
3. \(\frac{20}{27} \lambda\)
4. \(\frac{20}{27} \lambda\)

Answer: 3. \(\frac{20}{27} \lambda\)

Question 26. In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to : ( e hc = 1240 nm–V)

1. 2.0 V
2. 0.5 V
3. 1.0 V
4. 1.5 V

Answer: 3. 1.0 V

Question 27. 2 A particle of mass m moves in a circular orbit in a central potential field U(r) = 2. If Bohr’s quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as:

1. \(r_n \propto \sqrt{n}, E_n \propto \frac{1}{n}\)
2. \(r_n \propto \sqrt{n}, E_n \propto n\)
3. \(r_n \propto n^2, E_n \propto \frac{1}{n^2}\)
4. \(r_n \propto \mathrm{n}, \mathrm{E}_{\mathrm{n}} \propto \mathrm{n}\)

Answer: 2. \(r_n \propto \sqrt{n}, E_n \propto n\)

Question 28. A particle A of mass ‘m’ and charge ‘q’ is accelerated by a potential difference of 50 V. Another particle B of mass 4m and charge q is accelerated by a potential difference of 2500 V. The ratio of de–Broglie A wavelengths \(\mathrm{} \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}\) is close to

1. 14.14
2. 10.00
3. 0.07
4. 4.47

Answer: 1. 14.14

Question 29. An alpha-particle of mass m suffers 1 -1-dimensional elastic collision with a nucleus at rest of unknown mass. it’s scattered directly backward losing, 64% of its initial kinetic energy. The mass of the nucleus is:

1. 3.5 m
2. 1.5 m
3. 4 m
4. 2 m

Answer: 3. 4 m

Question 30. When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photocurrent is \(-\frac{V_0}{2}\) When the surface is illuminated by monochromatic light of frequency \(\frac{v}{2},\) the stopping potential is –V0. The threshold frequency for photoelectric emission is:

1. \(\frac{4}{3} v\)
2. 2v
3. \(\frac{3 v}{2}\)
4. \(\frac{5 v}{3}\)

Answer: 3. \(\frac{3 v}{2}\)

Question 31. In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapor and emerges with an energy of 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to:

1. 1700 nm
2. 220 nm
3. 2020 nm
4. 250 nm

Answer: 4. 250 nm

Chapter 3 Modern Physics Multiple Choice Questions Self Practice Paper

Question 1. Yellow light of 557 nm wavelength is incident on a cesium surface. It is found that no photoelectrons flow in the circuit when the cathode-anode voltage drops below 0.25V. Then the threshold wavelength for the photoelectric effect from cesium is

1. 577 nm
2. 653 nm
3. 734 nm
4. 191 nm

Answer: 2. 653 nm

Question 2. The helium atom emits a photon of wavelength 0.1 A. The recoil energy of the atom due to the emission of photons will be

1. 2.04 eV
2. 4.91 eV
3. 1.67 eV
4. 9.10 eV

Answer: 1. 2.04 eV

Question 3. Electrons with an energy of 80 keV are incident on the tungsten target of an X−ray tube. K shell electrons of tungsten have −72.5keV energy. X−rays emitted by the tube contain only

1. A continuous x−ray spectrum (bremsstrahlung) with a minimum wavelength of ~ 0.155å.
2. A continuous X-ray spectrum (bremsstrahlung) with all wavelengths.
3. The characteristic x−ray spectrum of tungsten.
4. A continuous x−ray spectrum (bremsstrahlung) with a minimum wavelength of ~ 0.155å and the characteristic x−ray spectrum of tungsten.

Answer: 4. A continuous X-ray spectrum (bremsstrahlung) with a minimum wavelength of ~ 0.155å and the characteristic X-ray spectrum of tungsten.

Question 4. Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has a wavelength (given in terms of the Rydberg constant R for the hydrogen atom) equal to

1. 9/(5R)
2. 36/(5R)
3. 18/(5R)
4. 4/R

Answer: 3. 18/(5R)

Question 5. For the case discussed above, the wavelength of light emitted in the visible region by He + ions after collisions with H atoms is

1. 6.5 × 10–7 m
2. 5.6 × 10–7 m
3. 4.8 × 10–7 m
4. 4.0 × 10–7 m

Answer: 3. 4.8 × 10–7 m

Question 6. Photoelectric effect experiments are performed using three different metal plates p, q, and r having work functions p = 2.0 eV, vq = 2.5 eV, and vr = 3.0 eV respectively. A light beam containing wavelengths of 550 nm, 450 nm, and 350 nm with equal intensities illuminates each of the plates. The correct -V graph for the experiment is [Take hc = 1240 eV nm

Answer: 1.

Question 7. If λCu is the wavelength of the Kα X-ray line of copper (atomic number 29) and λMo is the wavelength of the Kα X-ray line of molybdenum (atomic number 42), then the ratio λCu/λMo is close to

1. 1.99
2. 2.14
3. 0.50
4. 0.48

Answer: 2. 2.14

Question 8. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u 1 and u 2, respectively. If the ratio u 1: u2 = 2: 1 and hc = 1240 eV nm, the work function of the metal is nearly

1. 3.7 eV
2. 3.2 eV
3. 2.8 eV
4. 2.5 eV

Answer: 1. 3.7 eV

Question 9. Consider a hydrogen atom with its electron in the nth orbital. Electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is (hc = 1242 eV nm)

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 10. The orbital angular momentum for an electron revolving in an orbit is given by. This momentum for an s-electron will be given by –

1. \(+\frac{1}{2} \cdot \frac{h}{2 \pi}\)
2. Zero
3. \(\frac{\mathrm{h}}{2 \pi}\)
4. \(\sqrt{2} \cdot \frac{h}{2 \pi}\)

Answer: 2. Zero

Question 11. In Bohr’s model of the hydrogen atom, the centripetal force is provided by the Coulomb attraction between the proton and the electron. If a0 is the radius of the ground state orbit, m is the mass e is the charge of an electron and e0 is the vacuum permittivity, the speed of the electron is:

1. zero
2. \(\frac{\mathrm{e}}{\sqrt{\varepsilon_0 \mathrm{a}_0 \mathrm{~m}}}\)
3. \(\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{a}_0 \mathrm{~m}}}\)
4. \(\frac{\sqrt{4 \pi \varepsilon_0 \mathrm{a}_0 \mathrm{~m}}}{\mathrm{e}}\)

Answer: 3. \(\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{a}_0 \mathrm{~m}}}\)

Modern Physics Photoelectric Effect Hertz’s observations

The phenomenon of photoelectric emission was discovered in 1887 by Heinrich Hertz (1857–1894), during his electromagnetic wave experiments. In his experimental investigation on the production of electromagnetic waves through a spark discharge, Hertz observed that high voltage sparks across the detector loop were enhanced when the emitter plate was illuminated by ultraviolet light from an arc lamp.

Light shining on the metal surface somehow facilitated the escape of free, charged particles which we now know as electrons. When light falls on a metal surface, some electrons near the surface absorb enough energy from the incident radiation to overcome the attraction of the positive ions in the surface material.

After gaining sufficient energy from the incident light, the electrons escape from the surface of the metal into the surrounding space.

Hallwach’s And Lenard’s Observations

Wilhelm Hallwachs and Phillipp Lenard investigated the phenomenon of photoelectric emission in detail during 1886–1902.

Lenard (1862–1947) observed that when ultraviolet radiations were allowed to fall on the emitter plate of an evacuated glass tube enclosing two electrodes (metal plates), current flows in the circuit figure.

As soon as the ultraviolet radiations were stopped, the current flow also stopped. These observations indicate that when ultraviolet radiations fall on the emitter plate C, electrons are ejected from it which are attracted towards the positive, collector plate A by the electric field. The electrons flow through the evacuated glass tube, resulting in the current flow.

Thus, light falling on the surface of the emitter causes current in the external circuit. Hallwachs and Lenard studied how this photocurrent varied with collector plate potential and with the frequency and intensity of incident light.

Hallwachs, in 1888, undertook the study further and connected a negatively charged zinc plate to an electroscope. He observed that the zinc plate lost its charge when it was illuminated by ultraviolet light.

Further, the uncharged zinc plate became positively charged when it was irradiated by ultraviolet light. The positive charge on a positively charged zinc plate was found to be further enhanced when it was illuminated by ultraviolet light.

From these observations, he concluded that negatively charged particles were emitted from the zinc plate under the action of ultraviolet light. After the discovery of the electron in 1897, it became evident that the incident light caused electrons to be emitted from the emitter plate.

Due to the negative charge, the emitted electrons are pushed toward the collector plate by the electric field. Hallwachs and Lenard also observed that when ultraviolet light fell on the emitter plate, no electrons were emitted at all when the frequency of the incident light was smaller than a certain minimum value, called the threshold frequency. This minimum frequency depends on the nature of the material of the emitter plate.

It was found that certain metals like zinc, cadmium, magnesium, etc. responded only to ultraviolet light, having a short wavelength, to cause electron emission from the surface. However, some alkali metals such as lithium, sodium, potassium, cesium, and rubidium were sensitive even to visible light.

All these photosensitive substances emit electrons when they are illuminated by light. After the discovery of electrons, these electrons were termed photoelectrons. The phenomenon is called the photoelectric effect.

Properties Of Photons

• A photon is a packet of energy emitted from a source of radiation. Photons are carrier particles of electromagnetic interaction.
• Photons travel in straight lines with speed of light c = 3 × 108 m/s.
• The energy of photons is given as \(\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}=\mathrm{mc}^2\) where v is frequency, lamba is wavelength, h is Planck’s constant.
• The effective or motional mass of photon is given as \(m=\frac{E}{c^2}=\frac{h \nu}{c^2}=\frac{h}{\lambda c}\)
• The momentum of a photon is given as \(\mathrm{p}=\mathrm{mc}=\frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{h} v}{\mathrm{c}}=\frac{\mathrm{h}}{\lambda}\)
• The photon is a chargeless particle of zero rest mass
• Photons are electrically neutral. They are not deflected by electric and magnetic fields.
• If E is the energy of source in joule then number of photons emitted is \(\mathrm{n}=\frac{\text { total energy radiated }}{\text { energy of each photon }}=\frac{E}{h \nu}=\frac{E \lambda}{h c}\)
• The intensity of photons is defined as the amount of energy carried per unit area per unit time. or power carried per unit area Intesity \(\left(I_p\right)=\frac{\text { Energy }}{\text { area } \times \text { time }}=\frac{\text { Power }}{\text { area }}, I_p=n h \nu=\frac{N}{4 \pi r^2} P\)

where n = number of photons per unit area per unit time N = number of photons, P = power of source

For a point source \(\mathrm{I}_{\mathrm{p}}=\mathrm{nh} v=\frac{\mathrm{N}}{4 \pi \mathrm{r}^2} \mathrm{P}\)

For a line source \(\mathrm{I}=\mathrm{nh} v=\frac{\mathrm{N}}{2 \pi \mathrm{r} \ell} \mathrm{P}\)

Solved Examples

Example 1. Find the number of photons in 6.62 joules of radiation energy of frequency 10¹²Hz.
Solution: Number Of Photons \(n=\frac{E}{h \nu}=\frac{6.62}{6.62 \times 10^{-34} \times 10^{12}}=10^{22}\)

Example 2. Calculate the energy and momentum of a photon of wavelength 6600Å.
Solution: energy of photon \(\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{6600 \times 10^{-10}}=3 \times 10^{-19} \mathrm{~J}\)

Momentum Of Photon \(\mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{6.6 \times 10^{-34}}{6600 \times 10^{-10}}=10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{sec}\)

Important Terms Related To The Photoelectric Effect

When electromagnetic radiations of suitable wavelength are incident on a metallic surface then electrons are emitted, this phenomenon is called the photoelectric effect.

Photoelectron: The electron emitted in the photoelectric effect is called a photoelectron.

Photoelectric current: If current passes through the circuit in a photoelectric effect then the current is called photoelectric current.

Work function: The minimum energy required to make an electron free from the metal is called work function. It is constant for a metal and denoted by  or W. It is the minimum for Cesium. It is relatively less for alkali metals.

Work functions of some photosensitive metals

To produce a photoelectric effect only metal and light are necessary but for observing it the circuit is completed. Figure shows an arrangement used to study the photoelectric effect.

Here the plate (1) is called emitter or cathode and the other plate (2) is called collector or anode.

Saturation current: When all the photoelectrons emitted by the cathode reach the anode then the current flowing in the circuit at that instant is known as saturated current, this is the maximum value of photoelectric current.

Stopping potential: The minimum magnitude of the negative potential of the anode to the cathode for which the current is zero is called stopping potential. This is also known as cutoff voltage. This voltage is independent of intensity.

Retarding potential: Negative potential of the anode to the cathode which is less than the stopping potential is called retarding potential.

Observations: (Made By Einstein)

A graph between the intensity of light and photoelectric current is found to be a straight line as shown in the figure. Photoelectric current is directly proportional to the intensity of incident radiation. In this experiment, the frequency and retarding potential are kept constant.

A graph between photoelectric current and a potential difference between cathode and anode is found as shown in the figure.

In the case of saturation current, rate of emission of photoelectrons = rate of flow of photoelectrons, here, vs→ stopping potential and it is a positive quantity Electrons emitted from the surface of metal have different energies. Maximum kinetic energy of photoelectron on the cathode = eVs KEmax = eV s

Whenever the photoelectric effect takes place, electrons are ejected out with kinetic energies ranging from 0 to K.Emax

i.e. 0 ≤ KEc ≤ eVs

The energy distribution of photoelectrons is shown in the figure.

Stopping potential (Vs) eVs = hv – W (Work function W = hv0)

⇒ \(V_s=\frac{h\left(v-v_0\right)}{e}\)

If the intensity is increased (keeping the frequency constant) then the saturation current is increased by the same factor by which intensity increases. The stopping potential is the same, so the maximum value of kinetic energy is not affected.

If light of different frequencies is used then the obtained plots.

It is clear from the graph, that as v increases, stopping potential increases, which means the maximum value of kinetic energy increases.

Graphs between the maximum kinetic energy of electrons ejected from different metals and the frequency of light used are found to be straight lines of the same slope.

Graph between Kmax and v m₁, m₂, m₃: Three different metals.

It is clear from the graph that there is a minimum frequency of electromagnetic radiation that can produce a photoelectric effect, which is called threshold frequency.

vth = Threshold frequency

For photoelectric effect v>vth

For no photoelectric effect v < vth Minimum frequency for the photoelectric effect.

Threshold wavelength \(\begin{aligned}
& v_{\min }=v_{t h} \\
& \left(\lambda_{t h}\right) \rightarrow
\end{aligned}\) The maximum wavelength of radiation that can produce a photoelectric effect. \(\lambda \leq \lambda_{\text {th }}\) for photo electric effect Maximum wavelength for photoelectric effect \(\lambda_{\max }=\lambda_{\mathbb{E n}} .\) Now write the equation of a straight line from the graph.

We have \(\mathrm{K}_{\max }=\mathrm{Av}+\mathrm{B}\)

When \(v=v_{\text {th }}, K_{\max }=0\)

And B = – Avth

Hence [Kmax = A(v – vth)]

and A = tan 0 = 6.63 × 10–34 J-s (from experimental data) later on ‘A’ was found to be ‘h’.

It is also observed that the photoelectric effect is an instantaneous process. When light falls on the surface electrons start ejecting without taking any time.

Three Major Features Of The Photoelectric Effect Cannot Be Explained In Terms Of The Classical Wave Theory Of Light.

Intensity: The energy crossing per unit area per unit time perpendicular to the direction of propagation is called the intensity of a wave.

Consider a cylindrical volume with an area of cross-section A and length c Δt along the X-axis. The energy contained in this cylinder crosses the area A in time Δt as the wave propagates at speed c. The energy contained.

The intensity is \(\begin{aligned}
& \mathrm{U}=\mathrm{U}_{\mathrm{av}}(\mathrm{c} . \Delta \mathrm{t}) \mathrm{A} \\
& \mathrm{I}=\frac{\mathrm{U}}{\mathrm{A} \Delta \mathrm{t}}=\mathrm{U}_{\mathrm{av}} \mathrm{c} .
\end{aligned}\)

In terms of maximum electric field, \(\mathrm{I}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{C}.\)

If we consider light as a wave then the intensity depends upon the electric field. If we take work function W = Δ . A . t, then \(t=\frac{W}{I A}\)

so for the photoelectric effect, there should be a time lag because the metal has a work function. However, it is observed that the photoelectric effect is an instantaneous process. Hence, light is not of a wave nature.

The intensity problem: Wave theory requires that the oscillating electric field vector E of the light wave increases in amplitude as the intensity of the light beam is increased. Since the force applied to the electron is eE, this suggests that the kinetic energy of the photoelectrons should also be increased as the light beam is made more intense. However, observation shows that maximum kinetic energy is independent of the light intensity.

The frequency problem: According to the wave theory, the photoelectric effect should occur for any frequency of the light, provided only that the light is intense enough to supply the energy needed to eject the photoelectrons. However, observations show that there exists for each surface a characteristic cutoff frequency vth, for frequencies less than that, the photoelectric effect does not occur, no matter how intense is light beam.

The time delay problem: If the energy acquired by a photoelectron is absorbed directly from the wave incident on the metal plate, the “effective target area” for an electron in the metal is limited and probably not much more than that of a circle of diameter roughly equal to that of an atom. In classical theory, the light energy is uniformly distributed over the wavefront.

Thus, if the light is feeble enough, there should be a measurable time lag, between the impinging of the light on the surface and the ejection of the photoelectron. During this interval, the electron should be absorbing energy from the beam until it has accumulated enough to escape. However, no detectable time lag has ever been measured. Now, quantum theory solves these problems by providing the correct interpretation of the photoelectric effect.

Planck’s Quantum Theory:

The light energy from any source is always an integral multiple of a smaller energy value called the quantum of light. Hence energy Q = NE,

where E = hv and N (number of photons) = 1,2,3,…

Here energy is quantized. hv is the quantum of energy, it is a packet of energy called a photon.

⇒ \(E=h \nu=\frac{h c}{\lambda} \quad \text { and } \quad h c=12400 \mathrm{eVA}\)

Einstein’s Photon Theory

In 1905 Einste made a remarkable assumption about the nature of light; namely, that, under some circumstances, it behaves as if its energy is concentrated into localized bundles, later called photons. The energy E of a single photon is given by E = hv, If we apply Einstein’s photon concept to the photoelectric effect, we can write hv = W + K max, (energy conservation)

The equation says that a single photon carries an energy h into the surface where it is absorbed by a single electron. Part of this energy W (called the work function of the emitting surface) is used in causing the electron to escape from the metal surface.

The excess energy (hv – W) becomes the electron kinetic energy; if the electron does not lose energy by internal collisions as it escapes from the metal, it will still have this much kinetic energy after it emerges. Thus Kmax represents the maximum kinetic energy that the photoelectron can have outside the surface. There is complete agreement of the photon theory with the experiment.

Now IA = Nhv \(\mathrm{N}=\frac{\mathrm{IA}}{\mathrm{h} \nu}\) no. of photons incident per unit time on an area ‘A’ when the light of intensity ‘I’ is incident normally.

If we double the light intensity, we double the number of photons and thus double the photoelectric current; we do not change the energy of the individual photons or the nature of the individual photoelectric processes.

The second objection (the frequency problem) is met if Kmax equals zero, we have

⇒ \(\mathrm{h} \mathrm{v}_{\mathrm{th}}=\mathrm{W} \text {, }\)

Which asserts that the photon has just enough energy to eject the photoelectrons and none extra to appear as kinetic energy. If vth is reduced below th, h will be smaller than W and the individual photons, no matter how many of them there are (that is, no matter how intense the illumination), will not have enough energy to eject photoelectrons.

The third objection (the time delay problem) follows from the photon theory because the required energy is supplied in a concentrated bundle. It is not spread uniformly over the beam cross-section as in the wave theory.

Hence Einstein’s equation for the photoelectric effect is given by

⇒ \(\mathrm{h} v=\mathrm{h} v_{\mathrm{th}}+\mathrm{K}_{\max } \quad \mathrm{K}_{\max }=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{\mathrm{th}}}\)

Solved Examples

Example 3. In an experiment on photoelectric emission, the following observations were made;

1. Wavelength of the incident light = 1.98 × 10–7 m;
2. Stopping potential = 2.5 volt.

Find:

1. The kinetic energy of photoelectrons with maximum speed.
2. Work function and
3. Threshold frequency;

Solution:

Since vs = 2.5 V, Kmax = eVs

so, K max = 2.5 eV

Energy of incident photon 12400 eV E = 1980 = 6.26 eV W = E – K max = 3.76 eV

⇒ \(\mathrm{E}=\frac{12400}{1980} \mathrm{eV}=6.26 \mathrm{eV} \quad \mathrm{W}=\mathrm{E}-\mathrm{K}_{\max }=3.76 \mathrm{eV}\)

⇒ \(\begin{aligned}
& \mathrm{h} v_{\text {th }}=\mathrm{W}=3.76 \times 1.6 \times 10^{-19} \mathrm{~J} \\
& v_{\text {th }}=\frac{3.76 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} \approx 9.1 \times 10^{14} \mathrm{~Hz}
\end{aligned}\)

Example 4. A beam of light consists of four wavelengths 4000 Å, 4800 Å, 6000 Å, and 7000 Å, each of intensity 1.5 × 10–3 Wm–2. The beam falls normally on an area of 10–4 m2 of a clean metallic surface of work function 1.9 eV. Assuming no loss of light energy (i.e. each capable photon emits one electron) calculate the number of photoelectrons liberated per second.
Solution:

⇒ \(\begin{aligned}
& E_1=\frac{12400}{4000}=3.1 \mathrm{eV}, \quad E_2=\frac{12400}{4800}=2.58 \mathrm{eV} \quad E_3=\frac{12400}{6000}=2.06 \mathrm{eV} \\
& \text { and } \quad E_4=\frac{12400}{7000}=1.77 \mathrm{eV}
\end{aligned}\)

Therefore, light of wavelengths 4000 Å, 4800 Å, and 6000 Å can only emit photoelectrons.

