NCERT Class 11 Chemistry Chapter 2 Structure Of Atom Long Question And Answers

NCERT Solutions For Class 11 Chemistry Chapter 2 Structure Of Atom Long Question And Answers

Question 1. The second line of the Lyman series of H-atom coincides with the sixth line of the Paschen series of an ionic species ‘X. Identify ‘X. (Suppose the value of the Rydberg constant, R is the same in both cases)
Answer:

For the second line of the Lyman series of H-atom,

⇒ \(\bar{v}=R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\)

For the sixth line of the Paschen series of the species ‘X with atomic number Z, v \(=R Z^2\left(\frac{1}{3^2}-\frac{1}{9^2}\right)\)

Since the Second Line Of Lyman Seriea Coincides With The Sixth Line Of the Paschen Series Of The Species X We Can equate

⇒ \(R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=R Z^2\left(\frac{1}{3^2}-\frac{1}{9^2}\right)\)

⇒ \(\frac{8}{9}=Z^2 \times \frac{8}{81} \quad \text { or, } Z^2=9\)

Z = 3

∴ The Ionic Spec ies Would Be Li²⁺

Question 2. An element of atomic weight Z consists of two isotopes of mass number (Z-1) and (Z + 2). Calculate the % of the higher isotope.
Answer:

Let the % of the higher isotope [mass number (Z + 2) ] be x.

Hence other isotope [mass number (Z- 1) ] will be (100- x)

Average atomic weight (Z) \(=\frac{x(Z+2)+(100-x)(Z-1)}{100}\)

100Z = Zx + 2x+ 100Z- 100- Zx + x

Or, 3x = 100

or, x= 33.3%

Question 3. Show that the sum of energies for the transition from n = 3 to n = 2 and from n = 2 to n = 1 is equals to the energy of transition from n = 3 to n = 1 in the case of an H-atom. Are wavelength and frequencies of the emitted spectrum also additive to their energies?
Answer:

⇒  \(\begin{aligned}& \Delta E_{3 \rightarrow 2}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\& \Delta E_{2 \rightarrow 1}=R_H\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\& \Delta E_{3 \rightarrow 1}=R_H\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\end{aligned}\)

From equation (1), (2) and (3) we have,

⇒ \(\Delta E_{3 \rightarrow 2}+\Delta E_{2 \rightarrow 1}=R_H\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=\Delta E_{3 \rightarrow 1}\)

⇒ \(\text { Thus, } \Delta E_{3 \rightarrow 1}=\Delta E_{3 \rightarrow 2}+\Delta E_{2 \rightarrow 1}\)

Since E – hv hence frequencies are also additive but \(E=\frac{h c}{\lambda}\) and thus wavelengths are not additive

Question 4. The Schrodinger wave equation for the 2s electron of a hydrogen atom is, \(\psi_{2 s}=\frac{1}{4 \sqrt{2 \pi}}\left[\frac{1}{a_0}\right]^{3 / 2} \times\left[2-\frac{r}{a_0}\right] \times e^{-r / 2 a_0}\) Node is defined as the point where the probability of finding an electron is zero.
Answer: 

∴ \(\text { If } r=r_0, \psi_{2 s}^2=0\)

∴ \(\frac{1}{32 \pi}\left(\frac{1}{a_0}\right)^2\left(2-\frac{r_0}{a_0}\right)^2 e^{-r_0 / 2 a_0}=0\)

The only factor that can be zero in the above expression is \(\left(2-\frac{r_0}{a_0}\right)\)

∴ \(2-\frac{r_0}{a_0}=0 ; \quad \text { or, } r_0=2 a_0 \text {. }\)

Question 5. If the uncertainty in the position of a moving electron is equal to its DC Broglie wavelength, prove that Its velocity is completely uncertain.
Answer: Uncertainty in the position of the electron, Ax = λ.

λ = \(\frac{h}{p}\) [From de Broglie equation]

∴ \(p=\frac{h}{\lambda}=\frac{h}{\Delta x} \quad \text { or, } \Delta x=\frac{h}{p}\)

According to Heisenberg’s uncertainty principle

∴ \(\Delta x \cdot \Delta p  \frac{h}{4 \pi} \quad \text { or, } \frac{h}{p} \cdot \Delta p\frac{h}{4 \pi} \quad \text { or, } \quad \frac{\Delta p}{p}\frac{1}{4 \pi}\)

or, \(\frac{m \Delta v}{m v}  \frac{1}{4 \pi} \quad \text { or, } \quad \Delta v  \frac{v}{4 \pi}\)

The uncertainty in velocity is so large that its velocity Is uncertain.

Question 6. The electron revolving In the n-th orbit of the Be³⁺ ion has the same speed as that of the electron in the ground state of the hydrogen atom. Find the value of n.
Answer:

The velocity of an electron in the n-th orbit of hydrogen-like species is given by, \(v_n=\frac{Z}{n} \times v_1\)

[where v₁ = velocity of the electron in the 1st orbit of H-atom i.e., the velocity of the electron in the ground state of H-atom, and Z = Atomic number of hydrogen-like species]

Now for \(\mathrm{Be}^{3+} \text {-ion } Z=4 \text {, so } v_n=\frac{4}{n} \times v_1\)

But it is given that, v_n = v₁

∴ \(v_1=\frac{4}{n} v_1 \quad \text { or } n=4\)

Question 7. The mass number of an ion with a unit negative charge is 37. The number of neutrons present in the ion is 10.6% more than that of electrons. Identify the ion.
Answer:

Let the number of protons in the ion = x. Therefore, the number of electrons =x + 1 (y the ion contains a unit negative charge). Thus, the number of neutrons =37-x.

Number of neutrons Number of electrons

= 37- x – (x + 1) = 36-2x

Percent of excess neutrons as compared to electrons =\(\frac{(36-2 x) \times 100}{x+1}\)

Given= \(\frac{(36-2 x) \times 100}{x+1}\)

= 10.6

Or = \(\frac{36-2 x}{x+1}=\frac{106}{1000}\)

or, x = 17.04=17 [v the number of protons present in an atom or an ion cannot be a fraction]

Hence it is a chloride ion (Cl)

Question 8. Mention Heisenberg’s uncertainty principle. Calculate the uncertainty of velocity of an electron which has an uncertainty in the position of 1Å
Answer:

According to the Heisenberg uncertainty principle,

⇒ \(\Delta x \times \Delta p\frac{h}{4 \pi} \quad \text { or } \Delta x \times m \Delta v \frac{h}{4 \pi}\)

The uncertainty in position, Δx = 1Å  = 10^-10m

Thus  Δv

= \(\frac{h}{\Delta x \times m \times 4 \pi}\)

= \(\frac{6.626 \times 10^{-34}}{10^{-10} \times 9.1 \times 10^{-31} \times 4 \pi}\)

= \(5.794 \times 10^5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Question 9. Calculate the number of electrons which will together weigh one gram. Calculate the mass and charge of one mole of electrons.
Answer:

Mass of one electron = 9.11× 10^-31kg.

∴ 1g or 10 3 kg = \(\frac{1}{9.11 \times 10^{-31}} \times 10^{-3}\) electrons 9.11 ×10^-31 = 1.098 × 10²⁷ elctrons.

Mass of one electron = 9.11× 10^-31kg.

So, mass of one mole of elctrons = (9.11 ×10^-31kg) x (6.022 ×10²³) = 5.485 × 10^-7 kg Charge on one electron = 1.602 × 10^-31C

Thus, charge on one mole of elctrons = (1.602 × 10^-19C) × (6.022 × 10²³) = 9.65 × 10⁴C.

Question 10.

1. Calculate the total number of electrons present in one mole of methane.
2. Find
   1. The total number and
   2. The total mass of neutrons in 7 mg of 14C. (Assume that the mass of a neutron = 1.675 × 10^-27kg).
3. Find
   1. The total number and
   2. The total mass of protons in 34 mg of NH₃ at STP.

Will the answer change if the temperature and pressure are changed?
Answer:

1. 1 molecule of CH₄ contains =6 + 4 = 10 electrons. Thus, one mole or 6.022 × 10²³ molecules will contain

= 6.022 × 10²³× 10 = 6.022 × 10²⁴ electrons.

2.

1. 1 mol ¹⁴C-atom = 6.022 × 10²³

¹⁴C atoms = 14g

¹⁴C .One ¹⁴C -atom contains =14-6 = 8 neutrons.

∴ 14g or 14000 mg 14C = 8 × 6.022 × 10²³ neutrons

2.  Mass of one neutron = 1.675 × 10^-27kg.

So, the mass of 2.4088× 10^-21 k neutrons

= (2.4088 × 10²¹) × (1.675 × 10^-27) = 4.0347 × 10^-6 kg

3.

1. 1 mol NH₃  ≡ l7g NH₄ ≡ 6.022 × 10²³ molecules of NH₃.

Therefore, the number of protons present in 17g

NH₃ = (6.022 × 10²³) × (7 + 3) = 6.022 × 10²⁴

Number of protons in 34 mg or 0.034 g mass 6.022× 10²⁴

NH’₃ – Hg× 0.034g = 1.2044 × 10²²

2. Mass of a proton = 1.6725 × 10^-27kg

Mass of 1.2044 × 10²² protons

= (1.6725× 10^-27) × (1.2044 × 10²²) =2.014 × 10^-5kg

There is no effect of temperature and pressure.