Number of photoelectrons emitted per second = No. of photons incident per second)

⇒ \(=\frac{I_1 A_1}{E_1}+\frac{I_2 A_2}{E_2}+\frac{I_3 A_3}{E_3} \quad=I A\left(\frac{1}{E_1}+\frac{1}{E_2}+\frac{1}{E_3}\right)=\frac{\left(1.5 \times 10^{-3}\right)\left(10^{-4}\right)}{1.6 \times 10^{-19}}\left(\frac{1}{3.1}+\frac{1}{2.58}+\frac{1}{2.06}\right)\)

Example 5. A small potassium foil is placed (perpendicular to the direction of incidence of light) at a distance r (= 0.5 m) from a point light source whose output power P0 is 1.0W.

Assuming the wave nature of light how long would it take for the foil to soak up enough energy (= 1.8 eV) from the beam to eject an electron? Assume that the ejected photoelectron collected its energy from a circular area of the foil whose radius equals the radius of a potassium atom (1.3 × 10–10 m.

Solution: If the source radiates uniformly in all directions, the intensity  of the light at a distance r is given by \(\mathrm{I}=\frac{P_0}{4 \pi \mathrm{r}^2}=\frac{1.0 \mathrm{~W}}{4 \pi(0.5 \mathrm{~m})^2}=0.32 \mathrm{~W} / \mathrm{m}^2 \text {. }\)

The target area A is π(1.3 × 10–10 m)2 or 5.3 × 10–20 m2, so the rate at which energy falls on the target is given by P = πA = (0.32 W/m2) (5.3 × 10–20 m2) = 1.7 × 10–20 J/s.

If all this incoming energy is absorbed, the time required to accumulate enough energy for the electron to escape is \(\mathrm{t}=\left(\frac{1.8 \mathrm{eV}}{1.7 \times 10^{-20} \mathrm{~J} / \mathrm{s}}\right)\left(\frac{1.6 \times 10^{-19} \mathrm{~J}}{1 \mathrm{eV}}\right)=17 \mathrm{~s}.\)

Our selection of a radius for the effective target area was somewhat arbitrary, but no matter what reasonable area we choose, we should still calculate a “soak-up time” within the range of easy measurement. However, no time delay has ever been observed under any circumstances, the early experiments setting an upper limit of about 10–9 s for such delays.

Example 6. A metallic surface is irradiated with monochromatic light of variable wavelength. Above a wavelength of 5000 Å, no photoelectrons are emitted from the surface. With an unknown wavelength, the stopping potential is 3 V. Find the unknown wavelength.
Solution: using equation of photoelectric effect \(\mathrm{K}_{\max }=\mathrm{E}-\mathrm{W} \quad\left(\mathrm{K}_{\max }=\mathrm{eV}_{\mathrm{s}}\right)\)

Therefore \(3 \mathrm{eV}=\frac{12400}{\lambda}-\frac{12400}{5000}=\frac{12400}{\lambda}-2.48 \mathrm{eV} \text { or } \lambda=2262 \mathrm{~A}\)

Example 7. Illuminating the surface of a certain metal alternately with light of wavelengths λ1 = 0.35 λm and λ2 = 0.54 μm, it was found that the corresponding maximum velocities of photoelectrons have a ratio μ = 2. Find the work function of that metal.
Solution: Using the equation for two wavelengths

⇒ \(\begin{aligned}
&\frac{1}{2} m v_1^2=\frac{h c}{\lambda_1}-W\\
&\frac{1}{2} m v_2^2=\frac{h c}{\lambda_2}-W
\end{aligned}\)

Dividing Eq. 1 with Eq. 2 with v1 = 2v2, we have \(4=\frac{\frac{h c}{\lambda_1}-W}{\frac{h c}{\lambda_2}-W}\)

⇒ \(3 \mathrm{~W}=4\left(\frac{\mathrm{hc}}{\lambda_2}\right)-\left(\frac{\mathrm{hc}}{\lambda_1}\right)=\frac{4 \times 12400}{5400}-\frac{12400}{3500}=5.64 \mathrm{eV}\)

Example 8. Light described at a place by the equation E = (100 V/m) [sin (5 × 10 15 s–1) t + sin (8 × 1015 s–1)t] falls on a metal surface having work function 2.0 eV. Calculate the maximum kinetic energy of the photoelectrons.
Solution: The light contains two different frequencies. The one with a larger frequency will cause photoelectrons with the largest kinetic energy. This larger frequency is

⇒ \(v=\frac{\omega}{2 \pi}=\frac{8 \times 10^{15} \mathrm{~s}^{-1}}{2 \pi}\)

The maximum kinetic energy of the photoelectrons is

⇒ \(\begin{aligned}
\mathrm{K}_{\max } & =\mathrm{hu}-\mathrm{W} \\
& =\left(4.14 \times 10^{-15} \mathrm{eV}-\mathrm{s}\right) \times\left(\frac{8 \times 10^{15}}{2 \pi} \mathrm{s}^{-1}\right)-2.0 \mathrm{eV}=5.27 \mathrm{eV}-2.0 \mathrm{eV}=3.27 \mathrm{eV} .
\end{aligned}\)

Compton Effect

The scattering of a photon by an electron in which the wavelength of the scattered photon is greater than the wavelength of the incident photon is called the Compton effect. Conservation of energy gives h_v + m₀ c² = h_v’ + mc²

Conservation of momentum along y-axis gives \(0=\frac{h v^{\prime}}{c} \sin \theta-p \sin \theta\)

Conservation of momentum along x-axis gives \(\frac{\mathrm{h} v}{\mathrm{c}}=\frac{\mathrm{h} v^{\prime}}{\mathrm{c}} \cos \theta+\mathrm{p} \cos \phi\)

Compton shif \(\Delta \lambda=\lambda^{\prime}-\lambda=\frac{h}{m_0 c}(1-\cos \theta)\)

The quantity \(\lambda_{\mathrm{c}}=\frac{\mathrm{h}}{\mathrm{m}_0 \mathrm{c}}=2.42 \times 10^{12} \mathrm{~m}\) s called Compton wavelength of electron.

For maximum shift \(\theta=\pi \quad \text { so } \quad(\Delta \lambda)_{\max }=\frac{2 \mathrm{~h}}{\mathrm{~m}_0 \mathrm{c}}=4.48 \times 10^{-12} \mathrm{~m}\)

The Compton effect shows the quantum concept and particle nature of photons.

De-Broglie Wavelength Of Matter Wave

A photon of frequency v and wavelength λ has energy

⇒ \(E=h \nu=\frac{h c}{\lambda}\)

By Einstein’s energy-mass relation, E = mc2 the equivalent mass m of the photon is given by,

⇒ \(\begin{gathered}
m=\frac{E}{c^2}=\frac{h \nu}{c^2}=\frac{h}{\lambda c} \\
\lambda=\frac{h}{m c} \quad \text { or } \quad \lambda=\frac{h}{p}
\end{gathered}\)

Here p is the momentum of the photon. By analogy de-Broglie suggested that a particle of mass m moving with speed v behaves in some ways like waves of wavelength (called de-Broglie wavelength and the wave is called matter wave) given by,

⇒ \(\lambda=\frac{h}{m v}=\frac{h}{p}\)

where p is the momentum of the particle. Momentum is related to the kinetic energy by the equation, \(p=\sqrt{2 K m}\) and a charge q when accelerated by a potential difference V gains a kinetic energy K = qV. Combining all these relations Eq. (3), can be written as, \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{Km}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{qVm}}} \text { (de-Broglie wavelength) }\) de-Broglie wavelength for an electron If an electron (charge = e) is accelerated by a potential of V volts, it acquires a kinetic energy, K = eV Substituting the values of h, m, and q in Eq., we get a simple formula for calculating deBroglie wavelength of an electron.

⇒ \(\lambda(\text { in } \quad A)=\sqrt{\frac{150}{V(\text { in volts })}} \frac{12.27 }{\sqrt{V}}\)

The de-Broglie wavelength of a proton:

Velocity \(V_p=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}_{\mathrm{p}}}}\)
Momentum ,\(p_p=\sqrt{2 m_p e V}\)

So Wavelength is \(\lambda_p=\frac{h}{\sqrt{2 m_p e V}}=\frac{0.286}{\sqrt{V}} A\)

de-Broglie wavelength of a gas molecule: Let us consider a gas molecule at absolute temperature T. The Kinetic energy of a gas molecule is given by

⇒ \(K. E .=\frac{3}{2} k T \text {; }\) ,therefore \(\lambda_{\text {gas molecule }}=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}\)

Force Due To Radiation (Photon)

Each photon has a definite energy and a definite linear momentum. All photons of light of a particular wavelength λ have the same energy E = hc/λ and the same momentum p = h/λ. When light of intensity λ falls on a surface, it exerts force on that surface.

Assume the absorption and reflection coefficient of the surface to be ‘a’ and ‘r’ and assume no transmission. Assume the light beam falls on the surface of surface area ‘A’ perpendicularly as shown in the figure.

For calculating the force exerted by the beam on the surface, we consider the following cases

Case: 1

a = 1, r = 0

Initial momentum of the photon \(=\frac{\mathrm{h}}{\lambda}\)

Inal momentum of photon = 0

Change in momentum of photon = \(\frac{\mathrm{h}}{\lambda}\) \(\Delta \mathrm{P}=\frac{\mathrm{h}}{\lambda}\) energy incident per unit time = IA A no. of photons incident per unit time \(=\frac{I A}{h v}=\frac{I A \lambda}{h c}\)

Therefore total change in momentum per unit time \(=\mathrm{n} \Delta \mathrm{P}=\frac{\mathrm{IA} \lambda}{\mathrm{hc}} \times \frac{\mathrm{h}}{\lambda}=\frac{\mathrm{IA}}{\mathrm{c}}\)

Force on photons = total change in momentum per unit time \(=\frac{I A}{c}\)

force on plate due to photons(F) \(=\frac{\mathrm{IA}}{\mathrm{c}}\) pressure \(=\frac{F}{A}=\frac{I A}{c A}=\frac{I}{c}\)

Case : (2)

when r = 1, a = 0

Initial momentum of the photon \(=\frac{\mathrm{h}}{\lambda}\) (downward)

Final momentum of photon \(=\frac{\mathrm{h}}{\lambda}\) (upward)

Change in momentum \(=\frac{\mathrm{h}}{\lambda}+\frac{\mathrm{h}}{\lambda}=\frac{2 \mathrm{~h}}{\lambda}\)

∴ Energy incident per unit time = IA

Number of photons incidient per unit = \(\frac{\mathrm{IA} \lambda}{\mathrm{hc}}\)

∴ Total change in momentum per unit time \(=n \cdot \Delta P=\frac{I A \lambda}{h c} \cdot \frac{2 h}{\lambda}=\frac{2 I A}{C}\)

Force = total change in momentum per unit time

⇒ \(F=\frac{2 I A}{c}\) (upward on photons and downward on the plate)

Pressure \(P=\frac{F}{A}=\frac{2 I A}{c A}=\frac{2 I}{c}\)

Case 3

When o < r < 1 a + r = 1

change in momentum of a photon when it is reflected \(=\frac{2 h}{\lambda}\) upward

change in momentum of a photon when it is absorbed \(=\frac{\mathrm{h}}{\lambda}\) (upward)

No. of photons incident per unit time \(=\frac{\mathrm{IA} \lambda}{\mathrm{hc}}\)

No. of photons reflected per unit time \(=\frac{\mathrm{IA} \lambda}{\mathrm{hc}} \cdot \mathrm{r}\)

No. of photon absorbed per unit time \(=\frac{I A \lambda}{h c}(1-r)\)

force due to absorbed photon (Fa) \(=\frac{I A \lambda}{h c}(1-r) \cdot \frac{h}{\lambda}=\frac{I A}{c}(1-r) \quad \text { (downward) }\)

Force due to reflected photon (Fr) \(=\frac{I A \lambda}{h c} \cdot r \frac{2 h}{\lambda}=\frac{2 I A \lambda}{c}\)

Total force = Fa+F \((\text { downward })=\frac{1 A}{c}(1-r)+\frac{2 \mid A r}{c}=\frac{1 A}{c}(1+r)\)

Now Pressure \(P=\frac{I A}{c}(1+r) \times \frac{1}{A}=\frac{I}{c}(1+r)\)

Example 9. An electron is accelerated by a potential difference of 50 volts. Find the de-Broglie wavelength associated with it.
Solution: For an electron, the de-Broglie wavelength is given by,

Example 10. Find the ratio of De-Broglie wavelength of molecules of hydrogen and helium which are at temperatures 27ºC and 127ºC respectively.
Solution: de-Broglie wavelength is given by

Therefore \(\frac{\lambda_{\mathrm{H}_2}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_2} \mathrm{~T}_{\mathrm{H}_2}}}=\sqrt{\frac{4}{2} \cdot \frac{(127+273)}{(27+273)}}=\sqrt{\frac{8}{3}}\)

Example 11. A plate of mass 10 gm is in equilibrium in the air due to the force exerted by a light beam on a plate. Calculate the power of the beam. Assume the plate is perfectly absorbing.

Solution: Since the plate is in the air, so gravitational force will act on this

⇒ \(\begin{aligned}
\mathrm{F}_{\text {gravitational }} & =\mathrm{mg} \\
& =10 \times 10^{-3} \times 10 \\
& =10^{-1} \mathrm{~N}
\end{aligned}\)

for equilibrium force exerted by light beam should be equal to \(\text { gravitational }\)

⇒ \(\mathrm{F}_{\text {photon }}=\mathrm{F}_{\text {gravitational }}\)

Let the power of the light beam be P

Therefore \(\begin{aligned}
& F_{\text {photon }}=\frac{P}{C} \\
& P=3.0 \times 10^8 \times 10^{-1} \\
& P=3 \times 10^7 \mathrm{~W}
\end{aligned}\)

⇒ \(\frac{P}{c}=10^{-1}\)

Davisson And Germer Experiment

• The experiment demonstrates the diffraction of electron beam by crystal surfaces
• The experiment provides the first experimental evidence for the wave nature of the material particles

• The electrons are diffracted like X-rays. The Bragg’s law of diffraction are
  D sin Φ = nλ and 2d sinθ = nλ
  • where D = interatomic distance and d = interplanar distance
  • θ = angle between the scattering plane and the incident beam
  • Φ = scattering angle 2θ + Φ = 180º
• The electrons are produced and accelerated into a beam by an electron gun.
  • The energy of electrons is given as \(\mathrm{E}=\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}\)

The accelerated electron beam is made to fall on a Ni crystal. The scattered electrons are detected by a detector

The experimental results are shown in the form of polar graphs plotted between scattering angle Φ and intensity of scattered electron beam at different accelerating voltages. The distance of the curve from point O is proportional to the intensity of the scattered electron beam.

Important Results

• The intensity of scattered electrons depends on the scattering angle Φ
• The kink at Φ = 50° is observed at all accelerating voltage.
• The size of the link becomes maximum at 54 volts. \(\text { For } \phi=50^{\circ} \quad \text { For } \theta=\frac{180-\phi}{2}=65^{\circ}\)
• D = sinλ = nλ 2 d sinθ = nθ
• D = 2.15Å & n = 1 (for Ni)
• n = 1 and d = 0.91Å (for Ni)
  • ∴ λ = 2.15 sin 50 = 1.65Å
• λ = 2 × 0.91 sin65 = 1.65Å
• For V = 54 volt

de-Broglie wavelength

This value of λ is in close agreement with the experimental value. Thus this experiment verifies de Broglie’s hypothesis.

Solved Example

Example 12. An electron beam of energy 10 KeV is incident on metallic foil. If the interatomic distance is 0.55Å. Find the angle of diffraction.
Solution: \(\lambda=D \sin \phi \quad \text { and } \quad \lambda=\frac{12.27}{\sqrt{V}} A \text { so } \frac{12.27}{\sqrt{V}}=D \sin \phi\)

⇒ \(\frac{12.27 \times 10^{-10}}{\sqrt{10 \times 10^3}}=0.55 \times 10^{-10} \sin \phi\) \(\sin \phi=\frac{12.27}{0.53 \times 100}=0.2231\)

⇒ \(\text { or } \phi=\sin ^{-1}(0.2231) \approx 12.89^{\circ}\)

Thomson’s Atomic Model:

J.J. Thomson suggested that atoms are just positively charged lumps of matter with electrons embedded in them like raisins in a fruit cake. The total charge of the atom is zero and the atom is electrically neutral. Thomson’s model called the ‘plum pudding’ model is illustrated in the figure.

Thomson played an important role in discovering the electron, through a gas discharge tube by discovering cathode rays. His idea was taken seriously. But the real atom turned out to be quite different.

Rutherford’s Nuclear Atom:

Rutherford suggested that; “ All the positive charge and nearly all the mass were concentrated in a very small volume of the nucleus at the center of the atom. The electrons were supposed to move in circular orbits around the nucleus (like planets around the sun). The electron static attraction between the two opposite charges is the required centripetal force for such motion.

Hence \(\frac{m v^2}{r}=\frac{k Z e^2}{r^2}\) and total energy = potential energy + kineitc energy \(=\frac{-k Z e^2}{2 r}\)

Rutherford’s model of the atom, although strongly supported by evidence for the nucleus, is inconsistent with classical physics. This model suffers from two defects.

Regarding the stability of an atom: An electron moving in a circular orbit around a nucleus is accelerating and according to electromagnetic theory it should, therefore, emit radiation continuously and thereby lose energy.

If total energy decreases then radius increases as given by the above formula. If this happened the radius of the orbit would decrease and the electron would spiral into the nucleus in a fraction of a second. But atoms do not collapse. In 1913 an effort was made by Neils Bohr to overcome this paradox.

Regarding the explanation of line spectrum: In Rutherford’s model, due to continuously changing radii of the circular orbits of electrons, the frequency of revolution of the electrons must be changing. As a result, electrons will radiate electromagnetic waves of all frequencies, i.e., the spectrum of these waves will be ‘continuous’ in nature. But experimentally the atomic spectra are not continuous. Instead, they are line spectra.

The Bohr’s Atomic Model

In 1913, Prof. Niel Bohr removed the difficulties of Rutherford’s atomic model by the application of Planck’s quantum theory. For this, he proposed the following postulates

An electron moves only in certain circular orbits, called stationary orbits. In stationary orbits, electrons do not emit radiation, contrary to the predictions of classical electromagnetic theory.

According to Bohr, there is a definite energy associated with each stable orbit, and an atom radiates energy only when it makes a transition from one of these orbits to another. If the energy of an electron in the higher orbit is E2 and that in the lower orbit is E1, then the frequency of the radiated waves is given by

hv=E2-E1 or \(v=\frac{E_2-E_1}{h}\)

Bohr found that the magnitude of the electron’s angular momentum is quantized, and this magnitude for the electron must be an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\) The magnitude of the angular momentum is L = mvr for a particle with mass m moving with speed v in a circle of radius r. So, according to Bohr’s postulate, \(m v r=\frac{n h}{2 \pi} \quad(n=1,2,3 \ldots .)\)

Each value of n corresponds to a permitted value of the orbit radius, which we will denote by rn. The value of n for each orbit is called the principal quantum number for the orbit. Thus, \(m v_n r_n=m v r=\frac{n h}{2 \pi}\)

According to Newton’s second law, a radially inward centripetal force of magnitude \(F=\frac{m v^2}{r_n}\) is needed by the electron which is provided by the electrical attraction between the positive proton and the negative electron.

Thus \(\frac{m v_n^2}{r_n}=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r_n^2}\)

Solving Eqs. (2) and (3), we get 2

⇒ \(\begin{aligned}
& r_n=\frac{\varepsilon_0 n^2 h^2}{\pi m e^2} \\
& v_n=\frac{e^2}{2 \varepsilon_0 n h}
\end{aligned}\)

The smallest orbit radius corresponds to n = 1. We’ll denote this minimum radius, called the Bohr radius as a 0. Thus, \(\mathrm{a}_0=\frac{\varepsilon_0 \mathrm{~h}^2}{\pi m \mathrm{e}^2}\)

Substituting values of e0, h, p, m, and e, we get a 0 = 0.529 × 10–10 m = 0.529 Å

Eq. 4, in terms of a0 can be written as, r n = n2 a 0 or rn  n2

Similarly, substituting values of e, 0, and h with n = 1 in Eq. (v), we get

v1 = 2.19 × 106 m/s

This is the greatest possible speed of the electron in the hydrogen atom. Which is
approximately equal to c/137 where c is the speed of light in vacuum.
Eq. (v), in terms of v1, can be written as,

⇒ \(v_n=\frac{v_1}{n} \quad \text { or } \quad v_n \propto \frac{1}{n}\)

Energy levels: Kinetic and potential energies Kn and U n in the nth orbit are given by

⇒ \(\mathrm{K}_{\mathrm{n}}=\frac{1}{2} \mathrm{mv}_{\mathrm{n}}{ }^2=\frac{\mathrm{me}^4}{8 \varepsilon_0^2 \mathrm{n}^2 \mathrm{~h}^2} \quad \text { and } \quad \mathrm{U}_{\mathrm{n}}=-\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{e}^2}{\mathrm{r}_{\mathrm{n}}} \quad=-\frac{m e^4}{4 \varepsilon_0^2 \mathrm{n}^2 \mathrm{~h}^2}\) (assuming infinity as a zero potential energy level)

The total energy En is the sum of the kinetic and potential energies.

So, \(E_n=K_n+U_n=-\frac{m e^4}{8 \varepsilon_0{ }^2 n^2 h^2}\)

Substituting values of m, e, e0, and h with n = 1, we get the least energy of the atom in the first orbit, which is –13.6 eV. Hence, E1=-13.6ev and \(E_n=\frac{E_1}{n^2}=-\frac{13.6}{n^2} \mathrm{eV}\)

Substituting n = 2, 3, 4, …., etc., we get the energies of atoms in different orbits.

E2 = – 3.40 eV, E3 = – 1.51 eV, …. E∞ = 0

Solved Examples

Example 13. An -particle with kinetic energy 10 MeV is heading towards a stationary point-nucleus of atomic number 50. Calculate the distance of the closest approach.
Solution: Therefore \(T E_{\mathrm{A}}=\mathrm{TE}_{\mathrm{B}}\) \(10 \times 10^6 e=\frac{K \times(2 e)(50 e)}{r_0}\)

r0 = 1.44 × 10–14 m

r0 = 1.44 × 10–4 Å

Example 14. A beam of α-particles of velocity 2.1 × 107 m/s is scattered by a gold (z = 79) foil. Find out the distance of the closest approach of the α-particle to the gold nucleus. The value of charge/mass for α-particle is 4.8 × 107 C/kg.
Solution:

⇒ \(\frac{1}{2} m_\alpha V_\alpha{ }^2=\frac{K(2 e)(Z e)}{r_0}\)

⇒\(r_0=\frac{2 \mathrm{~K}\left(\frac{2 \mathrm{e}}{\mathrm{m}_\alpha}\right)(79 \mathrm{e})}{\mathrm{V}_\alpha^2}=\frac{2 \times\left(9 \times 10^9\right)\left(4.8 \times 10^7\right)\left(79 \times 1.6 \times 10^{-19}\right)}{\left(2.1 \times 10^7\right)^2} ; r_0=2.5 \times 10^{-14} \mathrm{~m}\)

Hydrogen-Like Atoms The Bohr model of hydrogen can be extended to hydrogen-like atoms, i.e., one electron atom, the nuclear charge is +ze, where z is the atomic number, equal to the number of protons in the nucleus.

The effect in the previous analysis is to replace e2 everywhere with ze2. Thus, the equations for, rn, vn, and E n are altered as under:

⇒ \(r_n=\frac{\varepsilon_0 n^2 h^2}{n m z e^2}=\frac{n^2}{z} a_0 \quad \text { or } \quad r_n \propto \frac{n^2}{z}\)

where a 0 = 0.529 Å (radius of the first orbit of H)

⇒ \(v_n=\frac{z e^2}{2 \varepsilon_0 n h}=\frac{z}{n} v_1 \quad \text { or } \quad v_n \propto \frac{z}{n}\)

where v 1= 2.19 × 106 m/s (speed of an electron in the first orbit of H)

⇒ \(E_n=-\frac{m z^2 e^4}{8 \varepsilon_0^2 n^2 h^2}=\frac{z^2}{n^2} E_1 \text { or } \quad E_n \propto \frac{z^2}{n^2}\)

Where E1 = –13.60 eV (energy of the atom in the first orbit of H)

Destinations valid for single electron system

Ground state: The lowest energy state of any atom or ion is called the ground state of the atom. Ground state energy of H atom = –13.6 eV Ground state energy of He+ Ion = –54.4 eV

Ground state energy of Li++ Ion = –122.4 eV

Excited State: The state of an atom other than the ground state is called its excited state.

n = 2 first excited state

n = 3-second excited state

n = 4 third excited state

n = n

0 + 1 n0th excited state

Ionization energy (IE.): The minimum energy required to move an electron from the ground state to n = ∞ is called the ionization energy of the atom or ion.

• The ionization energy of H atom = 13.6 eV
• Ionisation energy of He+ Ion = 54.4 eV
• Ionisation energy of Li++ Ion = 122.4 eV

Ionization potential (I.P.): The potential difference through which a free electron must be accelerated from rest such that its kinetic energy becomes equal to the ionization energy of the atom is called the ionization potential of the atom.

• I.P of H atom = 13.6 V
• I.P. of He+ Ion = 54.4 V

Excitation energy: Energy required to move an electron from the ground state of the atom to any other exited state of the atom is called excitation energy of that state.

Energy in the ground state of H atom = –13.6 eV

Energy in the first excited state of H-atom = –3.4 eV

1st excitation energy = 10.2 eV.

Excitation Potential: The potential difference through which an electron must be accelerated from rest so that its kinetic energy becomes equal to the excitation energy of any state is called the excitation potential of that state.

1st excitation energy = 10.2 eV.

1st excitation potential = 10.2 V.

Binding energy or Separation energy: Energy required to move an electron from any state to n = is called binding energy of that state. or energy released during the formation of an H-like atom/ion from n = ∞ to some particular n is called binding energy of that state. The binding energy of the ground state of H-atom = 13.6 eV.

Solved Examples

Example 15. First excitation potential of a hypothetical hydrogen-like atom is 15 volts. Find the third excitation potential of the atom.
Solution: Let the energy of the ground state = E0

⇒ \(E_0=-13.6 Z^2 e V \quad \text { and } \quad E_n=\frac{E_0}{n^2} \quad \Rightarrow \quad n=2, E_2=\frac{E_0}{4}\) Given \(\frac{E_0}{4}-E_0=15-\frac{3 E_0}{4}=15 \quad \text { for } \quad n=4, \quad E_4=\frac{E_0}{16}\) third excitation energy

⇒ \(=\frac{E_0}{16}-E_0=-\frac{15}{16} E_0=-\frac{15}{16} \cdot\left(\frac{-4 \times 15}{3}\right)=\frac{75}{4} \mathrm{eV}\)

Therefore Excitation potential is \(\frac{75}{4} V\)

Third excitation potential is \(\frac{75}{4} V\)

Emission spectrum of hydrogen atom:

Under normal conditions, the single electron in the hydrogen atom stays in the ground state (n = 1). It is excited to some higher energy state when it acquires some energy from external sources. But it hardly stays there for more than 10–8 seconds.

A photon corresponding to a particular spectrum line is emitted when an atom makes a transition from a state in an excited level to a state in a lower excited level or the ground level.

Let ni be the initial and nf the final energy state, then depending on the final energy state following series are observed in the emission spectrum of the hydrogen atom.

On Screen:

A photograph of spectral lines of the Lyman, Balmer, Paschen series of atomic hydrogen.

1, 2, 3….. represents the 1, 2 & 3 line of Lyman, Balmer, Paschen series.

The hydrogen spectrum (some selected lines)

Series limit: The line of any group having maximum energy of the photon and minimum wavelength of that group is called the series limit.

For the Lyman series nf = 1, for the Balmer series n f = 2, and so on.

Wavelength of Photon Emitted in De-excitation

According to Bohr when an atom makes a transition from a higher energy level to a lower level it emits a photon with energy equal to the energy difference between the initial and final levels. If E i is the initial energy of the atom before such a transition, Ef is its final energy after the transition, and the photon’s energy is \(h v=\frac{h c}{\lambda}\) then conservation of energy gives, \(h \nu=\frac{h c}{\lambda}=E_i-E_f\) (energy of emitted photon)

By 1913, the spectrum of hydrogen had been studied intensively. The visible line with longest wavelength, or lowest frequency is in the red and is called Hα, the next line, in the blue-green, is called Hb, and so on. In 1885, Johann Balmer, a Swiss teacher found a formula that gives the wavelengths of these lines. This is now called the Balmer series. The Balmer’s formula is,

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\)

Here, n = 3, 4, 5 …., etc. R = Rydberg constant = 1.097 × 107 m–1 andα is the wavelength of light/photon emitted during the transition, For n = 3, we obtain the wavelength of Hα line. Similarly, for n = 4, we obtain the wavelength of Hα line. For n = , the smallest wavelength (=3646 Å) of this series is obtained. Using the relation \(\mathrm{E}=\frac{\mathrm{hc}}{\lambda}\) we can find the photon energies corresponding to the wavelength of the Balmar series.

This formula suggests that, \(E_n=-\frac{R h c}{n^2}, n=1,2,3 \ldots . .\)

The wavelengths corresponding to other spectral series (Lyman, Paschen, (etc.) can be
represented by a formula similar to Balmer’s formula.

The wavelengths corresponding to other spectral series (Lyman, Paschen, (etc.) can be
represented by a formula similar to Balmer’s formula.

Lymen Series: \(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right), n=2,3,4 \ldots .\)

Paschen Serie: \(\frac{1}{\lambda}=R\left(\frac{1}{3^2}-\frac{1}{n^2}\right), n=4,5,6 \ldots . .\)

Brackett Series: \(\frac{1}{\lambda}=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right), n=5,6,7 \ldots . .\)

Pfund Series: \(\frac{1}{\lambda}=R\left(\frac{1}{5^2}-\frac{1}{n^2}\right), n=6,7,8\)

The Lyman series is in the ultraviolet and the Paschen. Brackett and Pfund’s series are in the infrared region

Solved Examples

Example 16. Calculate the wavelength and the frequency of the Hb line of the Balmer series for hydrogen.
Solution: The hb line of the Balmer series corresponds to the transition from n = 4 to n = 2 level. The corresponding wavelength for the Hβ line is, \(\frac{1}{\lambda}=\left(1.097 \times 10^7\right)\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\)

= 0.2056 × 107 = 4.9 × 10–7 m

⇒ \(v=\frac{c}{\lambda}=\frac{3.0 \times 10^8}{4.9 \times 10^{-7}}\)

⇒ \(v=\frac{c}{\lambda}=\frac{3.0 \times 10^8}{4.9 \times 10^{-7}}\)

= 6.12 × 1014 Hz

Example 17. Find the largest and shortest wavelengths in the Lyman series for hydrogen. In what region of the electromagnetic spectrum does each series lie?
Solution: The transition equation for the Lyman series is given by

⇒ \(\frac{1}{\lambda}=R\left[\frac{1}{(1)^2}-\frac{1}{n^2}\right] \quad n=2,3, \ldots . .\) for the largest wavelength, n = 2

⇒ \(\frac{1}{\lambda_{\max }}=1.097 \times 10^7\left(\frac{1}{1}-\frac{1}{4}\right)=0.823 \times 10^7\)

The shortest wavelength corresponds to n = infinity

Example 18. How many different wavelengths can be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number n?
Solution: From the nth state, the atom may go to (n – 1)th state, …., 2nd state, or 1st state. So there are (n – 1) possible transitions starting from the nth state. The atoms reaching (n – 1)th state may make (n – 2) different transitions. Similarly for other lower states. The total number of possible transitions is \((n-1)+(n-2)+(n-3)+\ldots \ldots \ldots \ldots 2+1=\frac{n(n-1)}{2}\)

Example 19. Find the kinetic energy potential energy and total energy in the first and second orbits of the hydrogen atom if the potential energy in the first orbit is taken to be zero.
Solution: \(\begin{array}{llll}
E_1=-13.60 \mathrm{eV} & \mathrm{K}_1=-\mathrm{E}_1=13.60 \mathrm{eV} & \mathrm{U}_1=2 \mathrm{E}_1=-27.20 \mathrm{eV} \\
\mathrm{E}_2=\frac{E_1}{(2)^2}=-3.40 \mathrm{eV} & \mathrm{K}_2=3.40 \mathrm{eV} & \text { and } & \mathrm{U}_2=-6.80 \mathrm{eV}
\end{array}\)

Now U 1 = 0, i.e., potential energy has been increased by 27.20 eV while kinetic energy will remain unchanged. So values of kinetic energy, potential energy, and total energy in the first orbit are 13.60 eV, 0, and 13.60 respectively and for the second orbit, these values are 3.40 eV, 20.40 eV, and 23.80 eV.

Example 20. A small particle of mass m moves in such a way that the potential energy U = ar2 where a is a constant and r is the distance of the particle from the origin. Assuming Bohr’s model of quantization of angular momentum and circular orbits, find the radius of n nth allowed orbit.
Solution: The force at a distance r is, \(\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dr}}=-2 \mathrm{ar}\)

Suppose r is the radius of the nth orbit. The necessary centripetal force is provided by the above force. Thus, \(\frac{m v^2}{r}=2 a r\)

Further, the quantization of angular momentum gives, \(m v r=\frac{n h}{2 \pi}\) Solving Eqs. 1 and 2 for r, we get \(r=\left(\frac{n^2 h^2}{8 a m \pi^2}\right)^{1 / 4}\)

Example 21. An electron is orbiting in a circular orbit of radius r under the influence of a constant magnetic field of strength B. Assuming that Bohr’s postulate regarding the quantization of angular momentum holds good for this electron, find

• The Allowed Values Of The Radius ‘R’ Of The Orbit.
• The Kinetic Energy Of The Electron In Orbit
• The potential energy of interaction between the magnetic moment of the orbital current
  due to the electron moving in its orbit and the magnetic field B.
• The total energy of the allowed energy levels.

Solution: Radius of circular path

⇒ \(\begin{aligned}
&r=\frac{m v}{B e}\\
&\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}
\end{aligned}\)

Solving these two equations, we get

⇒ \(r=\sqrt{\frac{n h}{2 \pi B e}} \quad \text { and } v=\sqrt{\frac{n h B e}{2 \pi m^2}}\)

⇒ \(K=\frac{1}{2} m v^2=\frac{n h B e}{4 \pi m}\)

⇒ \(M=i A=\left(\frac{e}{T}\right)\left(\pi r^2\right)=\frac{e v r}{2}=\frac{e}{2} \sqrt{\frac{n h}{2 \pi B e}} \sqrt{\frac{n h B e}{2 \pi m^2}}=\frac{\text { nhe }}{4 \pi m}\)

Now Potential energy U=-M.B

⇒ \(=\frac{\text { nheB }}{4 \pi m}\)

⇒ \(E=U+K=\frac{n h e B}{2 \pi m}\)

Calculation of recoil speed of atom on the emission of a photon momentum of photon \(=\mathrm{mc}=\frac{\mathrm{h}}{\lambda}\)

⇒ \(\mathrm{mv}=\frac{\mathrm{h}}{\lambda^{\prime}}\)

According to energy conservation

⇒ \(\frac{1}{2} m v^2+\frac{h c}{\lambda^{\prime}}=10.2 \mathrm{eV}\)

Since the mass of the atom is much larger than the photon Hence \(\frac{1}{2} m v^2\)

⇒ \(\begin{aligned}
& \frac{\mathrm{hc}}{\lambda^{\prime}}=10.2 \mathrm{eV} \Rightarrow \frac{\mathrm{h}}{\lambda^{\prime}}=\frac{10.2}{\mathrm{c}} \mathrm{eV} \\
& \mathrm{m} v=\frac{10.2}{\mathrm{c}} \mathrm{eV} \Rightarrow \quad v=\frac{10.2}{\mathrm{~cm}} \\
&
\end{aligned}\)

Recoil speed of atom \(=\frac{10.2}{\mathrm{~cm}}\)

X-RAYS

• X-rays were discovered by Wilhelm Roentgen in 1895. They are also called Roentgen rays.
• X-rays are produced by bombarding high-speed electrons on a target of high atomic weight and high melting point.
• The wavelength of X-rays lies between -rays and UV rays.
• The wavelength range for X-rays is 0.1 Å to 10Å.
• The frequency range for X-rays is 1016 Hz to 1018 Hz.
• The energy range for X-rays is 100 to 10000 eV.
• Hard X-rays: High-frequency X-rays are called hard X-rays.
• Hard X-rays have high penetration power
• The wavelength range is from 0.1A to 10A.
• They have a high frequency of 1018 Hz and a high energy of ~104 eV.
• Soft X-rays: Low-frequency X-rays are called soft X-rays.
• These have low penetrating power
• They have higher wavelengths (10Å to 100Å)
• They have a low frequency of 1016 Hz and a low energy of 102 eV.
• It was discovered by ROENTGEN. The wavelength of X-rays is found between 0.1 Å to 10 Å. These rays are invisible to the eye. They are electromagnetic waves and have speed c = 3 × 10 8 m/s in a vacuum. Its photons have energy around 1000 times more than the visible light.

When fast-moving electrons having the energy of the order of several KeV strike the metallic target then x-rays are produced.

Production of x-rays by coolidge tube:

The melting point, specific heat capacity, and atomic number of the target should be high. When voltage is applied across the filament then the filament on being heated emits electrons from it. Now for giving the beam shape of electrons, the collimator is used. Now when an electron strikes the target then x-rays are produced.

When electrons strike the target, some part of the energy is lost and converted into heat. Since the target should not melt or absorb heat at the melting point, the specific heat of the target should be high.

Here copper rod is attached so that the heat produced can go behind and can absorb heat and the target does not get heated very high. For more energetic electrons, accelerating voltage is increased. For more no. of photons, the voltage across the filament is increased. The X-rays were analyzed by mostly taking their spectrum

Continuous x-ray: When high energy electrons (accelerated by coolidge tube potential) strike the target element they are defeated by coulomb attraction of the nucleus & due to numerous glancing collisions with the atoms of the target, they lose energy which appears in the form of electromagnetic waves (bremsstrahlung or braking radiation) & the remaining part increases the kinetic energy of the colliding particles of the target. The energy received by the colliding particles goes into heating the target. The electron makes another collision with its remaining energy.

• When highly energetic electrons enter into target material, they are decelerated. In this process emission of energy takes place. The spectrum of this energy is continuous. This is also called bremsstrahlung.
• Continuous spectrum (v or lamba) depends upon the potential difference between filament and target.
• It does not depend upon the nature of the target material.
• If V is the potential difference & v is the frequency of emitted x-ray photon then.

Variation of frequency (v) and wavelength (λ) of x-rays with a potential difference is plotted as shown in the figure:

⇒ \(\mathrm{eV}=\frac{1}{2} \mathrm{~m} v^2=h v_{\max }=\frac{h c}{\lambda_{\text {min }}}\)

Variation of Intensity of x-rays with λ is plotted as shown in the figure:

The minimum wavelength corresponds to the maximum energy of the x-rays which in turn is
equal to the maximum kinetic energy eV of the striking electrons thus

⇒ \(\mathrm{eV}=\frac{1}{2} \mathrm{mv} v^2=\mathrm{h} v_{\max }=\frac{\mathrm{hc}}{\lambda_{\min }} \Rightarrow \lambda_{\min }=\frac{\mathrm{hc}}{\mathrm{eV}}=\frac{12400}{\mathrm{~V} \text { (involts) }} \mathrm{A} .\)

We see that cutoff wavelength λmin depends only on the accelerating voltage applied between the target and filament. It does not depend upon the material of the target, it is the same for two different metals (Z and Z’)

Solved Examples

Example 22. An X-ray tube operates at 20 kV. A particular electron loses 5% of its kinetic energy to emit an X-ray photon at the first collision. Find the wavelength corresponding to this photon.
Solution: Kinetic energy acquired by the electron is K = eV = 20 × 103 eV.

The energy of the photon = 0.05 × 20 = 103 eV = 103 eV.

⇒ \(\text { Thus, } \quad \frac{\mathrm{h} \nu}{\lambda}=10^3 \mathrm{eV}=\frac{\left(4.14 \times 10^{-15} \mathrm{eV}-\mathrm{s}\right) \times\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{10^3 \mathrm{eV}}=\frac{1242 \mathrm{eV}-\mathrm{nm}}{10^3 \mathrm{eV}}=1.24 \mathrm{~nm}\)

Characteristic X-rays

The sharp peaks obtained in graph are known as characteristic X-rays because they are characteristic of the target material. λ1, λ2, λ3, λ4, …….. = characteristic wavelength of material having atomic number Z is called characteristic x-rays, and the spectrum obtained is called a characteristic spectrum. If the target of atomic number Z’ is used then peaks are shifted.

Characteristic x-ray emission occurs when an energetic electron collides with a target and removes an inner shell electron from an atom, the vacancy created in the shell is filled when an electron from a higher level drops into it.

Suppose the vacancy created in the innermost K-shell is filled by an electron dropping from the next higher level L-shell then Kα characteristic x-ray is obtained. If a vacancy in the K-shell is filled by an electron from the M-shell, the Kβ line is produced, and so on.

similarly, Lα, Lβ,…..Mα, Mβ lines are produced.

Solved Examples

Example 23. Find which is the Kx and Kb

Solution: \(\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}, \quad \lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}\)

since the energy difference of Kx is less than Kb

⇒ \(\begin{aligned}
& \Delta \mathrm{E}_{\mathrm{k} \alpha}<\Delta \mathrm{E}_{\mathrm{k} \beta} \\
& \lambda_{\mathrm{k} \beta}<\lambda_{k \alpha}
\end{aligned}\)

Example 24. Find which is Kα and Lα
Solution:

1 is Kα and 2 is Lα

Properties Of X-Rays

• X-rays are electromagnetic waves of short wavelengths that travel in straight lines with the speed of light.
• They are chargeless and are not deflected in electric and magnetic fields.
• They cause fluorescence in many substances like zinc sulfide, cadmium tungstate, and barium platino-cyanide.
• They produce photochemical reactions and affect a photographic plate more severely than light.
• Like light, they show reflection, refraction, interference, diffraction, and polarization.
• They ionize the gases through which they pass.
• When they fall on matter they produce photoelectric effect and Compton effect.
• They are highly penetrating and can pass through many solids. Example: They pass through 1 mm thick.
• Aluminum sheet while being absorbed by a sheet of lead of the same thickness.
• The penetration power depends on the applied potential difference and atomic number of the cathode. It destroys tissues of animal bodies and white blood cells.

Absorption Of X-Rays

The intensity of the X-ray beam is defined as the amount of energy carried per unit area per sec perpendicular to the direction of the flow of energy.

• When a beam of X-rays with incident intensity l 0 passes through material then the intensity of emergent X-rays (I) is I = I 0 e–ex
• where u is absorption coefficient and x is thickness of medium
• The intensity of transmitted X-rays reduces exponentially with the thickness of the material.
• The absorption coefficient of the material is defined as the reciprocal of thickness after which the intensity of X-rays falls to \(\frac{1}{\mathrm{e}}\) times the original intensity.
• At \(\mu=\frac{1}{x} \quad \Rightarrow \quad \mathrm{I}=\mathrm{I}_0 / \mathrm{e}\)
• The absorption coefficient depends on the wavelength of X-rays (lamba), the atomic number (Z) of the material, and the density (p) of the material.
• Absorption coefficient \(\mu=C Z^4 \lambda^3 \rho\)
  1. \(\mu \propto \lambda^3\)
  2. \(\mu \propto \frac{1}{v^3}\)
  3. \(\mu \propto Z^4\)
  4. \(\mu \propto \rho\)
• The best absorber of X-rays is lead while the lowest absorption takes place in air.
• Half Thickness (X1/2): The thickness of a given sheet which reduces the intensity of incident X-rays to half of its initial value is called half thickness \(x=x_{1 / 2} \text { so } \quad x_{1 / 2}=\frac{0.693}{\mu}\left(\frac{I}{I_0}\right)=\left(\frac{1}{2}\right)^{x / x_{1 / 2}}\)
• For photographing human body parts BaSO4 is used.
• If the number of electrons striking the target is increased, the intensity of X-rays produced also increases.
• The patients are asked to drink BaSO4. the solution before X-ray examination because it is a good absorber of X-rays.