Question 11. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?
Answer:

Wavelength of light (A) = 4000pm

= 4000 × 10^-12m = 4 × 10^-9m

According to Planck’s quantum theory, the energy of a photon

⇒ \(E=N h v=N h \frac{c}{\lambda}\)

[N= no. of photons h= 6.626 × 10^-34 J-s , c = 3.0 ×10⁸ m-s^-1]

⇒  \(N=\frac{E \times \lambda}{h \times c}\)

= \(\frac{1 \mathrm{~J} \times\left(4 \times 10^{-9} \mathrm{~m}\right)}{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(3.0 \times 10^8\mathrm{~m} \cdot \mathrm{s}^{-1}\right)}\)

= \(2.012 \times 10^{16}\)

Question 12. A photon of wavelength 4 x 10-7m strikes on the metal surface, the work function of the metal is 2.13 eV. Calculate the energy of the photon (eV) the kinetic energy of the emission, and the velocity of the photoelectron (leV = 1.6020 × 10^-19J)
Answer:

Energy of a photon (E) = ,\(h v=\frac{h c}{\lambda}\)

=  \(\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right) \times\left(3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}{4 \times 10^{-7} \mathrm{~m}}\)

= \(4.97 \times 10^{-19} \mathrm{~J}\)

Or, energy of a photon \(=\frac{4.97 \times 10^{-19}}{1.602 \times 10^{-19}} \mathrm{eV}=3.10 \mathrm{eV}\)

The kinetic energy of emitted electron = energy of a photon- work function of a metal.

= (4.97 × 10^-19– 2.13 × 1.602 × 10^-19)J

= 1.56 × 10^-19

J = 0.97eV

Kinetic energy of photoelctron, \(\frac{1}{2} m v^2=1.56 \times 10^{-19} \mathrm{~J}\)

⇒ \(\text { or, } v^2=\frac{2 \times 1.56 \times 10^{-19}}{m}=\frac{2 \times 1.56 \times 10^{-19}}{9.108 \times 10^{-31}}=0.34 \times 10^{12}\)

Velocity of photoelectron, v = 5.83 × 10^-5m.s^-1

Question 13. A 25-watt bulb emits monochromatic yellow light of wavelength of 0.57 pm. Calculate the rate of emission of quanta per second.
Answer:

Energy emitted by the bulb = 25 watts = 25 J-s^-1

Energy ofa photon (E) = hv \(h \frac{c}{\lambda}\)

λ = 0.57μm

c= \(0.57 \times 10^{-6} \mathrm{~m}\)

c = \(3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

h = \(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\)

Substituting the values we get,

E= \(\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}{0.57 \times 10^{-6} \mathrm{~m}}=3.49 \times 10^{-19} \mathrm{~J}\)

No. of quanta emitted \(=\frac{25 \mathrm{~J} \cdot \mathrm{s}^{-1}}{3.49 \times 10^{-19} \mathrm{~J}}=7.16 \times 10^{19} \mathrm{~s}^{-1}\)

Question 14. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes a transition from an energy level with n = 4 to an energy level with n = 2?
Answer:

According to the Rydberg equation, wave number,

⇒ \(\bar{v}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

⇒  \(\bar{v}=109677\left[\frac{1}{(2)^2}-\frac{1}{(4)^2}\right] \mathrm{cm}^{-1}\)

[R = 109677 cm^-1 and n₂ = 4, n₁ = 2]cm^-1

Or, v = 109677 × \(\frac{3}{16}\) = 20564.43cm

λ = \(\frac{1}{\bar{\nu}}=\frac{1}{20564.43} \mathrm{~cm}\)

= 4.86 × 10^-5cm = 486 × 10^-7crn

= 486 × 10^-9nm

= 486nm.

Question 15. How much energy is needed to ionize an H-atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of the H-atom (energy required to remove the electron from n = 1 orbit).
Answer:

⇒  \(E_n=-\frac{21.76 \times 10^{-19}}{n^2} \mathrm{~J}\)

⇒\(E_1=-21.76 \times 10^{-19} \mathrm{~J}\)

⇒ \(E_5=-\frac{21.76 \times 10^{-19}}{(5)^2}\)

= \(-8.704 \times 10^{-20} \mathrm{~J} \text { and } E_{\infty}=0\)

⇒  \(\frac{\Delta E}{\Delta E^{\prime}}=\frac{8.704 \times 10^{-20} \mathrm{~J}}{21.76 \times 10^{-19} \mathrm{~J}}=4 \times 10^{-2}\)

Question 16. The energy associated with the first orbit in the hydrogen atom is -2.18 ×  10^-18J. atom^-1. What is the energy associated with the fifth orbit? Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.
Answer:

Energy of the first orbit (E1) = -2.18 ×  10^-18J-atom^-1

We know that the energy of n-th orbit (En) \(=E_1 \times \frac{Z^2}{n^2}\)

=\(-2.18 \times 10^{-18} \times \frac{1^2}{n^2}\)

For H-atom, Z= 1

Thus energy of 5th Orbit (e5) \(=-2.18 \times 10^{-18} \times \frac{1^2}{5^2}\)

= -8.72 × 10^-20 J atom

Bohr radius for n-th-orbit of H-atom (rn) = 0.529 × n²A

[For H-atom, Z = 1]

∴ Bohr radius of 5th orbit, r₅ = 0.529 × 5² Å

= 13.225Å

Question 17. Find the wave number for the longest wavelength transition In the Balmer series of atomic hydrogen.
Answer:

The equation used for explaining the line spectrum of hydrogen for \(\bar{v}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) Balmer series, n₁ = 2

As \(\bar{v}=\frac{1}{\lambda}, \lambda\) will be longest if n2 is minimum.

Thus, here n₂ = n₁ + 1

= 2+1

= 3 (R = 109677 cm^-1 )

So. \(\vec{v}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\)

Or, \(\bar{v}=109677\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \mathrm{cm}^{-1}\)

= \(109677 \times \frac{5}{36} \mathrm{~cm}^{-1}=1.5233 \times 10^4 \mathrm{~cm}^{-1}\)

Question 18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18 × 5^-11 erg.
Answer:

Given: energy of ground state (Ex) = -2.18 × 10¹ erg

Energy of n-th orbit, \(E_n=\frac{E_1}{n^2} \operatorname{erg}=\frac{-2.18 \times 10^{-11}}{n^2} \mathrm{erg}\)

The amount of energy required when an electron jumps from the last orbit to 5th orbit.

⇒ \(\Delta E=E_5-E_1=\frac{-2.18 \times 10^{-11}}{5^2}-\left(-2.18 \times 10^{-11}\right)\)

= \(2.18 \times 10^{-11}\left(1-\frac{1}{25}\right)\)

= 2.093 × 10¹¹ erg

=  2.093 × 10^-18 J

[1 erg = 10^-7 J ]

The amount of energy released when the electron returns from the 5th orbit to the list-orbit =2.093 × 10¹¹ erg

We know that, AE = hv \(=h \frac{c}{\lambda}\)

⇒ \(\frac{h c}{\Delta E}=\frac{\left(6.626 \times 10^{-27} \mathrm{erg} \cdot \mathrm{s}\right) \times\left(3 \times 10^{10} \mathrm{~cm} \cdot \mathrm{s}^{-1}\right)}{2.093 \times 10^{-11} \mathrm{erg}}\)

= 9.497 × 10^-6cm

=950 × 10^-8cm

= 950 Å

Question 19. The electron energy in the hydrogen atom is given by E_n = (-2.18 × 10^-18)/n²J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:

E₂ = \(-\frac{2.18 \times 10^{-18}}{(2)^2} \mathrm{~J}=-5.45 \times 10^{-19} \mathrm{~J}\)

The energy required to remove an electron completely from the orbit with

⇒ n \(n=2, \Delta E=E_{\infty 0}-E_2=0-\left(-5.45 \times 10^{-19} \mathrm{~J}\right)\)

= 5.45 × 10^-19J

⇒ \(=\frac{h c}{\Delta E}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right) \times\left(3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}{5.45 \times 10^{-19} \mathrm{~J}}\)

= \(3.647 \times 10^{-7} \mathrm{~m}=3.647 \times 10^{-5} \mathrm{~cm}\)

Question 20. Calculate the wavelength of an electron moving with a velocity of 2.05× 10^-7 m.s^-1
Answer:

The velocity of an electron ( v) = 2.05 × 10^-17 S^-1 According to de Broglie equation,

Wavelength \((\lambda)=\frac{h}{m v}\)

= \(\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{\left(9.108 \times 10^{-31} \mathrm{~kg}\right) \times\left(2.05 \times 10^7 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}\)

Thus, the wavelength of the electron will be 3.549 × 10^-11 m

Question 21. The mass of an electron is 9.1 x 10-31kg. If its K.E is 3.0 × 10^-25J, calculate its wavelength.
Answer:

Mass of an electron (m) = 9.1 × 10^-31 kg

⇒ \(\mathrm{KE}=\frac{1}{2} m v^2=3.0 \times 10^{-25} \mathrm{~J}\)

Velocity of an electron \((v)=\sqrt{\frac{2 \times \mathrm{KE}}{m}}=\sqrt{\frac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31} \mathrm{~kg}}}\)

= 812 m.s^-1

Wavelength of the moving electron \((\lambda)=\frac{h}{m v}\)

= \(\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(812 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}=8.967 \times 10^{-7} \mathrm{In}\)

Question 22.