Solved Examples

Example 25. The absorption coefficient of AI for soft X-rays is 1.73 per cm. Find the percentage of transmitted X-rays from a sheet of thickness 0.578 cm.
Solution: \(\begin{aligned}
& I=I_0 e^{-1 \dot{x}} \\
& \text { so } \quad \frac{I}{I_0}=e^{-\mu \alpha}=e^{-1.73 \times 0.578} \quad \text { or } \quad \frac{I}{I_0}=e^{-1}=\frac{1}{e}=\frac{1}{2.718}=37 \% \\
&
\end{aligned}\)

Example 26. When X-rays of wavelength 0.5Å pass through a 10 mm thick AI sheet then their intensity is reduced to one-sixth. Find the absorption coefficient for Aluminium.
Solution: \(\mu=\frac{2.303}{\mathrm{x}} \log \left(\frac{\mathrm{I}_0}{\mathrm{I}}\right)=\frac{2.303}{10} \log _{10} 6=\frac{2.303 \times 0.7781}{10}=0.198 / \mathrm{mm}\)

Diffraction Of R- Rays

• The diffraction of X-rays by a crystal was discovered in 1912 by Von Laue.
• The diffraction of X-rays is possible because interatomic spacings in a crystal is of the order of wavelength of X-rays.
• The diffraction of X-rays takes place according to Bragg’s law 2d sin= n.
• This helps to determine the crystal structure and wavelength of X-rays.

X-Rays Dose:

• The dose of X-ray is measured in terms of produced ions of free energy via ionization.
• These are measured in Roentgen
• Roentgen does not measure energy but it measures ionization power.
• The safe dose for the human body per week is one Roentgen.
• One Roentgen is the number of x-rays that emit (2.5 × 104 J) free energy through the ionization of 1 gram of air at NTP.

Uses Of X-Rays

• Surgery X-rays pass through flesh but are stopped by bones. So they are used to detect fractures, foreign bodies, and diseased organs. The photograph obtained is called a radiograph.
• Radiotherapy The X-rays can kill the diseased tissues of the body. They are used to cure skin
  disease, malignant tumors, etc.
• Industry To detect defects in motor tires, golf and tennis balls, wood, and wireless valves. Used to test the uniformity of insulating material and for detecting the presence of pearls in oysters.
• Scientific research is Used to study the structure of crystals, the structure and properties of atoms, arrangement of atoms and molecules in matter.
• Detective departments were Used at customs ports to detect goods like explosives, opium concealed in parcels without opening, and detection of precious metals like gold and silver in the bodies of smugglers.

Moseley’s Law:

Moseley measured the frequencies of characteristic x-rays for a large number of elements and plotted the square root of frequency against position number in the periodic table. He discovered that the plot is very close to a straight line not passing through the origin.

Moseley’s observations can be mathematically expressed as \(\sqrt{v}=a(Z-b)\) a and b are positive constants for one type of x-rays and all elements (independent of Z). Moseley’s Law can be derived based on Bohr’s theory of the atom, frequency of x-rays is given by

⇒ \(\sqrt{v}=\sqrt{C R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)} \cdot(Z-b)\) by using the formula \(\frac{1}{\lambda}=R z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) with modification for multi-electron system. b → known as screening constant or shielding effect, and (Z – b) is the effective nuclear charge. for Kx line n1=1,n2=2 \(\sqrt{v}=\sqrt{\frac{3 R C}{4}}(Z-b) \quad \Rightarrow \quad \sqrt{v}=a(Z-b)\)

Here \(a=\sqrt{\frac{3 R C}{4}} \text {, }\) b=1 for lines

Solved Examples

Example 27. Find in Z1 and Z2 which one is greater.

Solution: since \(\sqrt{v} \equiv \sqrt{c R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)} \cdot(Z-b)\)

If Z is greater then v will be greater, and lamba will be less.

therefore \(\lambda_1<\lambda_2\)

Dividing yields \(\sqrt{\frac{\lambda_{c_0}}{\lambda_x}}=\frac{Z_x-1}{Z_{c_0}-1}\)

⇒ \(\sqrt{\frac{178.9 \mathrm{pm}}{143.5 \mathrm{pm}}}=\frac{Z_{\mathrm{x}}-1}{27-1} .\)

Solving for the unknown, we find Zx = 30.0; the impurity is zinc

Example 29. Find the constants a and b in Moseley’s equation from the following data.

Element Z Wavelength of Kα X-ray
Mo 42 71 pm
Co 27 178.5 pm

Solution: Moseley’s equation is

⇒ \(\sqrt{v}=a(Z-b)\)

Thus, \(\sqrt{\frac{c}{\lambda_1}}=a\left(Z_1-b\right)\)

and \(\sqrt{\frac{c}{\lambda_2}}=a\left(Z_2-b\right)\)

From (1) and (2) ,\(\sqrt{c}\left(\frac{1}{\sqrt{\lambda_1}}-\frac{1}{\sqrt{\lambda_2}}\right)=a \quad\left(Z_1-Z_2\right)\)

Or, \(a=\frac{\sqrt{c}}{\left(Z_1-Z_2\right)}\left(\frac{1}{\sqrt{\lambda_1}}-\frac{1}{\sqrt{\lambda_2}}\right)\)

⇒ \(=\frac{\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)^{1 / 2}}{42-27}\left[\frac{1}{\left(71 \times 10^{-12} \mathrm{~m}\right)^{1 / 2}}-\frac{1}{\left(178.5 \times 10^{-12} \mathrm{~m}\right)^{1 / 2}}\right]=5.0 \times 10^7(\mathrm{~Hz})^{1 / 2}\)

Dividing 1 by 2

⇒ \(\sqrt{\frac{\lambda_2}{\lambda_1}}=\frac{Z_1-b}{Z_2-b} \text { or, } \quad \sqrt{\frac{178.5}{71}}=\frac{42-b}{27-b} \quad \text { or, } \quad b=1.37\)

Solved Miscellaneous Problems

Problem 1. Find the momentum of a 12.0 MeV photon.
Solution: \(p=\frac{E}{c}=12 \mathrm{MeV} / \mathrm{c}\)

Problem 2. Monochromatic light of wavelength 3000 Å is incident normally on a surface of area 4 cm 2. If the intensity of the light is 15 × 10–2 W/m2, determine the rate at which photons strike the surface.
Solution: Rate at which photons strike the surface

⇒ \(=\frac{\mathrm{IA}}{\mathrm{hc} / \lambda}=\frac{6 \times 10^{-5} \mathrm{~J} / \mathrm{s}}{6.63 \times 10^{-19} \mathrm{~J} / \text { photon }}=9.05 \times 10^{13} \text { photon } / \mathrm{s} \text {. }\)

Problem 3. The kinetic energies of photoelectrons range from zero to 4.0 × 10 –19 J when light of wavelength 3000 Å falls on a surface. What is the stopping potential for this light?
Solution: \(\mathrm{K}_{\max }=4.0 \times 10^{-19} \mathrm{~J} \times \frac{1 \mathrm{eV}}{1.6 \times 10^{-19} \mathrm{~J}}=2.5 \mathrm{eV}\)

Then, from eV s = K max , V s = 2.5 V

Problem 4. Find the de Broglie wavelength of a 0.01 kg pellet having a velocity of 10 m/s.
Solution: \(\lambda=\mathrm{h} / \mathrm{p}=\frac{6.63 \times 10^{-34} \mathrm{~J} s}{0.01 \mathrm{~kg} \times 10 \mathrm{~m} / \mathrm{s}}=6.63 \times 10^{-23} \mathrm{~A} \text {. }\)

Problem 5. Determine the accelerating potential necessary to give an electron a de Broglie wavelength of 1 Å, which is the size of the interatomic spacing of atoms in a crystal.
Solution: \(\mathrm{V}=\frac{\mathrm{h}^2}{2 \mathrm{~m}_0 \mathrm{e} \lambda^2}=151 \mathrm{~V}\)

Problem 6. Determine the wavelength of the second line of the Paschen series for hydrogen.
Solution: \(\frac{1}{\lambda}=\left(1.097 \times 10^{-3} A^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{5^2}\right) \quad \text { or } \quad \lambda=12,820 A \text {. }\)

Problem 7. How many different photons can be emitted by hydrogen atoms that undergo transitions to the ground state from the n = 5 state?
Solution: No possible transition from n = 5 are = 70. 10 Photons

Problem 8. An electron rotates in a circle around a nucleus with a positive charge Ze. How is the electrons’ velocity related to the radius of its orbit?
Solution: The force on the electron due to the nuclear provides the required centripetal force

⇒ \(\frac{1}{4 \pi \varepsilon_0} \frac{Z e . ~}{r^2}=\frac{\mathrm{mv}^2}{\mathrm{r}} \Rightarrow v=\sqrt{\frac{\mathrm{Ze}^2}{4 \pi \varepsilon_0 \cdot \mathrm{rm}}}\)

⇒ \(v=\sqrt{\frac{Z e^2}{4 \pi \varepsilon_0 \cdot \mathrm{rm}}} .\)

Problem 9. A H-atom in the ground state is moving with initial kinetic energy K. It collides head-on with a He + ion in the ground state kept at rest but free to move. Find the minimum value of K so that both the particles can excite to their first excited state.
Solution: Energy available for excitation \(=\frac{4 K}{5}\)

Total energy required for excitation = 10.2 ev + 40.8 eV = 51.0 ev

⇒ \(\frac{4 K}{5}=51 \quad \Rightarrow \quad k=63.75 \mathrm{eV}\)

Problem 10. A TV tube operates with a 20 kV accelerating potential. What are the maximum–energy X–rays from the TV set?
Solution: The electrons in the TV tube have an energy of 20 keV, and if these electrons are brought to rest by a collision in which one X-ray photon is emitted, the photon energy is 20 keV.v a(Z b)=

Problem 11. In the Moseley relation, which will have the greater value for the constant a for Kx or Kb transition?
Solution: A is larger for the K transitions than for the K transitions.

Problem 12. A He+ ion is at rest and is in the ground state. A neutron with initial kinetic energy K collides head-on with the He+ ion. Find the minimum value of K so that there can be an inelastic collision between these two particles.
Solution: Here the loss during the collision can only be used to excite the atoms or electrons. So according to quantum mechanics loss = {0, 40.8eV, 48.3eV, ……, 54.4eV}

⇒ \(E_n=-13.6 \frac{Z^2}{n^2} e V\)

Now according to Newtonion mechanics

Minimum loss = 0

the maximum loss will be for perfectly inelastic collision.

let v

0 is the initial speed of the neutron and vf is the final common speed.

so by momentum conservation mv0 = mvf + 4mv \(v_f=\frac{v_0}{5}\)

where m = mass of Neutron

mass of He+ ion = 4m

so the final kinetic energy of the system

⇒ \(\text { K.E. }=\frac{1}{2} m v_f^2+\frac{1}{2} 4 m v_f^2=\frac{1}{2} \cdot(5 m) \cdot \frac{v_0^2}{25}=\frac{1}{5} \cdot\left(\frac{1}{2} m v_0^2\right)=\frac{K}{5}\) so loss will be \(\left[0, \frac{4 K}{5}\right]\)

For inelastic collision, there should be at least one common value other than zero in set (1) and

since \(\frac{4 \mathrm{~K}}{5}>40.8 \mathrm{eV}\)

K > 51 eV minimum value of K = 51 eV.

Problem 13. A moving hydrogen atom makes a head-on collision with a stationary hydrogen atom. Before the collision, both atoms are in the ground state, and after the collision, they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation states?
Solution: Let K be the kinetic energy of the moving hydrogen atom and K’, the kinetic energy of combined mass after collision. From conservation of linear momentum,

⇒ \(p=p^{\prime} \text { or } \sqrt{2 \mathrm{Km}}=\sqrt{2 \mathrm{~K}^{\prime}(2 m)}\)

or K = 2K’ ……….1

From conservation of energy, K = K’ + E…………..2

Solving Eqs. (1) and (2), we get,\(\Delta \mathrm{E}=\frac{\mathrm{K}}{2}\)

Now the minimum value of E for a hydrogen atom is 10.2 eV. or E  10.2 eV’

⇒ \(\begin{aligned}
& \frac{K}{2} \geq 10.2 \\
& \mathrm{~K} \geq 20.4 \mathrm{eV}
\end{aligned}\)

Therefore, the minimum kinetic energy of moving hydrogen is 20.4 eV

Nuclear Physics

The branch of physics deals with the study of the nucleus.

1. Nucleus:

1. Discoverer: Rutherford
2. Constituents: Neutrons (n) and protons (p) [collectively known as nucleons]
   1. Neutron: It is a neutral particle. J. Chadwick discovered it. Mass of neutron, m n = 1.6749286 × 10–27 kg. 1.00866 amu
   2. Proton: It has a charge equal to +e. Goldstein discovered it. Mass of proton, m p = 1.6726231 × 10–27 kg 1.00727 amu p n ~ m m (c)
3. Representation:

⇒ \({ }_z X^A \quad \text { or } \quad{ }_Z^A X\)

where X symbol of the atom Z Atomic number = number of protons A Atomic mass number = total number of nucleons. = no. of protons + no. of neutrons.

Atomic mass number:

It is the nearest integer value of mass represented in a.m.u. (atomic mass unit). 1 a.m.u. = 12 [mass of one atom of 6C¹² atom at rest and in ground state] = 1.6603 × 10–27 kg 931.478 MeV/c2 mass of proton (m p ) = mass of neutron (mn) = 1 a.m.u.

Some definitions:

• Isotopes: The nuclei having the same number of protons but different numbers of neutrons are called isotopes.
• Isotones: Nuclei with the same neutron number N but different atomic numbers Z are called isotones.
• Isobars: The nuclei with the same mass number but different atomic numbers are called isobars.

Size of the nucleus: Order of 10–15 m (fermi) Radius of the nucleus; R = R0A1/3 where R 0 = 1.2 × 10–15 m (which is an empirical constant) A = Atomic mass number of atoms.

Density: \(\begin{aligned}
\text { density } & =\frac{\text { mass }}{\text { volume }} \cong \frac{A m_p}{\frac{4}{3} \pi R^3}=\frac{A m_p}{\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3}=\frac{3 m_p}{4 \pi R_0^3} \\
& =2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3
\end{aligned}\)

Nuclei of almost all atoms have almost the same density as nuclear density is independent of the mass number (A) and atomic number (Z).

Solved Examples

Example 1. Calculate the radius of 70Ge.
Solution: We have, R = R 0 A1/3 = (1.2 fm) (70)1/3 = (1.2 fm) (4.12) = 4.94 fm.

Example 2. Calculate the electric potential energy of interaction due to the electric repulsion between two nuclei of 12C when they ‘touch’ each other at the surface
Solution: The radius of a 12C nucleus is R = R 0 A1/3 = (1.2 fm) (12)1/3 = 2.74 fm.

The separation between the centers of the nuclei is 2R = 5.04 fm. The potential energy of the pair is

⇒ \(\begin{aligned}
\mathrm{U} & =\frac{\mathrm{q}_1 \mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}} \\
& =\left(9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2\right) \frac{\left(6 \times 1.6 \times 10^{-19} \mathrm{C}\right)^2}{5.04 \times 10^{-15} \mathrm{~m}} \\
& =1.64 \times 10^{-12} \mathrm{~J}=10.2 \mathrm{MeV}
\end{aligned}\)

Mass Defect

It has been observed that there is a difference between the expected mass and the actual mass of a nucleus.

⇒ \(\begin{aligned}
& M_{\text {expected }}=Z m_p+(A-Z) m_n \\
& M_{\text {ossered }}=M_{\text {trom }}-Z m_e
\end{aligned}\)

It is found that

⇒ \(M_{\text {observed }}<M_{\text {expecled }}\)

Hence, the mass defect is defined as

⇒ \(\begin{aligned}
& \text { Mass defect }=M_{\text {expected }}-M_{\text {observed }} \\
& \Delta m=\left[Z m_p+(A-Z) m_n\right]-\left[M_{\text {atom }}-Z m_e\right]
\end{aligned}\)

Binding Energy

It is the minimum energy required to break the nucleus into its constituent particles. or Amount of energy released during the formation of the nucleus by its constituent particles and bringing them from infinite separation. Binding Energy (B.E.) = Δmc² BE = Δm (in amu) × 931 MeV/amu = Δm × 931 MeV

Note: If binding energy per nucleon is more for a nucleus then it is more stable. For example

If \(\left(\frac{B \cdot E_1}{A_1}\right)>\left(\frac{B \cdot E_2}{A_2}\right)\) Then NUclues 1 Would Be More Stable

Example 3. The following data is available about 3 nuclei P, Q, And R. Arrange them in decreasing order of stability

Solution: \(\begin{aligned}
& \left(\frac{B \cdot E}{A}\right)_P=\frac{100}{10}=10 \\
& \left(\frac{B E}{A}\right)_Q=\frac{60}{5}=12 \\
& \left(\frac{B \cdot E}{A}\right)_R=\frac{66}{6}=11
\end{aligned}\)

The stability order is Q > R > P.

Example 4. The three stable isotopes \({ }_{10}^{20} \mathrm{Ne},{ }_{10}^{21} \mathrm{Ne} \text { and }{ }_{10}^{22} \mathrm{Ne}\) have respective abundances of 90.51% 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u, and 22.00 u respectively. Obtain the average atomic mass of neon.
Solution: \(\mathrm{m}=\frac{90.51 \times 19.99+0.27 \times 20.99+9.22 \times 22}{100}=20.18 u\)

Example 5 A nuclear reaction is given as A+B→C+D

1. The binding energies of A, B, C, and D are given as B₁, B₂, B₃, And B₄
2. Find the energy released in the reaction

Solution: (B₃ + B₄) – (B₁ + B₂)

Example 6 Calculate the binding energy of an alpha particle from the following data: 1 mass of 1H atom = 1.007826 u mass of neutron = 1.008665 u 4 mass of 2 He atom = 4.00260 u Take 1 u = 931 MeV/c2.
Solution: The alpha particle contains 2 protons and 2 neutrons. The binding energy is B

= (2 × 1.007826 u + 2 × 1.008665 u – 4.00260 u)c²

= (0.03038 u)c² = 0.03038 × 931 MeV

= 28.3 MeV. 56 56 26 Fe

Example 7. Find the binding energy of. Atomic mass is 55.9349 u and that of 1H is 1.00783 u. Mass of neutron = 1.00867 u. 5626Fe
Solution: The number of protons in = 26 and the number of neutrons = 56 – 26 = 30.
\({ }_{26}^{56} \mathrm{Fe}\)  The binding energy is\({ }_{26}^{56} \mathrm{Fe}\)

= [26 × 1.00783 u + 30 × 1.00867 u – 55.9349 u] c²

= (0.52878 u)c²

= (0.52878 u) (931 MeV/u) = 492 MeV.

Variation of binding energy per nucleon with mass number:

The binding energy per nucleon first increases on average and reaches a maximum of about 8.8 MeV for A = 56. For still heavier nuclei, the binding energy per nucleon slowly decreases as A increases.

Binding energy per nucleon is maximum for 26Fe56, which is equal to 8.75 MeV. Binding energy per nucleon is more for medium nuclei than for heavy nuclei. Hence, medium nuclei are highly stable.

1. The heavier nuclei being unstable tend to split into medium nuclei. This process is called Fission.
2. The Lighter nuclei being unstable tend to fuse into a medium nucleus. This process is called Fusion.

Radioactivity:

1. It was discovered by Henry Becquerel.
2. Spontaneous emission of radiations (αβγ) from unstable nuclei is called radioactivity. Substances that show radioactivity are known as radioactive substances.
3. Radioactivity was studied in detail by Rutherford.
4. In radioactive decay, an unstable nucleus emits α particle or β particle. After the emission of α or β the remaining nucleus may emit β-particle, and convert into a more stable nucleus.

α-particle:

It is a doubly charged helium nucleus. It contains two protons and two neutrons.

Mass of α-particle = Mass of ,\({ }_2 \mathrm{He}^4 \text { atom }-2 m_e \approx 4 m_p\)

Charge of α-particle = + 2 e.

β-particle:

It is a doubly charged helium nucleus. It contains two protons and two neutrons.

β Mass of β-particle = Mass of 2He4 atom – 2me 4 mp

Charge of β-particle = + 2 e

Antiparticle:

A particle is called an antiparticle of another if on collision both can annihilate (destroy completely) and convert into energy.

For example: electron ( – e, me ) and positron ( + e, m e ) are anti particles neutrino and antineutrino are anti particles.

γ-particle: They are energetic photons of energy of the order of Mev and having rest mass zero.

Radioactive Decay (Displacement Law):

α-decay:

⇒ \({ }_z X^A \quad \rightarrow \quad{ }_{z-2} Y^{A-4}+{ }_2 \mathrm{He}^4+Q\)

Q value: It is defined as energy released during the decay process.

Q value = rest mass energy of reactants – rest mass energy of products.

This energy is available in the form of an increase in the K.E. of the products

This energy is available in the form of an increase in the K.E. of the products.