1. Write the electronic configurations of the following ions: 

1. H^–
2. Xa^–
3. O^2-
4. F^–

2. What are the atomic numbers of elements whose outermost electrons are represented by 

1. 3s¹
2.  2p³
3. 3p⁵

3. Which atoms are indicated by the following configurations?

1. [He]2s¹
2. [Ne]3s²3p³
3. [Ar]4s²3d¹

Answer: 

1. 

1. H^–: Is²
2. Na⁺ : ls²2s²2p⁵
3. O^2--: ls²s²p⁶
4. F^–: ls²2s²2p⁶

2.

1. The configuration ofthe elementwill be ls²2s²2p⁶3s¹; its atomic numberwill be 11.
2. The configuration of the element will be 1s²2s²2p³; its atomic number will be 7.
3. The configuration of the element will be ls²2s²2p⁶3s²3p⁵; its atomic number will be 17.

3.

1. Lithium (Li), (Z = 3)
2. Phosphorous (P), (Z =15)
3. Scandium (Sc), (Z= 21).

Question 23. An atomic orbital has n = 3. What are the possible values of 1 and m1? List the quantum numbers (m{ and 1) of electrons for 3d orbital. Which of the following orbitals are possible? 1 p, 2s. 2p and 3f
Answer:

1. If n = 3 , then 1 = 0, 1, 2

When I = 0, m₁ = 0; when l = 1 , m₁ = -1, 0, +1 ;

When l = 2, m₁= -2, -1, 0, +1, +2

2. For 3d-subshell, n = 3 and 1 = 2; for l = 2, m₁= -2,-1,+1,+2

3. lp is not possible because for p-subshe,l=1. When n = 1, l cannot be 1. 2s is possible because for subshell, 1 = 0, when n = 2, 1 can be 0. 2p is possible because for p-subshell,l =1, when n = 2,l can be 1. 3f is not possible because for f-subshell, l = 3. When n = 3,   l cannot be 3.

Question 24. Calculate the energy required for the process \(\mathrm{He}^{+}(g) \rightarrow \mathrm{He}^{2+}(g)+e\) The ionization energy of the H-atom in the ground state is 2.18 ×10^-18J.atom^-1
Answer:

The energy of an electron residing in the nth-orbit of a hydrogen-like atom/ion is En \(=-\frac{2 \pi^2 m Z^2 e^4}{n^2 h^2}\)

Ionization enthalpy for h- atom = E∞ – E₁

= 0\(-\left[-\frac{2 \pi^2 m e^4 \times 1^2}{1^2 \times h^2}\right]\)

Or, \(\frac{2 \pi^2 m e^4}{h^2}=2.18 \times 10^{-18} \mathrm{~J}\)

IE = 2. 18 ×  10^-18J

Again, ionisation enthalpy of He+ ion = E∞ – E₁

= 0 \(-\left[-\frac{2 \pi^2 m e^4 \times 2^2}{1^2 \times h^2}\right]\)

Since Z = 2

= \(4 \times \frac{2 \pi^2 m e^4}{h^2}=4 \times 2.18 \times 10^{-18} \mathrm{~J}=8.72 \times 10^{-18} \mathrm{~J}\)

∴ Energy required for the process = 8.72 ×  10^-18J

Question 25. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each radiation and energy difference between two excited states.
Answer:

λ₁ = 589nm = 589× 10^-9m

∴ Frequency, v1 \(=\frac{c}{\lambda_1}=\frac{3.0 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{589 \times 10^{-9} \mathrm{~m}}\)

λ₂ = 589.6nm = 589.6 × 10^-9 m

∴ Frequency (v2) \(=\frac{c}{\lambda_2}=\frac{3.0 \times 10^8}{589.6 \times 10^{-9}}=5.088 \times 10^{14} \mathrm{~s}^{-1}\)

The difference in energy (AE) = E₁-E₂ = h(v₁-v₂)

= 6.626 ×  10^-34× (5.093- 5.088) × 10¹⁴

= 3.313 ×  10^-22J

Question 26. The ejection of the photoelectron from the silver metal In the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Answer:

The energy of incident radiation, E = hv = work function of a metal + Kinetic energy of photoelectrons.

or, \(B=h v=h \frac{c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right) \times\left(3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}{\left(256.7 \times 10^{-9} \mathrm{~m}\right)}\)

Or, E = 7.74 ×  10^-19J

= 4.03eV

Since leV = 1.602 ×  10^-19J)

The potential applied provides the Kinetic energy to the electron. Thus, the kinetic energy of the electron =0.35eV. So, the work function of silver metal = (4.83- 0.35)eV

= 4.48eV.