Let, Mx = mass of atom ZXA

My = mass of atom Z – 2YA – 4

MHe = mass of atom 2He4

Q value = [(M x – Zme) – {(M y – (Z – 2) me) + (M He – 2me)}]c2 = [Mx – My – MHe ] c2

Considering the actual number of electrons in α-decay

Q value = [Mx – (My + 2me) – (MHe – 2me)]c2 = [Mx – My – MHe ] c2

Calculation of kinetic energy of final products:

As atom X was initially at rest and no external forces were acting, so final momentum also had to be zero. Hence both Y and α-particle will have the same momentum in magnitude but in opposite directions.

⇒ \(\mathrm{p}_\alpha{ }^2=\mathrm{p}_{\mathrm{Y}}{ }^2 \quad 2 m_\alpha \mathrm{T}_\alpha=2 \mathrm{~m}_{\mathrm{Y}} \mathrm{T}_{\mathrm{Y}} \text { (Here we are representing } \mathrm{T} \text { for kinetic energy) }\)

⇒ \(\begin{aligned}
& \mathrm{Q}=\mathrm{T}_{\mathrm{y}}+\mathrm{T}_\alpha \quad \mathrm{m}_\alpha \mathrm{T}_\alpha=\mathrm{m}_{\mathrm{Y}} \mathrm{T}_{\mathrm{Y}} \\
& \mathrm{T}_\alpha=\frac{\mathrm{m}_{\mathrm{Y}}}{\mathrm{m}_\alpha+\mathrm{m}_{\mathrm{Y}}} \mathrm{Q} ; \quad \mathrm{T}_{\mathrm{Y}}=\frac{\mathrm{m}_\alpha}{\mathrm{m}_\alpha+\mathrm{m}_{\mathrm{Y}}} \mathrm{Q} \\
& \mathrm{T}_\alpha=\frac{\mathrm{A}-4}{\mathrm{~A}} \mathrm{Q} ; \quad \mathrm{T}_{\mathrm{Y}}=\frac{4}{\mathrm{~A}} \mathrm{Q} \\
&
\end{aligned}\)

From the above calculation, one can see that all the α-particles emitted should have the same kinetic energy. Hence, if they are passed through a region of a uniform magnetic field having a direction perpendicular to the velocity, they should move in a circle of the same radius.

⇒ \(r=\frac{m v}{q B}=\frac{m v}{2 e B}=\frac{\sqrt{2 K m}}{2 e B}\)

Experimental Observation:

Experimentally it has been observed that all the α-particles do not move in a circle of the same radius, but they move in `circles having different radii.

This shows that they have different kinetic energies. But it is also observed that they follow circular paths of some fixed values of radius i.e. the energy of emitted α-particles is not the same but it is quantized.

The reason behind this is that all the daughter nuclei produced are not in their ground state but some of the daughter nuclei may be produced in their excited states and they emit photons to acquire their ground state.

The only difference between Y and Y* is that Y* is excited and Y is in ground state. Let, the energy of emitted γ-particles be E.

Therefore \(\begin{aligned}
& \mathrm{Q}=\mathrm{T}_\alpha+\mathrm{T}_{\mathrm{Y}}+\mathrm{E} \\
& \mathrm{Q}=\left[\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}-\mathrm{M}_{\mathrm{He}}\right] \mathrm{c}^2 \mathrm{~T}_\alpha+\mathrm{T}_{\mathrm{Y}}=\mathrm{Q}-\mathrm{E}
\end{aligned}\)

β- decay

⇒ \({ }_z X^A \longrightarrow{ }_{z+1} Y^A+{ }_{-1} e^0+Q\)

–1e0 can also be written as –1β0.

Here also one can see that through momentum and energy conservation, we will get

⇒ \(T_e=\frac{m_Y}{m_e+m_Y} Q; T_Y=\frac{m_e}{m_e+m_Y} Q\)

as m e << m Y, we can consider that all the energy is taken away by the electron. From the above results, we will find that all the β-particles emitted will have the same energy and hence they have the same radius if passed through a region of perpendicular magnetic field. However, experimental observations were completely different.

On passing through a region of uniform magnetic field perpendicular to the velocity, it was observed that β-particles take circular paths of different radii having a continuous spectrum.

To explain this, Paulling has introduced the extra particles called neutrino and antineutrino (antiparticles of neutrino). \(\bar{v}\) → antineutrino, v → neutrino

Properties of antineutrino (\(\bar{v}\)) & neutrino(v):

They are like photons having rest mass = 0 speed = c

Energy, E = mc²

They are chargeless (neutral)

They have spin quantum number \(s= \pm \frac{1}{2}\)

Considering the emission of antineutrino, the equation of β – decay can be written as

⇒ \({ }_2 X^A \longrightarrow{ }_{z+1} Y^A+{ }_{-1} e^0+Q+\bar{v}\)

Production of antineutrino along with the electron helps to explain the continuous spectrum because the energy is distributed randomly between electrons and it also helps to explain the spin quantum number balance (p, n, and ± e each has spin quantum number ± 1/2).

During β – decay, inside the nucleus a neutron is converted to a proton with emission of an electron and antineutrino.

Let \(\begin{aligned}
& \mathrm{n} \rightarrow \mathrm{p}^{+}{ }_{-1} \mathrm{e}^0+\bar{v} \\
& \mathrm{M}_{\mathrm{x}}=\text { mass of atom } \mathrm{Z}^{\mathrm{A}} \\
& \mathrm{M}_{\mathrm{y}}=\text { mass of atom } \\
& \mathrm{m}_{\mathrm{e}+1}=\text { mass of electron }
\end{aligned}\)

Q value = [(M X – Zme) – {(MY – (z + 1) me) + me}] c2 = [MX – MY] c2

Considering an actual number of electrons.

Q value = [M X – {(MY – me) + me}] c2 = [MX – MY] c

Comparison Of Properties Of α, β, And γ Radiations

Solved Example

Example 8. Consider the beta decay \({ }_{198} \mathrm{Au} \rightarrow{ }^{198} \mathrm{Hg}^{\star}+\beta^{-}+\bar{v}\) where 198Hg* represents a mercury nucleus in an excited state at energy 1.088 MeV above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic mass of 198Au is 197.968233 u and that of ¹⁹⁸Hg is 197.966760 u.
Solution: The product nucleus ¹⁹⁸Hg is formed in its ground state, the kinetic energy available to the electron and the antineutrino is Q = [m(¹⁹⁸Au) – m(¹⁹⁸Hg)]c².

As ¹⁹⁸Hg has energy 1.088 MeV more than ¹⁹⁸Hg in ground state, the kinetic energy actually available is Q = [m(¹⁹⁸Au) – m(¹⁹⁸Hg)]c2 – 1.088 MeV 931 MeV = (197.968233 u – 197.966760 u)\(\left(931 \frac{\mathrm{MeV}}{\mathrm{u}}\right)-1.088 \mathrm{MeV}\) = 1.3686 MeV – 1.088 MeV = 0.2806 MeV. This is also the maximum possible kinetic energy of the electron emitted.

β+ – decay:

In B+ decay, inside a nucleus, a proton is converted into a neutron, positron, and neutrino.

As mass increases during the conversion of a proton to a neutron, hence it requires energy for β+ decay to take place, β+ decay is a rare process. It can take place in the nucleus where a proton can take energy from the nucleus itself.

Q value = [(M X– Zme) – {(M Y – (Z – 1) me) + me}] c² = [MX – MY – 2me] c²

Considering the actual number of electrons.

Q value = [M X – {(MY + me) + me}] c²

= [MX – MY – 2me] c

Solved Examples

Example 9. Calculate the Q-value in the following decays:

⇒ \(\begin{aligned}
& { }^{19} \mathrm{O} \rightarrow{ }^{19} \mathrm{~F}+\mathrm{e}^{-}+\bar{v} \\
& { }^{25} \mathrm{Al} \rightarrow{ }^{25} \mathrm{Mg}+\mathrm{e}^{+}+\mathrm{v} .
\end{aligned}\)

The atomic masses needed are as follows

⇒ \(\begin{array}{cccc}
{ }^{19} \mathrm{O} & { }^{19} \mathrm{~F} & { }^{25} \mathrm{Al} & { }^{25} \mathrm{Mg} \\
19.003576 \mathrm{u} & 18.998403 \mathrm{u} & 24.990432 \mathrm{u} & 24.985839 \mathrm{u}
\end{array}\)

Solution: The Q- the value of β‾-decay is

Q = [m(¹⁹O) – m(¹⁹F)]c²

= [19.003576 u – 18.998403 u ] (931 MeV/u) = 4.816 MeV

The Q-value of β+ -decay is

Q = [m(25Al) – m(25Mg) – 2me]c²

⇒ \(=\left[24.99032 \mathrm{u}-24.985839 \mathrm{u}-2 \times 0.511 \frac{\mathrm{MeV}}{\mathrm{c}^2}\right] \mathrm{c}^2\)

= (0.004593 u) (931 MeV/u) – 1.022 MeV

= 4.276 MeV – 1.022 MeV = 3.254 MeV

Pair Production And Pair Annihilation

The collision of γray photons by a nucleus & production of an electron-positron pair is known as pair production.

The rest mass of each of the electron & the positron is 9.1 × 10 –31 kg. so, the rest mass energy of each of them is E₀ = m₀ c² = (9.1 × 10–31) (3 × 108)²
= 8.2 × 10–14 joule
= 0.51 MeV
Hence for pair production, the energy of γ-photon must be at least 2 × 0.51 = 1.02 MeV.

K Capture :

It is a rare process that is found only in a few nuclei. In this process the nucleus captures one of the atomic electrons from the K shell. A proton in the nucleus combines with this electron and converts itself into a neutron. A neutrino is also emitted in the process and is emitted from the nucleus.

⇒ \(p+{ }_{-1} e^0 \rightarrow n+v\)

If X and Y are atoms then the reaction is written as:

⇒ \({ }_z X^A \rightarrow{ }_{z-1} Y^A+v+Q+\)

If X and Y are taken as the nucleus, then the reaction is written as :

⇒ \({ }_z X^A+{ }_{-1} e^0 \rightarrow{ }_{z-1} Y^A+v\)

Note:

• Nuclei having atomic numbers from Z = 84 to 112 show radioactivity.
• Nuclei having Z = 1 to 83 are stable (only a few exceptions are there)
• Whenever a neutron is produced, a neutrino is also produced.
• Whenever a neutron is converted into a proton, an antineutrino is produced.
• It is usually accompanied by x-ray emission.

Uses Of Radioactive Isotopes

1. In Medicine

1. Co⁶⁰ for the treatment of cancer
2. Na²⁴ for circulation of blood
3. I¹³¹ for thyroid
4. Sr⁹⁰ for treatment of skin & eye
5. Fe⁵⁹ for location of brain tumor
6. Radiographs of castings and teeth

2. In Industries

For detecting leakage in water and oil pipe lines for investigation of wear & tear, study of plastics & alloys, and thickness measurement.

3. In Agriculture

1. C¹⁴ to study the kinetics of plant photosynthesis.
2. P³² to find the nature of phosphate which is best for given soil & crop
3. Co⁶⁰ for protecting potato crops from earthworms.
4. Sterilization of insects for pest control.

Question 4. In Scientific research

• K⁴⁰ to find the age of meteorites
• S³⁵ in factories

5. Carbon dating

• It is used to find the age of the earth and fossils
• The age of the earth is found by Uranium disintegration and fossil age by disintegration of C 14.
• The estimated age of the earth is about 5 × 109 years.
• The half-life of C14 is 7500 years.

6. As Tracers

• A very small quantity of radio isotope present in any specimen is called a tracer.
• This technique is used to study complex biochemical reactions, in the detection of cracks, blockage, etc., tracing sewage or silt in the sea.

7. In Geology

• For dating geological specimens like ancient rocks, and lunar rocks using Uranium
• For dating archaeological specimens, and biological specimens using C14.

9. Nuclear Stability:

The figure shows a plot of neutron number N versus proton number Z for the nuclides found in nature. The solid line in the figure represents the stable nuclides. For light-stable nuclides, the neutron number is equal to the proton number so the ratio N/Z is equal to 1.

The ratio N/Z increases for the heavier nuclides and becomes about 1.6 for the heaviest stable nuclides. The points (Z, N) for stable nuclides fall in a rather well-defined narrow region. There are nuclides to the left of the stability belt as well as to the right of it.

The nuclides to the left of the stability region have excess neutrons, whereas, those to the right of the stability belt have excess protons.

These nuclides are unstable and decay with time according to the laws of radioactive disintegration. Nuclides with excess neutrons (lying above the stability belt) show β− decay while nuclides with excess protons (lying below the stability belt) show β+ decay and K – capture.

Nuclear Force:

1. Nuclear forces are attractive and are responsible for keeping the nucleons bound in a nucleus despite repulsion between the positively charged protons.
2. It is the strongest force within nuclear dimensions (F n = 100 Fe)
3. It is short range force (acts only inside the nucleus)
4. It acts only between neutron-neutron, neutron-proton, and proton-proton i.e. between nucleons.
5. It does not depend on the nature of nucleons.
6. An important property of nuclear force is that it is not a central force. The force between a pair of nucleons is not solely determined by the distance between the nucleons. For example, the nuclear force depends on the directions of the spins of the nucleons.
7. The force is stronger if the spins of the nucleons are parallel (i.e., both nucleons have m s = + 1/2 or – 1/2) and is weaker if the spins are antiparallel (i.e., one nucleon has m s = + 1/2 and the other has m s = – 1/2). Here m s is a spin quantum number.

Radioactive Decay: Statistical Law:

• (Given by Rutherford and Soddy)
• Rate of radioactive decay λN
• where N = number of active nuclei = λN
• where λ= decay constant of the radioactive substance.
• The decay constant is different for different radioactive substances, but it does not depend on the amount of substance and time. Sλ unit of λ is s–1 If λ1 = λ2 then the first substance is more radioactive (less stable) than the second one. For the case, if A decays to B with decay constant λ

Rate of radioactive decay of \(\mathrm{A}=-\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}\)

Number of nuclei decayed (i.e. the number of nuclei of B formed)

⇒ \(\begin{aligned}
& \mathrm{N}^{\prime}=\mathrm{N}_0-\mathrm{N} \\
&=\mathrm{N}_0-\mathrm{N}_0 \mathrm{e}^{-2 .} \\
& \mathrm{N}^{\prime}=\mathrm{N}_0\left(1-\mathrm{e}^{-2.1}\right) \\
& \mathrm{N}^{\prime}=\mathrm{N}_0-\mathrm{N}
\end{aligned}\)

Half-Life (T_1/2):

It is the time in which number of active nuclei becomes half. N = N 0 e–λt

After one half-life, \(N=\frac{N_0}{2}\)

⇒ \(\frac{N_0}{2}=N_0 e^{-\lambda t} \Rightarrow t=\frac{\ln 2}{\lambda} \Rightarrow \frac{0.693}{\lambda}=t_{1 / 2}\)

⇒ \(t_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda}\)

Number of nuclei present after n half-lives i.e. after a time t = n t1/2

⇒  \(\begin{aligned}
& N=N_0 e^{-\lambda t} \quad=N_0 e^{-\lambda n t 1 / 2} \quad=N_0 e^{-\lambda n \frac{\ln 2}{\lambda}} \\
& =N_0 e^{\ln z^{(-n)}} \quad=N_0(2)^{-n}=N_0(1 / 2)^n \quad=\frac{N_0}{2^n} \\
&
\end{aligned}\)

⇒ \(\left\{n=\frac{t}{t_{1 / 2}}\right.\) It may be a fraction, need not to be an integer}

Or \(\mathrm{N}_0 \xrightarrow[\text { half life }]{\text { after ist }} \frac{N_0}{2} \xrightarrow{2} N_0\left(\frac{1}{2}\right)^2 \xrightarrow{3} N_0\left(\frac{1}{2}\right)^3 \cdots \ldots \ldots \ldots \ldots . . . . \xrightarrow{n} N_0\left(\frac{1}{2}\right)^n\)

Solved Examples

Example 10. A radioactive sample has 6.0 × 1018 active nuclei at a certain instant. How many of these nuclei will still be in the same active state after two half-lives?
Solution: In one half-life the number of active nuclei reduces to half the original number. Thus, in two half-lives the number is reduced to \(\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)\) of the original number. The number of remaining active nuclei is, therefore, \(6.0 \times 10^{18} \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)=1.5 \times 10^{18}.\)

Example 11. The number of 238U atoms in an ancient rock equals the number of 206Pb atoms. The half-life of decay of 238U is 4.5 × 10 9 y. Estimate the age of the rock assuming that all the 206Pb atoms are formed from the decay of 238U.
Solution: Since the number of 206Pb atoms equals the number of 238U atoms, half of the original 238U atoms have decayed. It takes one half-life to decay half of the active nuclei. Thus, the sample is 4.5 × 109 y old.

Activity:

Activity is defined as the rate of radioactive decay of nuclei It is denoted by A or R A = N

If a radioactive substance changes only due to decay then

⇒ \(A=-\frac{d N}{d t}\)

As in that case \(\begin{aligned}
& N=N_0 e^{-2 t} \\
& A=\lambda N=\lambda N_0 e^{-\lambda t} \quad \Rightarrow \quad A=A_0 e^{-\lambda t}
\end{aligned}\)

SI Unit of activity: becquerel (Bq) which is the same as 1 DPS (disintegration per second) The popular unit of activity is curie which is defined as 1 curie = 3.7 × 1010 DPS (which is the activity of 1 gm Radium)

Solved Examples

Example 12. The decay constant for the radioactive nuclide 64Cu is 1.516 × 10–5 s–1. Find the activity of a sample containing 1 μg of 64Cu. The atomic weight of copper = 63.5 g/mole. Neglect the mass difference between the given radioisotope and normal copper.
Solution: 63.5 g of copper has 6 × 1023 atoms. Thus, the number of atoms in 1 g of Cu is \(\mathrm{N}=\frac{6 \times 10^{23} \times 1 \mu \mathrm{g}}{63.5 \mathrm{~g}}=9.45 \times 10^{15}\)

The Activy = N

= (1.516 × 10–5 s–1) × (9.45 × 1015) = 1.43 × 1011 disintegrations/s

Activity after n half-lives \(\frac{A_0}{2^n}\)

Example 13. The half-life of a radioactive nuclide is 20 hours. What fraction of the original activity will remain after 40 hours?
Solution: 40 hours means 2 half-lives.

Thus, \(A=\frac{A_0}{2^2}=\frac{A_0}{4} \quad \text { or, } \quad \frac{A}{A_0}=\frac{1}{4} \text {. }\)

So fourth of the original activity will remain after 40 hours.

Specific activity: The activity per unit mass is called specific activity.

Average Life:

⇒ \(T_{\text {arg }}=\frac{\text { sum of ages of all the nuclei }}{N_0}=\frac{\int_0^{\infty} \lambda N_0 e^{-\lambda t} d t \cdot t}{N_0}=\frac{1}{\lambda}\)

Solved Examples

Example 14. The half-life of 198Au is 2.7 days. Calculate (a) the decay constant, (b) the average-life, and (c) the activity of 1.00 mg of 198Au. Take the atomic weight of 198Au to be 198 g/mol.
Solution: The half-life and the decay constant are related as

⇒ \(\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda} \text { or, } \lambda=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{2.7 \text { days }}=\frac{0.693}{2.7 \times 24 \times 3600 \mathrm{~s}}=2.9 \times 10^{-6} \mathrm{~s}^{-1} \text {. }\)

The average-life is \(t_{a v}=\frac{1}{\lambda}=3.9 \text { days. }\)

The activity is A =¹⁹⁸N. Now,¹⁹⁸g of ¹⁹⁸Au has 6 × 1023 atoms.

The number of atoms in 1.00 mg of ¹⁹⁸Au is

⇒ \(\mathrm{N}=6 \times 10^{23} \times \frac{1.0 \mathrm{mg}}{198 \mathrm{~g}}=3.03 \times 10^{18} .\)

Thus, A = λN = (2.9 × 10–6 s–1) (3.03 × 10 18)
= 8.8 × 1012 disintegrations/s

⇒  \(=\frac{8.8 \times 10^{12}}{3.7 \times 10^{10}} \mathrm{Ci}=240 \mathrm{Ci} .\)

Example 15. Suppose, the daughter nucleus in a nuclear decay is itself radioactive. Let λp and λd be the decay constants of the parent and the daughter nuclei. Also, let N p and N d be the number of parent and daughter nuclei at time t. Find the condition for which the number of daughter nuclei becomes constant.
Solution: The number of parent nuclei decaying in a short time interval t to t + dt is  p N p dt. This is also the number of daughter nuclei decaying during the same time interval is dN ddt. The number of the daughter nuclei will be constant if

⇒ \(\lambda_{\mathrm{p}} \mathrm{N}_{\mathrm{p}} \mathrm{dt}=\lambda_{\mathrm{d}} \mathrm{N}_{\mathrm{d}} \mathrm{dt} \quad \text { or, } \quad \lambda_{\mathrm{p}} \mathrm{N}_{\mathrm{p}}=\lambda_{\mathrm{d}} \mathrm{N}_{\mathrm{d}} \text {. }\)

Example 16. A radioactive nucleus can decay by two different processes. The half-life for the first process is t 1 and that for the second process is t2. Show that the effective half-life t of the nucleus is given by \(\frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_2}\)
Solution: The decay constant for the first process is \(\lambda_1=\frac{\ln 2}{t_1}\) and for the second process it is \(\lambda_2=\frac{\ln 2}{t_1}\) The probability that an active nucleus decays by the first process in a time interval dt is t1dt. Similarly, the probability that it decays by the second process is t2dt. The probability that it either decays by the first process or by the second process is t1dt + t2dt. If the effective decay constant is , this probability is also equal to tdt. Thus \(\begin{aligned}
& \lambda \mathrm{dt}=\lambda_1 \mathrm{dt}+\lambda_2 \mathrm{dt} \\
& \lambda=\lambda_1+\lambda_2 \\
& \frac{1}{\mathrm{t}}=\frac{1}{\mathrm{t}_1}+\frac{1}{\mathrm{t}_2} .
\end{aligned}\)

Nuclear Fission:

In nuclear fission heavy nuclei of A, above 200, break up into two or more fragments of comparable masses. The most attractive bid, from a practical point of view, to achieve energy from nuclear fission is to use ₉₂U²³⁵ as the fission material.

The technique is to hit a uranium sample with slow-moving neutrons (kinetic energy 0.04 eV, also called thermal neutrons).

A ₉₂U²³⁵ nucleus has a large probability of absorbing a slow neutron and forming a ₉₂U²³⁵ nucleus. This nucleus then fissions into two or more parts. A variety of combinations of the middle-weight nuclei may be formed due to the fission. For example, one may have

⇒ \({ }_{92} \mathrm{U}^{235}+{ }_0 \mathrm{n}^1 \rightarrow{ }_{92} \mathrm{U}^{236} \rightarrow \mathrm{X}+\mathrm{Y}+2_0 \mathrm{n}^1, \quad \text { or } \quad{ }_{92} \mathrm{U}^{235}+{ }_0 \mathrm{n}^1 \rightarrow{ }_{92} \mathrm{U}^{236} \rightarrow \mathrm{X}^{\prime}+\mathrm{Y}^{\prime}+3_0 \mathrm{n}^1\)

And several other combinations.

On average 2.5 neutrons are emitted in each fission event.

• Mass lost per reaction ≈ 0.2 a.m.u.
• In nuclear fission, the total B.E. increases and excess energy is released.
• In each fission event, about 200 MeV of energy is released a large part of which appears in the form of kinetic energies of the two fragments. Neutrons take away about 5MeV

⇒ \(\begin{aligned}
& { }_{92}^{235} U+{ }_{\circ} n^1 \rightarrow{ }_{92}^{236} U \rightarrow{ }_{56}^{141} B a+{ }_{36}^{92} K r+3{ }_o n^1+\text { energy } \\
& Q \text { value }=\left[\left(M_U-92 m_e+m_n\right)-\left\{\left(M_{\text {Ba }}-56 m_e\right)+\left(M_{K r}-36 m_e\right)+3 m_n\right\}\right] c^2 \\
& =\left[\left(M_U+m_n\right)-\left(M_{B a}+M_{K r}+3 m_n\right)\right] c^2
\end{aligned}\)

A very important and interesting feature of neutron-induced fission is the chain reaction. For working on nuclear reactors refer to your textbook.

Reproduction Factor

The ratio, of several fission produced by a given generation of neutrons to the number of fission of the preceding generation, is the reproduction factor or multiplication factor. It is the measure of the growth rate of the neutrons in the reactor. It is denoted by K.

For K = 1, the operation of the reactor is said to be critical.

If K > 1, then the reaction rate and the reactor power increase exponentially.

Unless the factor K is brought down very close to unity, the reactor will become supercritical and can even explode.

Nuclear Fusion (Thermo Nuclear Reaction):

The fusion reaction in the sun is a multi-step process in which the hydrogen is burned into helium. Thus, the fuel in the sun is the hydrogen in its core. The proton-proton (p, p) cycle by which this occurs is represented by the following sets of reactions

⇒ \({ }_1^1 \mathrm{H}+{ }_1^1 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+\mathrm{e}^{+}+v+0.42 \mathrm{MeV}\)

⇒ \(\begin{aligned}
& \mathrm{e}^{+}+\mathrm{e}^{-} \rightarrow \gamma+\gamma+1.02 \mathrm{MeV} \\
& { }_1^2 \mathrm{H}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+\gamma+5.49 \mathrm{MeV} \\
& { }_2^3 \mathrm{He}+{ }_2^3 \mathrm{He} \rightarrow{ }_2^4 \mathrm{He}+{ }_1^1 \mathrm{H}+{ }_1^1 \mathrm{H}+12.86 \mathrm{MeV}
\end{aligned}\)

For the fourth reaction to occur, the first three reactions must occur twice, in which case two light helium nuclei unite to form an ordinary helium nucleus. If we consider the combination 2(1) + 2(2) + 2(3) +(4), the net effect is

⇒ \(\begin{aligned}
& 4_1^1 \mathrm{H}+2 \mathrm{e}^{-} \rightarrow_2^4 \mathrm{He}+2 v+6 \gamma+26.7 \mathrm{MeV} \\
& \text { or }\left(4{ }_1^1 \mathrm{H}+4 \mathrm{e}^{-}\right) \rightarrow\left({ }_2^4 \mathrm{He}+2 \mathrm{e}^{-}\right)+2 v+6 \gamma+26.7 \mathrm{MeV}
\end{aligned}\)

Thus, four hydrogen atoms combine to form an atom with a release of 26.7 MeV of energy.
Note: In the case of fission and fusion, Δm = Δmatom = Δnucleus.

1. These reactions take place at ultra-high temperatures ( Δ107 to 109). At high pressure, it can take place at low temperatures also. For these reactions to take place nuclei should be brought upto 1 fermi distance which requires very high kinetic energy.
2. The energy released per mole in fusion exceeds the energy liberated in the fission of heavy nuclei.
3. The energy released per reaction in fission exceeds the energy liberated in the fusion of heavy nuclei.

Solved Example

Example 17. Calculate the energy released when three alpha particles combine to form a 12C nucleus. The 4 atomic mass of 2 He is 4.002603 u.
Solution: The mass of a 12C atom is exactly 12 u.

The energy released in the reaction \(3\left({ }_2^4 \mathrm{He}\right) \rightarrow{ }_6^{12} \mathrm{C}\)

⇒ \(\left[3 \mathrm{~m}\left({ }_2^4 \mathrm{He}\right)-\mathrm{m}\left({ }_6^{12} \mathrm{C}\right)\right] \mathrm{c}^2 \quad=[3 \times 4.002603 \mathrm{u}-12 \mathrm{u}](931 \mathrm{MeV} / \mathrm{u})=7.27 \mathrm{MeV} .\)

Solved Miscellaneous Problems

Problem 1. A radioactive sample decays with an average-life of 20 ms. A capacitor of capacitance 100 μF is charged to some potential and then the plates are connected through a resistance R. What should be the value of R so that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant in time?
Solution: The activity of the sample at time t is given by \(A=A_0 e^{-\lambda t}\)

Where λ is the decay constant and A0 is the activity at time t = 0 when the capacitor plates are connected. The charge on the capacitor at time t is given by

⇒ \(\mathrm{Q}=\mathrm{Q}_0 \mathrm{e}^{-U C R}\)

where Q 0 is the charge at t = 0 and C = 100 F is the capacitance. Thus, \(\frac{Q}{A}=\frac{Q_0}{A_0} \frac{e^{-t / C R}}{e^{-\lambda t}}\)

It is independent of t if \(\lambda=\frac{1}{\mathrm{CR}} \quad \text { or, } \quad \mathrm{R}=\frac{1}{\lambda \mathrm{C}}=\frac{\mathrm{t}_{\mathrm{av}}}{\mathrm{C}}=\frac{20 \times 10^{-3} \mathrm{~s}}{100 \times 10^{-6} \mathrm{~F}}=200 \Omega \text {. }\)

Problem 2. A factory produces a radioactive substance A at a constant rate R which decays with a decay constant to form a stable substance. Find the no. of nuclei of A and Number of nuclei of B, at any time t assuming the production of A starts at t = 0. Also, find out the maximum number of nuclei of ‘A’ present at any time during its formation.
Solution: Factory \(\underset{\text { const. rate }}{\mathrm{R}} \mathrm{A} \underset{\text { decay }}{\lambda} \mathrm{B}\)

Let N be the number of nuclei of A at any time therefore \(\frac{d N}{d t}=R-\lambda N \quad \int_0^N \frac{d N}{R-\lambda N}=\int_0^t d t\)

On solving we will get N = R/λ(1 – e -λt)

Number of nuclei of B at any time t, N B = R t – N A = Rt – R/λ(1 – e -λt) = R/λ (λt – 1 + e -λt).

Maximum number of nuclei of ‘A’ present at any time during its formation = R/λ.

Problem 3. Consider two deuterons moving towards each other at equal speeds in a deutron gas. What should be their kinetic energies (when they are widely separated) so that the closest separation between them becomes 2fm? Assume that the nuclear force is not effective for separations greater than 2 fm. At what temperature will the deuterons have this kinetic energy on average?
Solution: As the deuterons move, the electrostatics repulsion will slow them down. The loss in kinetic energy will be equal to the gain in electrostatics potential energy.

At the closest separation, the kinetic energy is zero and the potential energy is \(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\) If the initial kinetic energy of each deuteron is K and the closest separation is 2fm, we shall have

⇒ \(2 \mathrm{~K}=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0(2 \mathrm{fm})}=\frac{\left(1.6 \times 10^{-19} \mathrm{C}\right)^2 \times\left(9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2\right)}{2 \times 10^{-15} \mathrm{~m}} \quad \text { or } \quad \mathrm{K}=5.7 \times 10^{-14} \mathrm{~J} \text {. }\)

If the temperature of the gas is T, the average kinetic energy of the random motion of each nucleus will be 1.5 kT. The temperature needed for the deuterons to have the average kinetic energy of 5.7 × 10–14 J will be given by 1.5 kT = 5.7 × 10–14 J

Where K= or where k = Botzmann constant \(\mathrm{T}=\frac{5.7 \times 10^{-14} \mathrm{~J}}{1.5 \times 1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}}=2.8 \times 10^9 \mathrm{~K} .\)

Principles Of Communication Introduction

Communication is the process of transmission of information. The teacher transfers data to the students.

To be successful, the sender and receiver must understand a common language. The electronic communication system requires a source of information, a transmitting medium (channel), and a receiver.

The modern communication system has progressed in all three basic components. The information processing and sorting before communication, the channels like optical fiber, space (satellites), and processing through computers before being delivered.

Elements Of A Communication System

As pointed out communication system has three essential stages. Transmitter, medium, or channel and receiver.

The information may be a message, speech, picture, audio video, or group. The information has to be transformed into electrical signals. The transmitter processes and encodes the information to make it suitable for transmission through the channel as well as for reception.

This encoding process is modulation. Signal propagating through the channel may get distortion and we say noise is added to it due to channel imperfections. So the receiver receives a corrupt version.

The receiver has to process the corrupt signal to a recognizable form of the original signal for delivery to the user. There are two modes of communication.

Point-to-point communication example telephony where there is one transmitter and one receiver.

Broadcast mode involves one transmitter and a large number of receivers Example: Radio, TV, etc. Also on the basis that the signals used may be analog or digital, it may be mentioned that The signals both digital and analog are usually of low frequency and hence cannot be transmitted as such These signals require higher frequency waves on which these can ride over called carrier waves This process is called modulation.

Basic Terminology Used In Communication Systems

Let us acquaint ourselves with the basic terminology used in communications.

Transducer: It is a device that converts one form of energy into the other. However, in communication systems, we have to convert all types of signals into electrical. So an electric transducer is a device that converts a physical signal (or variable) such as pressure, temperature, and force displacement. Light sound, etc. into a corresponding electrical signal.

Signal: Information in an electric form suitable for transmission is called a signal. If the signal is in the form of a continuous variation of voltage or current, it is called an analog signal as shown in Figure 2. The variation has to be single-valued. Since all periodic functions may be broken into sine and cosine components and hence sine wave is a fundamental analogue signal.

Sound and picture signals in TV are analog. In digital signal Figure 2 the voltage has only two values either low (0) or high (1). Thus a digital signal is a discontinuous signal.

There are several coding schemes suitable for digital communication systems. Binary coded decimal (BCD). American standard code for information Interchange (ASCII) is the most popular digital code to represent numbers, letters, and certain characters.

Noise: The unwanted signals generated inside and outside the system are referred to as noise.

Transmitter: It processes the incoming signal to make it suitable for transmission through the channel.

Receiver: The receiver extracts the relevant information from the received signal.

Attenuation: The loss in the signal power or strength during transmission is called attenuation.

Amplification: Increasing signal strength by converting other energy (DC or other frequency) into signal energy using electronic circuitry is called amplification.

Range: It is the largest distance between the source and receiver where the signal of sufficient strength is received.

Bandwidth (BW): The portion of the spectrum occupied by the signal or frequency range over which equipment operates is called bandwidth.

Modulation: Superimposing low-frequency messages over high-frequency carrier waves is called modulation. It is AM (amplitude modulated), FM (frequency modulated or PM phase modulated) depending on which quantity, amplitudes, frequency, or phase of the carrier wave varies with the signal.

Demodulation: The process of retrieving information from received waves is called demodulation.

Repeater: These are used to increase the range. It receives a signal from the transmitter amplifies it and retransmits after amplification. It may be at different frequencies. It is thus a combination of receiver and transmitter. Figure (3) shows how the range is extended beyond the mountain. A satellite station is also a sort of repeater.

Antenna or Aerial:

Bandwidth Of Signals

1. Analog signals: The communication signals are of varied nature and have a different range of frequencies example speech signals have a frequency range of 300 Hz to 3100 Hz. Thus the bandwidth for speech is (3100 – 300) = 2800 Hz = 2.8 Hz. The audio range is 20 to 20 kHz the video signals have a typical bandwidth of 4.2 MHz. A TV signal has both audio and video so its bandwidth is 6 MHz.
2. Digital Signals: Digital signals are in the form of rectangular waves. However, a rectangular wave may be constructed using harmonic sine and or cosine waves of frequencies v, 2v, 3v. This implies infinite bandwidth but for practical purposes, higher harmonics can be neglected. No doubt the received waves are distorted versions of the original wave but information is not lost and the rectangular signal is more or less received.

Band – Widths of various signals used in communication

Bandwidth Of Transmission Medium

The transmission media also have different bandwidths Example widely used coaxial cable has a bandwidth of ~750 MHz and operates below 18 GHz.

Free space communication has a wide range from 100 kHz to 10 GHz Optical fiber operates at Terahertz frequencies and has quite a wide range from 1 THz to 1000 THz (microwaves to UV). It provides a bandwidth of ~100 GHz.

Bandwidths of Transmission Channel/ Media used in communications

Space Communication: Propagation Of Electromagnetic Waves In Space

In space radio communication, an antenna at the transmitter radiates electromagnetic waves that travel through space and reach the receiving antenna at the other end. During propagation not only does the signal diminish but several factors influence the propagation.

The atmosphere is a highly dynamic system whose properties change not only with elevation but also with seasons. The propagation characteristics also depend on the frequency band. The following frequency bands are used in radio communication.

1. Medium Frequency Band (MF) 300 – 3000 kHz
2. High Frequency Band (HF) 3.0 – 30 MHz
3. Very High Frequency Band (VHF) 30 – 300 MHz
4. Ultra High-Frequency Band (UHF) 300 – 3000 MHz
5. Super High-Frequency Band (SHF) 3.0 – 30 GHz

The following are the modes of communication through space:

Ground Wave Transmission

For efficient signal radiation by antenna, its height should be in multiples of (/4) where  is wavelength of CW used. For high i,e. lower frequencies, the antenna size is large and these have to be located near to the ground.

In standard AM broadcast, ground base towers are used for broadcast. The waves propagate parallel to the ground. The propagation is called ground wave propagation. The wave induces a current in the ground where it passes.

Therefore it gets attenuated. Understandably, the higher is frequency greater the attenuation. Therefore range depends on power and frequency. The used frequency is less than a few MHz.

Sky Wave Transmission

The radio waves having a frequency of 2 to 30 MHz where propagating up the sky are reflected by the ionosphere and return to the earth. These waves used for communication, are known as sky waves.

At a height 65 to 400 km above the earth, the atmospheric gas absorbs ultraviolet and other high energetic cosmic radiations coming from the sky and subsequently ionizes there creating an ionic layer called the ionosphere.

Even in single reflection wave can cover about 4000 km distance, hence sky wave propagation is long-range transmission. Round the globe communication.

Is possible using skywave propagation? A few terms are important in skywave propagation

Virtual Height: The reflection of sky waves is through gradual bending like total reflection in the formation of a mirage. Hence virtual height is the height through which the angle of incidence is calculated to send waves.

Critical Frequency (fc): The maximum frequency reflected when beamed straight towards the layer. Frequency f > FC is not reflected

Maximum Usable Frequency (MUF): Radio waves when sent at an angle θ, the maximum usable frequency (MUF) is MUF = fc/cosθ ….…(1)

Skip Distance: The smallest distance of the receiver from the transmitter where a wave of a given frequency reaches. It follows from Figure 5 that the higher is angle of incidence on the ionic layer more is the skip distance.

Fading: Due to the arrival of many signals at the receiver emitted from the source simultaneously but reaching the receiver in time delay due to taking different paths diminished signal due to destructive interference. This is fading.

Space Wave Propagation

In space wave mode of propagation, the waves travel in a straight line from the transmitter to the receiving antenna i.e., communication is in the line of sight (LOS). The curvature of the earth limits the range.

As pointed out earlier this communication is above 40 MHz. At these frequencies, antennae are smaller and may be placed at heights many wavelengths above the ground.

To increase the range to the desired level, a satellite is used as a repeater which effectively increases antenna height to the height of the satellite. TV, mobile, and radar systems are examples.

The range versus antenna height relation may easily be determined using the geometry of

⇒ \(\begin{aligned}
& \Delta H M O, H O^2=H^2+M^2 \\
& \left(R+h_t\right)^2=d^2+R^2, R=\text { radius of the earth } \\
& R^2+h^2+2 R h_t=d_t^2+R^2 \\
& 2 R h_t+h^2=d_t^2, h<<R
\end{aligned}\)

⇒ \(\mathrm{d}_{\mathrm{T}}=\sqrt{2 R h_{\mathrm{t}}}\)

Similarly \(d_R=\sqrt{2 R h_R}\)……………………….2

Hence range \(d_m=\left(d_t+d_R\right)=\sqrt{2 R h_t}+\sqrt{2 R h_R}\)………………………..3

In Case the Satellite is Located At a height then

⇒ \(d_m=d_R=d=\sqrt{2 R h}\)…………………………….4

Area Covered \(\pi d^2=2 \pi R h\)…………………………………..5

Solved Examples

Example 1. In which frequency range, space waves are normally propagated

1. HF
2. VHF
3. UHF
4. SHF

Answer: 3. UHF

Example 2. An antenna behaves as a resonant circuit only when its length is

1. \(\frac{\lambda}{2}\)
2. \(\frac{\lambda}{4}\)
3. λ
4. \(\frac{\lambda}{2} \text { or integral multiple of } \frac{\lambda}{2}\)

Answer: 4. \(\frac{\lambda}{2} \text { or integral multiple of } \frac{\lambda}{2}\)

Question 3. The process of superimposing signal frequency (i.e. audio wave) on the carrier wave is known as

1. Transmission
2. Reception
3. Modulation
4. Detection

Answer: 3. Carrier + signal → modulation.

Question 4. Long-distance short-wave radio broadcasting uses

1. Ground wave
2. Ionospheric wave
3. Direct wave
4. Skywave

Answer: 3. Direct wave

Question 5. The maximum distance up to which TV transmission from a TV tower of height h can be received is proportional to

1. \(h^{1 / 2}\)
2. h
3. \(h^{3 / 2}\)
4. h2

Answer: \(\mathrm{d}=\sqrt{2 \mathrm{hR}} \Rightarrow \quad \mathrm{d} \propto \mathrm{h}^{1 / 2}\)

Example 6. A transmitting antenna at the top of a tower has a height of 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given the radius of Earth 6.4 × 106 m.
Solution:

⇒ \(\begin{aligned}
& \mathrm{d}_{\mathrm{m}}=\sqrt{2 \times 64 \times 10^5 \times 32}+\sqrt{2 \times 64 \times 10^5 \times 50} \mathrm{~m} \\
& =64 \times 10^2 \times \sqrt{10}+8 \times 10^3 \times \sqrt{10} \mathrm{~m} \\
& =144 \times 10^2 \times \sqrt{10} \mathrm{~m}=45.5 \mathrm{~km}
\end{aligned}\)

Example 7. In a communication system, noise is most likely to affect the signal

1. At the transmitter
2. In the channel or the transmission line
3. In the information source
4. At the receiver

Answer: 2. In the channel or the transmission line

Example 8. The waves used in telecommunication are

1. IR
2. UV
3. Microwave
4. Cosmic rays

Answer: 3. UV In telecommunication microwaves are used.

Example 9. Television signals on earth cannot be received at distances greater than 100 km from the transmission station. The reason behind this is that

1. The receiver antenna is unable to detect the signal at a distance greater than 100 km
2. The TV program consists of both audio and video signals
3. The TV signals are less powerful than radio signals
4. The surface of the earth is curved like a sphere

Answer: 4. The surface of the earth is curved like a sphere

Example 10. AM is used for broadcasting because

1. It is more noise-immune than other modulation systems
2. It requires less transmitting power compared with other systems
3. Its use avoids receiver complexity
4. No other modulation system can provide the necessary bandwidth faithful transmission

Answer: 3. Its use avoids receiver complexity

Example. 11 Range of frequencies allotted for commercial FM radio broadcasts is

1. 88 to 108 MHz
2. 88 to 108 kHz
3. 8 to 88 MHz s
4. 88 to 108 GHz

Answer: 1. 88 to 108 MHz

Example 12. At which of the following frequencies, the communication will not be reliable beyond the horizon

1. 1KHz
2. 1 MHz
3. 10 GHz
4. 100 GHz

Answer: 2. MHz travel is a line of sight.

Example 13. Modulation is used to

1. Reduce the bandwidth used
2. Isolate transmission of different users
3. Ensure transmission of intelligence to long-distance
4. To reduce the size of the antenna to a useable range

Answer: 1. Band width is reduced; 2 FM

Example 14. AM is used for broad costing because

1. The signal-to-noise ratio is low
2. It needs less transmission power
3. The AM receiver system is simple width complexity
4. No other modulation provides faithful bandwidth

Answer: 3. The AM receiver system is simple width complexity

Example 15. UHF frequencies normally propagate via

1. Ground wave
2. Skywave
3. Surface-wave
4. Space waves

Answer: 4. UHF travels as a space wave

Example 16. A microwave link operates at a central frequency of 10 GHz and 2% is used for telephone channels. If the telephone is allotted a bandwidth of 8 kHz the number of channels that can be operated simultaneously is

1. 12.5 × 105
2. 12.5 × 103
3. 2.5 × 107
4. 2.5 × 103

Answer: 3. Band width available \(\frac{2}{100} \times 10^9\) one channel needs 8 kHz No of channels operating \(=\frac{2 \times 10^7}{8.5 \times 10^3}=2.5 \times 10^4\)

Modulation And It’s Necessity

The signals to be transmitted (audio, video, or data) are low-frequency signals and there are inherent difficulties in transmitting these signals directly as discussed below.