Question 27. Emission transitions in the Paschcn series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 ×  10^-15(Hz) [1/3²-1/n²]. Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Answer:

⇒ \(v=\frac{c}{\lambda}=\frac{3.0 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{1285 \times 10^{-9} \mathrm{~m}}=3.29 \times 10^{15}\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\)

⇒ \(2.33 \times 10^{14}=3.29 \times 10^{15}\left(\frac{1}{9}-\frac{1}{n^2}\right)\)

Or = \(\quad \frac{2.33 \times 10^{14}}{3.29 \times 10^{15}}=\frac{1}{9}-\frac{1}{n^2} \text { or, } 0.071=\frac{1}{9}-\frac{1}{n^2}\)

⇒ \(\text { or, } \frac{1}{n^2}=\frac{1}{9}-0.071 \text { or, } \frac{1}{n^2}=0.040 \text { or, } n^2=25\)

or, n – 5. Therefore, radiation corresponding to 1285 nm belongs to the infrared region.

Question 28. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Answer:

No. of protons + no. of neutrons = mass number = 81

Let, the number of protons present in an atom = x

Number of neutrons present \(=x+\frac{31.7}{100} \times x\)

As given in the question, x + 1.317x = 81

or, 2.317x = 81 or, x = 34.96

= 35

Thus, number of protons = 35 l.e., atomic number = 35

The element with atomic number 35 is Br

∴ The Symbol is \({ }_{35}^{81} \mathrm{Br} \text {. }\)

Question 29. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate the dying frequency of emission, distance traveled by this radiation in the 30s energy of quantum, and the number of quanta present if it produces 21 of energy.
Answer:

A = 616nm = 616 ×  10^-19 m, c = 3 ×  10^-8m.s^-1

Radiation Frequency

⇒ \(v=\frac{c}{\lambda}=\frac{3 \times 10^8}{616 \times 10^{-9}}=4.87 \times 10^{14} \mathrm{~s}^{-1}\)

Distance (s) traveled by the radiation in 30 s = ext =3 x 108m-s-1 x 30 s = 9 ×  10⁹m.

Energy of quantum

E = hv = 6.626 × 10^-34 × 4.87 ×  10¹⁴

= 3.23 × 10^-19J.

Number of quanta = \(\frac{\text { total energy }}{\text { energy of each quantum }}\)

= \(\frac{2 \mathrm{~J}}{3.23 \times 10^{-19} \mathrm{~J}}=6.19 \times 10^{18}\)

Question 30. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 ×  10^-18J from the radiations of 600 nm, calculate the number of photons received by the detector.
Answer:

Wavelength λ= 600nm = 600 ×  10^-9m.

Energy of phyton \(=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{600 \times 10^{-9}}\)

= 3.313 ×  10^-19 J.

Number of photons detected by the detector

= \(\frac{\text { total energy received by the detector }}{\text { energy of each photon }}\)

= \(\frac{3.15 \times 10^{-18} \mathrm{~J}}{3.313 \times 10^{-19} \mathrm{~J}}=9.15 \approx 10 \text { photons. }\)

Question 31. Lifetimes of the molecules in the excited states are often measured by using a pulsed radiantly source of duration nearly in the nano-second range. If the radiation source has a duration of 2 ns and the number of photons emitted during the pulse source is 2.5×  10¹⁵, calculate the energy of the source.
Answer:

Frequency emission (v)=  \(=\frac{1}{\text { time period }}=\frac{1}{2 \times 10^{-9} \mathrm{~s}}\)

= 0.5×  10⁹s^-1

The energy of emission =Nhv

= 2.5 × 10¹⁵ × 6.626 ×10^-34 × 0.5 × 10⁹

= 8.2825 × 10¹⁰ J.

Question 32. The work function for the cesium atom is 1.9 eV. Calculate

1. The threshold wavelength and
2. The threshold frequency of the radiation. If the cesium element is irradiated with a wavelength of 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer:

The work function of cesium ( w₀) = h₀= 1.9eV

∴ Threshold frequency, v0 \(=\frac{1.9 \mathrm{eV}}{h}=\frac{1.9 \times 1.602 \times 10^{-19}}{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}\)

Or, v₀= 4.59 ×  10¹⁴ s^-1

[since leV = 1.602 × 10^-19J]

Again, threshold wavelength (A0) = c/v₀

= \(\frac{3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{4.59 \times 10^{14} \mathrm{~s}^{-1}}\)

= \(6.536 \times 10^{-7} \mathrm{~m}\)

1. Thus threshold wavelength of cesium = 6.536 × 10^-7m

= 654 × 10^-9m = 654nm

2.Threshold frequency = 4.59 × 10¹⁴s^-1Kinetic energy (KE) of the emitted electron

⇒ \(\frac{1}{2} m v^2=h v-w_0=\left(h v-h v_0\right)=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\)

Or, Kinetic energy (KE)

= \(\left(6.626 \times 10^{-34}\right)\left(3 \times 10^8\right)\left(\frac{1}{500 \times 10^{-9} \mathrm{~m}}-\frac{1}{654 \times 10^{-9} \mathrm{~m}}\right)\)

Or, KE = 9.36 × 10^-20J; So, the Kinetic energy of the emitted electron is 9.36 × 10^-20J.