So these signals are superimposed in any of the properties of high-frequency waves called carriers. When superimposed in amplitude i.e. resultant wave amplitude varies with the signal and we have amplitude modulated wave (AM).

When superimposed in frequency, the frequency of the resultant wave varies with the signal, we have a frequency-modulated (FM) wave similarly if superimposed in phase the phase of the resultant wave varies with the signal and we have a phase-modulated wave (PM). we now outline difficulties in low-frequency transmission.

Size Of Antenna Or Aerial

As pointed out earlier the minimum size of the antenna / aerial required is \(\frac{\lambda}{4}\) For transmitting a 20 kHz signal we require minimum antenna size \(\ell=\frac{\lambda}{4}=\frac{1}{4} \times \frac{3 \times 10^8}{20 \times 10^3} 3.75 \mathrm{~km}.\) This long antenna is not possible.

If instead the signal (or baseband signal) is modulated with CW of frequency 10 MHz the size required is \(\ell=\frac{1}{4} \times \frac{3 \times 10^8}{20 \times 10^3} 7.5 \mathrm{~m}.\) This is practically possible.

Power Radiated

The power radiated by the antenna is determined by length to wavelength ratio (l/λ). More power is radiated at high frequencies making communication better. The power transmitted varies as frequency square (f²)

Mixing Of Signals From Different Transmitters

If signals are transmitted at the baseband (original frequency band) then the receiver shall receive a signal from many transmitters simultaneously and they get mixed. If these signals are modulated on different carrier frequencies, the mixing is avoided and the receiver shall get the desired signal by tuning at that CW frequency.

In digital communication carrier waves are in the form of pulses the modulation may be pulse amplitude, pulse duration, and pulse width modulation as illustrated in Figure 9 The flow chart of Figure 10 below summarizes various types of modulations.

Properties Of Amplitude Modulation:

Frequency spectrum: In amplitude modulated (AM) wave the amplitude of the carrier wave varies with the signal i.e. signal is superimposed in amplitude. If the carrier wave is C(t) = AC sin ACt and the message signal is m(t) = Am sin ωmt then the modulated signal will be

1. Cm(t) = (VC + Vm sin ωmt) sin ωCt
2. Cm(t) = (VC + Vm sin ωmt) sin ωCt

⇒ \(=V_c\left(1+\frac{V_m}{V_c} \sin \omega_m t\right) \sin \omega_c t\)

Where \(\mu=\frac{V_m}{V_c}\) is called modulation index.

⇒ \(C_m(t)=V_c \sin \omega_m t+\left(\frac{V_m}{V_c}\right) \sin \omega_{c t} V_c \sin \omega_c t\)

= VC sin wCt + wVC sin with sin cwt

It is generally less than 1 ( < 1)

⇒ \(C_m(t)=V_c \sin \omega_c t+V_c\left(\frac{1}{2} \cos \left(\omega_c-\omega_m\right) t-\frac{1}{2} \cos \left(\omega_c+\omega_c\right) t\right.\)

Equation (8) shows three sets of angular frequencies original carrier viz., (ωc), (ωc – wm) known as lower sideband and (ωc + ωm) upper sideband of frequencies. The amplitude-modulated frequency spectrum. (f = w/2π)

It follows that if broadcasted bands are sufficiently separated so that sidebands do not overlap, different stations can operate without any interference.

Modulation Index: The ratio of change in amplitude of the carrier wave to amplitude of the original carrier wave is called modulation index or modulation factor, m a
\(m_a=\frac{k V_m}{V_C} \text {, }\) actor k determines the maximum change in amplitude for the given amplitude of the modulating wave.

On amplitude modulation, the maximum and minimum amplitudes are Amax and Amin then the maximum change in amplitude is (Vmax – VC) and so \(m_{\mathrm{a}}=\frac{V_{\max }-V_C}{V_C}\)

For example if VC = V and \(V_m=\frac{V}{2}\) Then \(m_a=V+\frac{V}{2}-V=V / 2=50 \%\)

⇒ \(m_a=V+\frac{V}{2}-V=V / 2=50 \%\)

If VC = VC and Vm = V then Vmax = 2V

If VC = V, Vm = 3/2V then Vmax = 5/2 V

In this case, the carrier is overmodulated> 100% it may be noted that the modulation factor ma, determines the strength and quality of the signal transmitted. An audio signal is generally AM modulated and hence higher is modulation the stronger and clearer will be the signal.

If u = 1 modulation index may be expressed in terms of maximum and minimum amplitude, Vmax and Vmin

If the carrier wave is modulated with many signals then the total modulation index is given by

Bandwidth Required: The required bandwidth is from the lower sideband to the upper side band hence,

⇒ \(\Delta f=\frac{\left(\omega_{\mathrm{c}}+\omega_{\mathrm{m}}\right)}{2 \pi}-\frac{\omega_{\mathrm{c}}-\omega_{\mathrm{m}}}{2 \pi}=\frac{2 \omega_{\mathrm{m}}}{2 \pi}=2 f_m\)

Power in AM Wave: Power dissipated in any circuit having resistance R and supplied at rms voltage Vrms, is given by \(\left(\frac{V_{\mathrm{ms}}^2}{\mathrm{R}}\right)\) The total power is the sum of the power in side bands plus the power in a carrier wave. Total power transmitted

⇒ \(\begin{aligned}
& \text { = } P_{\text {Total }}=P_{\text {LSB }}+P_{\text {USB }}+P_{c w} \\
& =\left(\frac{m_a V_{\mathrm{C}}}{2 \times \sqrt{2}}\right)^2 \frac{1}{R}+\left(\frac{m_a V_{\mathrm{C}}}{2 \times \sqrt{2}}\right)^2 \frac{1}{R}+\left(\frac{V_{\mathrm{C}}}{\sqrt{2}}\right)^2 \frac{1}{R}=\frac{V_{\mathrm{C}}^2}{2 R}\left(1+\frac{m_{\mathrm{a}}^2}{2}\right)
\end{aligned}\)

The ratio of power transmitted to carrier power

⇒ \(\frac{\frac{P_{\text {Total }}}{P_{\mathrm{CW}}}=\frac{V_{\mathrm{C}}^2}{2 R}\left(1+\frac{m_{\mathrm{a}}^2}{2}\right)}{\left(\frac{V_{\mathrm{C}}^2}{2 R}\right)=\left(1+\frac{m_{\mathrm{a}}^2}{2}\right)}\)

Fraction of power transmitted in the sideband

⇒ \(\frac{\frac{P_{S B}}{P_{\text {Total }}}=\frac{1}{R}\left(\frac{m_a V_{\mathrm{C}}}{2 \sqrt{2}}\right)^2}{\frac{V_{\mathrm{C}}^2}{2 R}\left(1+\frac{m_{\mathrm{a}}^2}{2}\right)=\left(\frac{m_a^2 / 2}{1+m_a^2 / 2}\right)}\)

Distortion-free maximum power transfer. For distortion-free transmission ma = 0. so

Production Of Amplitude Modulated Wave

To add the signal to the amplitude of CW the voltage signal is mixed up in a mixer and sampled by a square law device. The output is filtered by a band pass filter centered around the carrier frequency. The signal is amplified by a power amplifier before transmission.

Detection Of Amplitude Modulated Signal

The signal is received by the receiving antenna. As a received signal is weak due to attenuation in the channel, it has to be amplified. The detection of high frequency is difficult so, it is changed to a low frequency called intermediate frequency, (IF).

This converted signal is detected and amplified. The AM wave is rectified and rejects the lower part of the AM wave. Finally envelope is detected (filtering CW component). The entire process

Limitations of AM wave Transmission

AM communication has the following problems

1. Reception is noisy and becomes audio noise caused by various
2. Efficiency is low, machines get mixed
3. The operating range is small
4. The audio quality is poor.

Solved Examples

Example 17. An AM wave has 1800 watts of total power content, for 100% modulation the carrier should have power content equal to

1. 1000 watt
2. 1200 watt
3. 1500 watt
4. 1600 watt

Answer: \(P_t=P_c\left(1+\frac{m_a^2}{2}\right) ; \text { Here } m_a=1\)

Question 18. A TV tower has a height of 75m. What is the maximum distance and area up to which this TV transmission can be received? Take the radius of the earth as 6.4 × 10 6 m.
Solution: \(d=\sqrt{2 R h}=\sqrt{2 \times 6.4 \times 10^6 \times 75}=3.1 \times 10^4 \mathrm{~m}=31 \mathrm{~km}\)

Question 19. A TV tower has a height of 100 m. How much population is covered by the TV broadcast if the average population density around the tower is 1000 km –2? Given: radius of earth = 6.37 × 106 m.
Solution: h = 100m, R = 6.37 × 106 m,

Average population density = 1000 km–2 = 1000(103)–2m–2 =10–3m–2 2hR Distance up to which the transmission could be viewed, d =

Total area over which transmission could be viewed = πd2 = 2πhR

Population covered = 10–3 × 2πhR = 10–3 × 2 × 3.14 × 100 × 6.37 × 106 = 40 lakh

Example. 20 What is the modulation index of an overmodulated wave

1. 1
2. Zero
3. < 1
4. > 1

Solution 4. When ma > 1 then the carrier is said to be over-modulated.

Example. 21 Which of the following is the disadvantage of FM over AM

1. Larger bandwidth requirement
2. Larger noise
3. Higher modulation power
4. Low efficiency

Solution: 1. Frequency modulation requires a much wider channel (7 to 15 times) as compared to AM.

Example 22. When the modulating frequency is doubled, the modulation index is halved, and the modulating voltage constant of the modulation system is

1. Amplitude modulation
2. Phase modulation
3. Frequency modulation
4. All of the above

Solution: 3. Frequency modulation

Example 23. Indicate which one of the following systems is digital

1. Pulse position modulation
2. Pulse code modulation
3. Pulse width modulation
4. Pulse amplitude modulation

Solution: 2. Pulse code modulation

Example 24. In an FM system, a 7 kHz signal modulates a 108 MHz carrier so that the frequency deviation is 50 kHz. The carrier swing is

1. 7.143
2. 8
3. 0.71
4. 350

Solution: 1. 7.143

Example 25. The sinusoidal carrier voltage of frequency 1.5 MHz and amplitude 50 V is amplitude modulated by the sinusoidal voltage of frequency 10 kHz producing 50% modulation. The lower and upper sideband frequencies in kHz are

1. 1490,1510
2. 1510,1490
3. \(\frac{1}{1490}, \frac{1}{1510}\)
4. \(\frac{1}{1510}, \frac{1}{1490}\)

Solution: Here, fc = 1.5 MHz = 1500 kHz, fm = 10 kHz

• Lower sideband frequency = fc = fm = 1500 kHz – 10 kHz = 1490 kHz
• Upper sideband frequency = fc + fm = 1500 kHz + 10 kHz = 1510 kHz

Frequency Modulated Wave

Most of the noises affect the amplitude of the signal and hence to noise ratio is greatly improved if the amplitude of CW remains unaffected. This is what is done in frequency modulation where the frequency of CW varies.

Analysis

Consider a voltage signal vm = Vm cos wants to be frequency modulated on a carrier voltage wave vc = Vc cos (ωct + ω0).

Where ωm = 2πfm, ωc = 2ωfc are respectively angular frequencies of the signal and the carrier waves and Vm and Vc are their amplitude ω0 is the initial phase of the carrier. The instantaneous phase of carrier wave (CW).

⇒ \(\phi(t)=\omega \mathrm{ct}+\theta_0\)

The angular frequency of the modulated wave shall be

ω = ωc + k Vm cos ωmt

Where k is the frequency conversion factor which is constant. The phase of the FM wave at any instant shall be

⇒ \(\phi(\mathrm{t})=\int \omega \mathrm{dt}=\int\left(\omega_{\mathrm{c}}+\mathrm{kV} \mathrm{m}_{\mathrm{m}} \cos \omega_{\mathrm{m}} \mathrm{t}\right) \mathrm{dt}\)

⇒  \(\phi(t)=\omega_c t+\frac{k V_m}{\omega_m} \sin \omega_m t\)

Hence equation of FM voltage wave is \(v_{F M}(t)=V_c \sin \left[\omega_c t+\frac{k V_m}{\omega_m} \sin \omega_m t\right]\)

The instantaneous frequency of an FM wave is given by

⇒ \(\frac{1}{2 \pi} \frac{\partial \phi(t)}{d t}\)

Therefore \(f=\frac{1}{2 \pi} \omega_c+\frac{k V_m}{2 \pi} \cos \omega_m t\)

The maximum and minimum frequencies are obviously,

⇒ \(\begin{aligned}
& f_{\max }=f_c+\frac{k V_m}{2 \pi} \\
& f_{\min }=f_c-\frac{k V_m}{2 \pi}
\end{aligned}\)

The maximum change in frequency from the mean value is called frequency deviation

⇒ \(f_d=\left(f_{\max }-f_c\right)=\left(f_c-f_{\min }\right)=\frac{k V_m}{2 \pi}\)

The total variation of frequency from the maximum to the minimum is called carrier swing. It is twice the frequency deviation.

⇒ \(C S=2 f d=\frac{k V_m}{\pi}\)

The frequency modulation index mf is defined as the ratio of frequency deviation to the modulation frequency

⇒ \(m_f=\frac{f_d}{f_m}=\frac{\omega_d}{\omega_m}=\frac{k V_m}{\omega_m}\)

The equation of FM wave becomes vFM = VC sin (ct + mf sin mt) ……….(24) (c) Frequency spectrum

FM Side Bands:

Equation (23) may be expanded and trigonometric manipulations shall show that there are a series of sidebands \(\left(f_c \pm f_m\right),\left(f_c \pm 2 f_m\right),\left(f_c \pm 3 f_m\right)\), etc. with decreasing amplitudes. Side bands are equally spaced on either side of carrier frequency FC

Frequency bands in use

As pointed out frequency modulation (FM) has a better quantity of transmission with large bandwidth. The manmade noises and atmosphere changes do not affect transmission quality. Also, the fidelity is good for the transmission of music. The frequency bands in use are:

1. 88 to 108 MHz FM Radio
2. 47 to 230 MHz VHF TV
3. 470 to 960 MHz UHF TV

Digital Communication: Data Transmission & Retrieval

Digital communication ensures less noise and less error communication. Hear, carrier is a digital pulsating wave in binary codes 0 and 1. The analog signal is digitized. There are many encoding steps: source coding channel coding, etc. A typical digital communication system is shown in

There are normally three steps converting the signal into pulses of the same height and negligible width quantization and Coding quantized pulses following some rule.

Modem

Is short-term used for modulators and demodulators? As seen in Figure 1 5 modulator and demodulator are needed for two-way communication and a single modem unit serves the purpose.

Modulation Used

In digital communication modulation techniques used are shown in Figure 9 which is quite illustrative.

Modern communication systems commonly use frequency-shifting keys.

Optical Communication

Typical optical communication system. It uses optical frequency as a carrier so it has the following advantages

1. There is no electromagnetic interference
2. Enormous channel capacity
3. Requires optical fiber as communication channel
4. Mostly used in LAN (Local area networking)
5. Setup for digital communication (Block diagram)

Optical Fibre

Principle: Light travels in an optical fiber through total internal reflection from opposite walls. For total reflection light must travel from the sensor to the rarer medium and the angle of incidence from the densor to the rarer interface should be greater than the critical angle, \(\theta_c=\sin ^{-1}\left(\frac{\mu_{\text {rarer }}}{\mu_{\text {dens }}}\right)\) u’s are corresponding refractive indices

Optical fiber consists of an inner transparent cylindrical core of refractive index 1 surrounded by a transparent cylindrical cladding of refractive index u2 (< u1).

The core-cladding system is secured against mechanical shocks by multilayered polyester nylons, etc., Let a light ray be an incident making an angle u with the axis of the fiber as shown in

The ray incident on the core face is refracted into core and falls on the core-cladding interface at an angle of u1.

Using Snells law

u0 sin u = 1 sin r, r = angle of refraction ………….(25)

Where u0 is the refractive index of the outer medium for air u0 = 1. For the ray to be reflected from the corecladding interface.

Since r = (90 – θc), hence the value of the angle θ for total reflection at the core the clad interface is

⇒ \(\begin{aligned}
\mu_0 \sin \theta & =\mu_1 \sin \left(90-\theta_c\right)=\mu_1 \cos \theta_c=\mu_1 \sqrt{1-\sin ^2 \theta_c .} \\
\text { or } & \sin \theta=\mu_1 \sqrt{1-\left(\frac{\mu_2}{\mu_1}\right)^2}=\sqrt{\mu_1^2-\mu_2^2}
\end{aligned}\)

When 0 increases, r increases but  decreases therefore value of the angle given by Equation (27) is the maximum permissible value and it is called the maximum angle of acceptance

⇒ \(\theta_{\mathrm{m}}=\sin ^{-1}\left(\sqrt{\mu_1^2-\mu_2^2}\right)\)

The quantity sin gives the light-gathering capacity of the fiber. It is called numerical aperture. (NA)

⇒ \(\begin{aligned}
\quad & N A=\mu_0 \sin \theta_m=\sqrt{\mu_1^2-\mu_2^2} \\
\text { or } \quad \text { NA } & =\sqrt{\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}
\end{aligned}\)

Since \(\begin{aligned}
& \left(\mu_1-\mu_2\right) \text { is small so } \mu_2=\mu_1 \\
& =\mu_1 \sqrt{2\left(\frac{\mu_1-\mu_2}{\mu_1}\right)}=\mu_1 \sqrt{2 \Delta} \\
& \Delta=\frac{\mu_1-\mu_2}{\mu_1}
\end{aligned}\) is called fractional change in refractive index.

Satellite Communications

In the age of IT explosion where enormous data need to be transmitted and received, there occurs a need for higher frequency bands and more channels. This is possible only with satellite communication.

As discussed earlier, the signal from the transmitting station is sent to communication satellite equipped with transmitting and receiving systems known as radio Transponder (RT).

The signal transmitted and received by satellite is called up-link whereas as transmitted by satellite and received at the ground is called down the link to avoid confusion the frequencies of up and downlinks are kept different.

The commonly used satellite system consists of three geostationary satellites located on the vertices of an equilateral triangle having verticals on a geostationary orbit to cover entire globe space.

Most of the satellites are in geostationary orbit yet two more orbits are used for communication satellites. These are polar circular orbits near the earth about 1000 km, high, their inclination is 90º.

The other is a highly elliptical orbit inclined at 63º to fulfill the needs of high-altitude regions. Commonly known as the 63º slot Finally figure 18 shows the summary of various communication systems.

Example 26. If f0 and ff represent the carrier wave frequencies for amplitude and frequency modulations respectively, then

1. f0 > ff
2. f0 < ff
3. f0 = ff
4. f0 ≥ ff

Answer: f0 < ff

Example 27. The frequency of an FM transmitter without signal input is called

1. Lower sideband frequency
2. Upper sideband frequency
3. Resting frequency
4. None of these

Solution: 3. Resting frequency

Example 28. What type of modulation is employed in India for radio transmission

1. Amplitude modulation
2. Frequency modulation
3. Pulse modulation
4. None of these

Solution: 1. Amplitude modulation

Example 29. While tuning in a certain broadcast station with a receiver, we are actually

1. Varying the local oscillator frequency
2. Varying the frequency of the radio signal to be picked up
3. Tuning the antenna
4. None of these

Solution: 1. Varying the local oscillator frequency

Example 30. Consider telecommunication through optical fibers. Which of the following statements is not true

1. Optical fibers may have homogeneous core with suitable cladding
2. Optical fibers can be of graded refractive index
3. Optical fibers are subject to electromagnetic interference from outside
4. Optical fibers have extremely low transmission loss

Answer: 3. Optical fibers are subject to electromagnetic interference from outside

Example 31. The phenomenon by which light travels in an optical fiber is

1. Reflection
2. Refraction
3. Total internal reflection
4. Transmission

Answer: 3. Total internal reflection

Example 32. Consider an optical communication system operating at λ = 800 nm. Suppose, only 1% of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommodated for transmitting audio signals requiring a bandwidth for transmitting audio signals requiring a bandwidth of 8 kHz?

1. 4.8 × 108
2. 48
3. 6.2 × 108
4. 4.8 × 105

Answer: 2. 48

Example 33. A ground receiver station receives a signal at (i) 5 MHz and is transmitted from a ground transmitter at a height of 300 m, located at a distance of 100 km from the receiver station. The signal is coming via. Radius of earth = 6.4 × 106 m. Nmax of isophere = 1012m3.