Or, \(\frac{1}{2} m v^2=9.36 \times 10^{-20}\)

Or, \(v^2=\frac{9.36 \times 10^{-20} \times 2}{9.108 \times 10^{-31}} \text { or, } v=4.53 \times 10^5\)

Hence, the velocity of the emitted electron is 4.53 × 10⁵m.s^-1

Question 33. The following results are observed when sodium metal is irradiated with different wavelengths. Calculate

1. Threshold wavelength and
2. Planck’s constant.
   

λ(nm) – v × 10^-5(cm.s^-1)

1. 500 – 2.55
2. 450 – 4.35
3. 400- 5.35

Answer:

1. Let threshold wavelength = λ0nm = λ₀ × 10^-9m

Again, the Kinetic energy of the emitted electron.

⇒ \(\left(\frac{1}{2} m v^2\right)=h\left(v-v_0\right)=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\) …………………(1)

Putting the given values in (1) we get

⇒ \(\frac{1}{2} m\left(2.55 \times 10^5\right)^2=\frac{h c}{10^{-9}}\left(\frac{1}{500}-\frac{1}{\lambda_0}\right)\) …………………(2)

⇒ \(\frac{1}{2} m\left(4.35 \times 10^5\right)^2=\frac{h c}{10^{-9}}\left(\frac{1}{450}-\frac{1}{\lambda_0}\right)\)…………………(3)

⇒ \(\frac{1}{2} m\left(5.35 \times 10^5\right)^2=\frac{h c}{10^{-9}}\left(\frac{1}{400}-\frac{1}{\lambda_0}\right)\) …………………(4)

Dividing (3) by (2) we get \(\frac{\lambda_0-450}{450 \lambda_0} \times \frac{500 \lambda_0}{\lambda_0-500}=\left(\frac{4.35}{2.55}\right)^2\)

Or, λ₀ = 530.88=531 nm

2. Substituting the value of λ0 in (4) we have

⇒ \(\frac{1}{2} \times\left(9.108 \times 10^{-31}\right) \times\left(5.35 \times 10^5\right)^2\)

=\(\frac{h \times 3 \times 10^8}{10^{-9}}\left(\frac{1}{400}-\frac{1}{531}\right) \quad \text { or, } h=7.045 \times 10^{-34}\)

Value of Planck’s constant obtained = 7.045 ×  10^-34 J.s

Question 34. If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5× 10⁷ m.s^–1, calculate the energy with which it is bound to the nucleus.
Answer:

The energy of an incident photon

⇒\(\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right) \times\left(3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}{\left(150 \times 10^{-12} \mathrm{~m}\right)}\)

= 1.3252 × 10^-15 J

= 13.252 × 10^-16 J

Kinetic energy of emitted  electron \(\left(\frac{1}{2} m v^2\right)\)

= \(\frac{1}{2} \times\left(9.108 \times 10^{-31} \mathrm{~kg}\right) \times\left(1.5 \times 10^7 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)^2\)

= 1.025× 10^-16 J

So, the energy with which the electron was bound to the

nucleus =(13.252 × 10^-16 – 1.025 × 10^-16 )J

= 12.227 × 10^-16 J = 7.632 ¹ 10³eV

Question 35. Calculate the wavelength for the emission transition if it starts from the orbit having a radius of 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:

For a 1 electron system, the radius of the n-th orbit \(=\frac{52.9 n^2}{Z}\) pm

The radius of the orbit from which the transition of the electron occurs

= 1.3225nm = 1322.5pm \(=\frac{52.9 n_1^2}{Z}\)

The radius ofthe orbit to which the electron is added.

⇒ \(r_2=211.6 \mathrm{pm}=\frac{52.9 n_2^2}{Z}\)

⇒  \(\text { So, } \frac{r_1}{r_2}=\frac{1322.5}{211.6}=\frac{n_1^2}{n_2^2} \text { or, } \frac{n_1}{n_2}=2.5\)

When n₁ = 5 and n₂ = 2, the equation obtained for n1 and n2 is satisfied. Thus, the transition occurs from n = 5 to n = 2 and belongs to the Balmer series.

∴ Wave number (v) = 109677 \(\times\left(\frac{1}{2^2}-\frac{1}{5^2}\right)\)

= 2.3 × 10⁴cm^-1. and wavelength \((\lambda)=\frac{1}{\bar{v}}=\frac{1}{2.3 \times 10^4} \mathrm{~cm}\)

= 4.35 × 10^-5cm

= 435nm

Thus, it lies in the visible region.