1. Space wave
2. Skywave propagation
3. Satellite transponder
4. All of these

Answer: 4. All of these

Example 34. The antenna current of an AM broadcast transmitter modulated by 50% is 11 A. The carrier current is

1. 10.35 A
2. 9.25 A
3. 10 A
4. 5.5 A

Solution: 1. 10.35 A

Example 35. If several sine waves with modulation indices n 1, n2, n3,………modulate a carrier wave, then the total modulation index (n) of the wave is

1. n1 + n2…….. + 2 (n1 + n2…)
2. \(\sqrt{n_1-n_2+n_3 \ldots \ldots}\)
3. \(\sqrt{n_1^2+n_2^2+n_3^2 \ldots \ldots . .}\)
4. None Of These

Answer: 3. \(\sqrt{n_1^2+n_2^2+n_3^2 \ldots \ldots. .}\)

Example 36. A transmitter supplies 9 kW to the when unmodulated. The power radiated when modulated to 40% is

1. 5 kW
2. 9.72 kW
3. 10 kW
4. 12 kW

Solution: 2. 9.72 kW

Example 37. In an FM system, a 7 kHz signal modulates a 108 MHz carrier so that the frequency deviation is 50 kHz. The carrier swing is

1. 7.143
2. 8
3. 0.71
4. 350

Solution: 1. 7.143

Example 38. The modulation index of an FM carrier having a carrier swing of 200 kHz and a modulating signal of 10 kHz is

1. 5
2. 10
3. 20
4. 25

Solution 2. 10

Phase Velocity

It is defined as the velocity with which the peak of a sinusoidal pattern is moving. consider wave: cos [Ψ (tx)] cos (kx – wt)

there is peak at Ψ = 0 or θ = 0, so kx – wt = 0

⇒ \(x-w / k t=0 \quad \Rightarrow \quad V p=\frac{w}{k}\)

So for a particular frequency (w) related to time phase velocity is the rate at which the phase of the wave propagates in space.

In the transmission line group of waves travel out with one particular phase velocity is VP = W/K (depending upon frequency ‘w’ oscillation per sec)

⇒ \(V_{\text {group }}=\frac{\mathrm{uw}}{\mathrm{dK}} \quad \omega=\text { wave’s angular frequency }\)

⇒\(V_F=\frac{\text { Ratio of phase velocity }}{\text { velocity of light }}\)

MUF (maximum usable frequency): It is the maximum frequency for a given angle of incidence which gets reflected from the ionosphere. It depends on the angle of incidence.

⇒ \(\text { MUF }=\frac{\text { critical frequency }}{\cos \theta}\)

Relative Permittivity (Dielectric Constant)

The velocity of propagation of a signal in a transmission line is determined mainly by the permittivity of the dielectric material used to construct the line. Permittivity is a measure of the ability of the dielectric material to maintain a difference in electric charge over a given distance.

⇒ \(\varepsilon_{\mathrm{r}}=\frac{\mathrm{C}^2}{\mathrm{~V}_{\mathrm{P}}^2} \quad \Rightarrow \quad \varepsilon_{\mathrm{r}}=\frac{1}{(\text { velocity factor })^2} \quad \Rightarrow \quad \text { velocity factor }=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}}}}\)

where εr = Relative permittivity (dielectric constant)

C = Velocity of light in free space (3 × 108 m/s)

VP = Velocity of propagation (m/s)

Role Of Ionosphere In Radio – Communication

The ionosphere plays a great role in broadcasting, ship and air-craft communication, and navigation by reflecting the radio signals back to the receivers. However, its effectiveness depends on the frequency of the transmitted signal.

This is critical because the behavior of the ionosphere often shows marked differences between day and night. Moreover, it is known for changing its behavior during different seasons. The ionosphere refractive index as represented by the Appleton – Hartee equation is.

⇒ \(n=\left(1-\frac{f_p^2}{f^2}\right)^{1 / 2} \simeq 1-\frac{1}{2} \frac{f_p^2}{f^2}=1-\frac{40.3 N}{f^2}\)

where

fp: electron plasma frequency in Hz

N: electron number density in m–3 from (1), the magnitude of phase velocity can be derived as

⇒ \(V_p=\frac{w}{k}=\frac{w}{n} \cong C\left(1+\frac{40.3 N}{f^2}\right)\)

fP: Depending on the refractive index and frequency the group velocity can be obtained by the equation

⇒ \(vg=\frac{\partial w}{\partial K}=\frac{C}{\left(\frac{\partial(n f)}{\partial f}\right)} \cong C\left(1-\frac{40.3 \mathrm{~N}}{f^2}\right)\)

when the velocity of the wave varies with the frequency, the medium is known as a dispersive medium. because of this dispersion, the idea of group velocity is introduced to represent the velocity of the crest of a group of interfering waves.

⇒ \(\begin{aligned}
& V_p \text { (phase velocity) }=\frac{\omega}{K} \\
& V g \text { (group velocity) }=\frac{d \omega}{d K}
\end{aligned}\)

⇒ \(\begin{aligned}
& =\frac{d}{d k}\left(v_p k\right)=v_p+k \frac{d v_p}{d k} \\
& =v_p+\frac{2 \pi}{\lambda} \times \frac{d v_p}{d(2 \pi / \lambda)} \\
& =v_p+\frac{1}{\lambda} \times \frac{d v_p}{-\frac{1}{\lambda^2} d \lambda}
\end{aligned}\)

⇒ \(\begin{aligned}
& V_g=V p-\lambda \frac{d v_p}{d \lambda} \\
& \omega=2 \pi v=2 \pi \times \frac{E}{h}=\frac{2 \pi m c^2}{h} \\
& \omega=\frac{2 \pi c^2}{h} m_0 \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
\end{aligned}\)

V = velocity of the particle

⇒ \(K=\frac{2 \pi}{\lambda}=\frac{2 \pi}{h} m v=\frac{2 \pi v}{h} m_0 \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\)

⇒ \(\begin{aligned}
& V_g=\frac{\omega}{K}=\frac{c^2}{V} \\
& V_p V_g=C^2 \quad\left(V_p>c, V g<c\right) \\
& V_g=\frac{d \omega}{d k}=\frac{d \omega / d v}{d k / d v}=V
\end{aligned}\)

Experimental result \(\mu=\left(1-\frac{f_p^2}{f^2}\right)^{1 / 2}=1-\frac{f_p^2}{2 f^2}=1-\frac{40.3 N}{f^2}\)

⇒ \(V_p=\frac{c}{\mu}=C\left(1-\frac{40.3 N}{f^2}\right)^{-1}\)

⇒ \(V_p=C\left(1+\frac{40.3 N}{f^2}\right)\)

VpVg = c2 (=refractive index, vp = phase velocity, vg = group velocity)

⇒ \(V_g=\frac{c^2}{c\left(1+\frac{40.3 N}{f^2}\right)}\)

⇒ \(V g=c\left(1-\frac{40.3 N}{f^2}\right)\)

fp = (N = electron density, fp = electron plasma frequency)

MUF (most useable frequency) \(=\frac{f_c}{\cos i}\)

(Here fc = critical frequency for normal incidence , i = angle of incidence)

⇒ \(\mu=\sqrt{\varepsilon_{\mathrm{r}} \mu_{\mathrm{r}}}\)

Velocity factor \(=V F=\frac{1}{\mu}=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}} \mu_{\mathrm{r}}}}\)

Usually ur=1

therefore \(V F=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}}}}\)

Chapter 5 Principles Of Communication Multiple Choice Questions Exercise-1

Question 1. A digital signal –

1. Is less reliable than analog signal
2. Is more reliable than analog signal
3. Is equally reliable as the analog signal
4. Signal reliability meaning

Answer: 2. Is more reliable than analog signal

Question 2. Modern communication systems use

1. Analog circuits
2. Digital circuits
3. Combination of analog and digital circuits
4. Discrete circuits

Answer: 2. Digital circuits

Question 3. The audio signal –

1. Can be sent directly over the air for large distance
2. Cannot be sent directly over the air for large distance
3. Possess very high frequency
4. Posses very high intently

Answer: 2. Cannot be sent directly over the air for large distance

Question 4. The process of changing some characteristic of a carrier wave per the intensity of the signal is called –

1. Amplification
2. Rectification
3. Modulation
4. Normalization

Answer: 3. Modulation

Question 5. If a carrier wave of 1000 kHz is used to carry the signal, minimum the length of the transmitting antenna will be equal to –

1. 300 M
2. 150 M
3. 75 M
4. 750 M

Answer: 3. 75 M

Question 6. The type of modulation used for continuous wave and analog signal are –

1. One only
2. Two only
3. Three only
4. None of these

Answer: 3. Three only

Question 7. In amplitude modulation what changes in the carrier wave –

1. Only the amplitude is changed but the frequency remains the same
2. Both the amplitude and frequency change equally
3. The amplitude and frequency change unequally
4. Only phase changes

Answer: 1. Only the amplitude is changed but the frequency remains the same

Question 8. Modulation factor determines –

1. Only the strength of the transmitted signal
2. Only the quality of the transmitted signal
3. Both the strength and quality of the signal
4. None of the above

Answer: 3. Only the quality of the transmitted signal

Question 9. The degree of modulation is kept –

1. At any value
2. Less than 100%
3. Greater than 100 %
4. May be at any value from 100% to 200%

Answer: 2. Less than 100%

Question 10. If the maximum and minimum voltage of an am wave are vmax. And vmin. Respectively the modulation factor–

1. \(m=\frac{v_{\text {max. }}}{V_{\text {max. }}+V_{\text {min. }}}\)
2. \(m=\frac{v_{\text {min. }}}{V_{\text {max. }}+V_{\text {min. }}}\)
3. \(m=\frac{v_{\text {max. }}+V_{\text {min. }}}{V_{\text {max. }}-V_{\text {min } .}}\)
4. \(m=\frac{v_{\text {max. }}-V_{\text {min. }}}{V_{\text {max. }}+V_{\text {min. }}}\)

Answer: 4. \(m=\frac{v_{\text {max. }}-V_{\text {min. }}}{V_{\text {max. }}+V_{\text {min. }}}\)

Question 11. The am wave contains three frequencies, viz :

1. \(\frac{f_c}{2}, \frac{f_c+f_s}{2}, \frac{f_c-f_s}{2}\)
2. 2Fc, 2(fc + fs), 2(fc –fs)
3. Fc,(fc + fs), (fc – fs)
4. Fc, fc, fc

Answer: 3. Fc,FC + fs), (fc – fs)

Question 12. In a wave, carrier power is given by

1. \(p_c=\frac{2 e_c^2}{r}\)
2. \(p_c=\frac{e_c^2}{r}\)
3. \(p_c=\frac{e_c^2}{2 r}\)
4. \(p_c=\frac{e_c^2}{\sqrt{2} r}\)

Answer: 3. \(p_c=\frac{e_c^2}{2 r}\)

Question 13. The fraction of total power carried by side bands is given by\

1. \(\frac{p_s}{p_t}=m^2\)
2. \(\frac{p_s}{p_t}=\frac{1}{m^2}\)
3. \(\frac{p_s}{p_t}=\frac{2+m^2}{m^2}\)
4. \(\frac{p_s}{p_t}=\frac{m^2}{2+m^2}\)

Answer: 4. \(\frac{p_s}{p_t}=\frac{m^2}{2+m^2}\)

Question 14. Which of the following is/are the limitations of amplitude modulation?

1. Clear reception
2. High efficiency
3. Small operating range
4. Good audio quality

Answer: 3. Small operating range

Question 15. To avoid noise the frequency above which transmission of electrical energy is practical?

1. 0.2 khz
2. 2Khz
3. 20 Khz
4. 200Khz

Answer: 3. 20 Khz

Question 16. What type of modulation is employed in India for radio transmission?

1. Pulse modulation
2. Frequency modulation
3. Amplitude modulation
4. None of these

Answer: 3. Amplitude modulation

Question 17. For a carrier frequency of 100 kHz and a modulating frequency of 5 khz what is the bandwidth of a transmission–

1. 5 kHz
2. 10Khz
3. 20 kHz
4. 200 kHz

Answer: 2. 10Khz

Question 18. Which one of the following subsystems is used for the satellite’s orbit position and altitude?

1. Thrust subsystem
2. Power subsystem
3. Antenna subsystem
4. Stabilization subsystem

Answer: 1. Thrust subsystem

Question 19. Intelsat satellite works as a:

1. Transmitter
2. Repeater
3. Absorber
4. None of these

Answer: 2. Repeater

Question 20. Intelsat satellite is used for:

1. In-house radio communication
2. Intercontinental communication
3. Radar communication
4. None of the above

Answer: 2. Intercontinental communication

Question 21. A geosynchronous satellite is :

1. Located at a height of 35,860 km to ensure global coverage
2. Appears stationary over the Earth’s magnetic pole
3. Not stationary at all, but orbits the earth 24 hrs
4. Motionless in space (except for its spin)

Answer: 3. Not stationary at all, but orbits the earth 24 hrs

Question 22. The frequency band used for radar relay systems and television –

1. Uhf
2. Vlf
3. Vhf
4. Ehf

Answer: 1. Uhf

Question 23. Fading applies to :

1. Troposcatter propagation
2. Ionospheric propagation
3. Faraday rotation
4. Atmospheric storms

Answer: 1. Troposcatter propagation

Question 24. When microwave signals follow the curvature of the earth, this is known as :

1. Window
2. The faraday effect
3. Ionospheric reflection
4. Ducting

Answer: 4. Ducting

Question 25. In which of the regions of the earth’s atmosphere temperature decreases with height?

1. Ionosphere
2. Stratosphere
3. Troposphere
4. Mesosphere

Answer: 3. Troposphere

Question 26. Major parts of a communications system are :

1. Transmitter and modulator receiver
2. Receiver demodulator and communication channel
3. Transmitter and communication channel
4. Transmitter, receiver, and communication channel

Answer: 4. Transmitter, receiver, and communication channel

Question 27. Communication channels may consist of :

1. Transmission line
2. Optical fiber
3. Free space
4. All of the above

Answer: 4. All of the above

Question 28. The basic components of a transmitter are :

1. Message signal generator and antenna
2. Modulator and antenna
3. Signal generator and modulator
4. Message signal generator, modulator, and antenna

Answer: 4. Message signal generator, modulator, and antenna

Question 29. The message signal can be :

1. Analog only
2. Digital only
3. Analog and digital
4. Analog or digital

Answer: 4. Analog or digital

Question 30. A microphone converts

1. Sound signals into electrical signals
2. Electrical signals into sound signals
3. Both and above
4. Neither nor (2)

Answer: 1. Sound signals into electrical signals

Question 31. A loudspeaker converts

1. Electrical signals into sound signals
2. Sound signals into electrical signals
3. Both and above
4. Neither nor (2)

Answer: 1. Electrical signals into sound signals

Question 32. Which is more advantageous?

1. Analog data communication
2. Digital data communication?
3. Analog data communication
4. Digital data communication
5. Both are equally good
6. Depends on the application

Answer: 2. Digital data communication

Question 33. The message signal is usually of :

1. Audio frequency range
2. Radiofrequency range
3. Audio or radio frequency range
4. Mixture of both

Answer: 1. Audio frequency range

Question 34. Modulation is the phenomenon of :

1. Superimposing the audio frequency signal over a carrier wave
2. Separating the audio frequency signal from the carrier wave
3. Separating carrier wave from the modulated wave
4. Any of (1),(2),above

Answer: 1. Superimposing the audio frequency signal over a carrier wave

Question 35. In amplitude modulation, carrier wave frequencies are:

1. Lower compared to those in frequency modulation
2. Higher compared to those in frequency modulation
3. Same as in frequency modulation
4. Lower sometimes and higher sometimes than those in frequency modulation

Answer: 1. Lower compared to those in frequency modulation

Question 36. The transmission media can be :

1. Guided only
2. Unguided only
3. Both and
4. Neither nor (2)

Answer: 3. Both and

Question 37. A 1000 kHz carrier is modulated with 800 hz audio signals. What are the frequencies of the first pair of sidebands:

1. 1000.8 khz, 999.2 khz
2. 999.2 kHz, 998.4 kHz
3. 1001.6 khz, 1000.8 khz
4. 1000 kHz, 800 Hz

Answer: 1. 1000.8 kHz, 999.2 kHz

Question 38. In an amplitude-modulated wave, for an audio frequency of 500 cps, the appropriate carrier frequency will be:

1. 50 C/s
2. 100 C/s
3. 500 C/s
4. 50000 C/s

Answer: 4. 50000 C/s

Question 39. In a.m., The total modulation index should not exceed one or else :

1. The system will fail
2. Distortion will result
3. The amplifier will be damaged
4. None of the above

Answer: 2. Distortion will result

Question 40. Electromagnetic waves are caused polarized because it –

1. Undergoes reflection
2. Undergoes refraction
3. Transverse nature
4. Longitudinal in nature

Answer: 3. Transverse nature

Question 41. The velocity of electromagnetic waves in a dielectric medium ( r = 4) is –

1. 3 × 108 Metre/second
2. 1.5 × 108 meter/second
3. 6 × 108 Metre/second
4. 7.5 ×107 metre/second

Answer: 4. 7.5 ×107 metre/second

Question 42. An ‘antenna’ is :

1. Inductive
2. Capacitive
3. Resistive above its resonance frequency
4. None of the above

Answer: 1. Inductive

Question 43. The characteristic impedance of a loss-less transmission line is given by

1. \(z_0=\sqrt{l c}\)
2. \(z_0=\sqrt{l / c}\)
3. \(z_0=\sqrt{c / l}\)
4. Z0 = lc

Answer: 2. \(z_0=\sqrt{l / c}\)

Question 44. The q of a resonant transmission line is

1. \(q=\frac{\omega}{l r}\)
2. \(q=\frac{\omega r}{l}\)
3. \(q=\frac{l}{r}\)
4. \(q=\frac{\omega l}{r}\)

Answer: 3. \(q=\frac{l}{r}\)

Question 45. The distance between consecutive maxima and minima on a transmission line is given by –

1. λ/2
2. 2λ
3. λ
4. λ/4

Answer: 4. λ/4

Question 46. Through which mode of propagation, the radiowaves can be sent from one place to another –

1. Ground wave propagation
2. Sky wave propagation
3. Space wave propagation
4. All of them

Answer: 4. All of them

Question 47. The frequencies of electromagnetic waves employed in space communication vary over a range of –

1. 104 Hz to 107 Hz
2. 104 Hz to 1011 Hz
3. 1 Hz to 104 Hz
4. 1 Hz to 1011 Hz

Answer: 2. 104 Hz to 1011 Hz

Question 48. The wavelength of electromagnetic waves employed for space communication lies in the range of-

1. 1 Mm to 30 m
2. 1Mm to 300 m
3. 1 Mm to 3 km
4. 1 Mm to 30 km

Answer: 4. 1 Mm to 30 km

Question 49. The radiowaves of frequency 300 MHz to 3000 MHz belong to –

1. High-frequency band
2. Very high frequency band
3. Ultra high-frequency band
4. Super high frequency band

Answer: 3. Ultra high frequency band

Question 50. The maximum range of ground or surface wave propagation depends on –

1. The frequency of the radiowaves only
2. The power of the transmitter only
3. Both of them
4. None of them

Answer: 3. Both of them

Question 51. In which frequencies range space waves are normally propagated –

1. Hf
2. Vhf
3. Uhf
4. Shf

Answer: 3. Uhf

Question 52. For television broadcasting, the frequency employed is normally –

1. 30 – 300 M hz
2. 30 – 300 G hz
3. 30 – 300 K Hz
4. 30 – 300 Hz

Answer: 1. 30 – 300 M hz

Question 53. The sound waves after being converted into electrical waves are not transmitted as such because –

1. They travel with the speed of sound
2. The frequency is not constant
3. They are heavily absorbed by the atmosphere
4. The height of the antenna has to be increased several times

Answer: 3. They are heavily absorbed by the atmosphere

Question 54. The process of superimposing signal frequency (i.e. Audio wave) on the carrier wave is known as –

1. Transmission
2. Reception
3. Modulation
4. Detection

Answer: 3. Modulation

Question 55. In an amplitude-modulated wave for an audio frequency of 500 cycles/second, the appropriate carrier frequency will be –

1. 50 Cycles/sec
2. 100 Cycles/sec
3. 500 Cycles/sec
4. 50,000 Cycles/sec

Answer: 4. 50,000 Cycles/sec

Question 56. The tv. The transmission tower in Delhi has a height of 240 m. The distance up to which the broadcast can be received, (taking the radius of earth to be 6.4 x 106 m) is –

1. 100 Km
2. 60 Km
3. 55. Km
4. 50 Km

Answer: 3. 55. Km

Question 57. Radio waves of constant amplitude can be generated with –

1. Filter
2. Rectifier
3. Amplifier
4. Oscillator

Answer: 4. Oscillator

Question 58. The range of frequencies allotted for commercial FM radio broadcasts is –

1. 88 To 108 MHz
2. 88 To 108 kHz
3. 8 To 88 MHz
4. 88 To 108 GHz

Answer: 1. 88 To 108 MHz

Question 59. Intel set satellite works as a –

1. Transmitter
2. Receiver
3. Absorber
4. Repeater

Answer: 4. Repeater

Question 60. The space waves that are affected seriously by atmospheric conditions are

1. MF
2. Hf
3. Vhf
4. Uhf

Answer: 4. Uhf

Question 61. A sky wave with a frequency of 55 MHz is incident on the d-region of Earth’s atmosphere at 45°. The angle of refraction is (electron density for d-region is 400 electron/cc.) –

1. 60°
2. 45°
3. 30°
4. 15°

Answer: 2. 45°

Question 62. Which of the following is not a transducer?

1. Loudspeaker
2. Amplifier
3. Microphone
4. Human ear

Answer: 2. Amplifier

Question 63. Am is used for broadcasting because:

1. It is more noise-immune than other modulation systems
2. It requires less transmitting power compared with other systems
3. Its use avoids receiver complexity
4. No other modulation system can provide the necessary bandwidth faithful transmission.

Answer: 3. Its use avoids receiver complexity

Question 64. If μ1 and μ2 are the refractive indices of the materials of the core and cladding of an optical fiber, then the loss of light due to its leakage can be minimized by having

1. μ1 > μ2
2. μ1 < μ2
3. μ1 = μ2
4. None of these

Answer: 1. μ1 > μ2

Question 65. An antenna behaves as a resonant circuit only when its length is

1. \(\frac{\lambda}{2}\)
2. \(\frac{\lambda}{4}\)
3. λ
4. \(\frac{\lambda}{2} \text { or integral multiple of } \frac{\lambda}{2}\)

Answer: 4. \(\frac{\lambda}{2} \text { or integral multiple of } \frac{\lambda}{2}\)

Question 66. The electromagnetic waves of frequency 2 MHz to 30 MHz are

1. In-ground wave propagation
2. In sky wave propagation
3. In microwave propagation
4. In satellite communication