Question 36. If the velocity of the electron in Bohr’s first orbit is 2.19 ×  10⁶m.s^-1, calculate the de Broglie wavelength associated with it.
Answer:

v = \(2.19 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

m = \(9.108 \times 10^{-31} \mathrm{~kg}\)

λ = \(\frac{h}{m u}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{\left(9.108 \times 10^{-31} \mathrm{~kg}\right) \times\left(2.19 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}\)

= \(3.32 \times 10^{-10} \mathrm{~m}=332 \times 10^{-12} \mathrm{~m}\)

= 332pm

Question 37. The velocity associated with a proton moving in a j potential difference of 1000 V is 4.37 × 10^-3m.s^-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Answer:

Velocity of hockey ball =4.37 × 105m.s^-1 , mass = 0.1kg

∴ Wavelength (λ) = \(\frac{h}{m v}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{0.1 \mathrm{~kg} \times 4.37 \times 10^5 \mathrm{~m} \cdot \mathrm{s}^{-1}}\)

= \(1.52 \times 10^{-38} \mathrm{~m}\)

Question 38. If the position of the electron is measured within an accuracy of ± 0.002nm, calculate the uncertainty In the momentum of the electron. Suppose the momentum of the electron Is h/(4xm × 0.05)nm, is there any problem in defining this value?
Answer:

Given, Ax = 0.002nm = 2 x 10-3nm = 2 x 10-12m According to Heisenberg’s uncertainty principle.

⇒ \(\Delta x \times \Delta p=\frac{h}{4 \pi} \quad \text { or, } \Delta p=\frac{h}{4 \pi \Delta x}\)

= \(\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 2 \times 10^{-12}}=2.638 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}\)

Again Momentum of the electron

= \(=\frac{1}{4 \pi \times 0.05 \mathrm{~nm}}=\frac{n}{4 \pi \times 5 \times 10^{-11} \mathrm{~m}} \)

= \(\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{4 \times 3.14 \times 5 \times 10^{-11} \mathrm{~m}}\)

= \(1.055 \times 10^{-24} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}\)

The value of uncertainty cannot be greater than the actual momentum. Thus the actual magnitude of momentum cannot m be defined in as reality

Question 39. The quantum numbers of six electrons are given below. Arrange them in order of increasing energy. list if any of this combinationÿ) has/ have the same energy.

• n = 4, l = 2, m₁=-2 m_s =-1/2
• n = 4, l = 2, m₁=1 m_s =+1/2
• n = 4, l = 2, m₁=0 m_s =+1/2
• n = 4, l =2, m₁=-2 m_s =-1/2
• n = 4, l = 2, m₁=-2 m_s =+1/2
• n = 4, l = 2, m₁=-2 m_s =+1/2

Answer: The orbital occupied by the electrons that are designated by the given sets of quantum numbers are,

1. 4d
2. 3d
3. 4p
4. 3d
5. 3p
6. 4p

So, increasing the order of their energies will be:

5<2<4<6=3<1

Question 40. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of the He+ spectrum?
Answer:

For Balmer transition, n = 4 to n = 2, for He+ spectrum, the Rydberg equation is,

⇒ \(\bar{v}=\frac{1}{\lambda}=R Z^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R \times 2^2 \times \frac{3}{16}=\frac{3 R}{4}\)

∴  He, Z=2

For Hydrogen spectrum

⇒ \(\bar{v}=\frac{1}{\lambda}=R Z^2\left(\frac{1}{n^2}-\frac{1}{n_2^2}\right)=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

According To The Question,

⇒ \(R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{3 R}{4}\)

or, \(\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{3}{4}\)

The equation (1) will be true if n = 1 and n -2.

For hydrogen, the spectrum transition is from n = 2 to n = 1.

Question 41. Which of the following subshells have no real existence?

1. 2d
2. 3f
3. 4g
4. 5d

Answer:

In the case of d -subshell, l = 2. In the second shell ( n = 2), the values of l are 0 and 1. So there cannot be any d -d-subshell In the third shell. Therefore, we can say that there Is no real existence of a 2d subshell.

In the case of f-subshell, l = 3. In the third shell (n = 3), the values of l are 0, 1, and 2. Since there cannot be any f-subshell In this shell, there is no real existence of 3f- subshell.

ln case of g -subshell, l = 4. In the fourth shell (n = 4), the values of l are 0, 1, 2, and 3. Since there cannot be any g subshell in this shell, there is no real existence of a 4g subshell.

In the case of d -subshell, l = 2. In the fifth shell (n = 5), the values of l are 0, 1, 2, 3 and 4. Therefore 5d sub-shell has real existence.

Question 42. An electron is described by magnetic quantum no.m = +3. Indicate the lowest possible value of ‘n ’for this electron. (tv) n = 4,1=0
Answer:

For an electron having magnetic quantum number m = +3, the lowest possible value of azimuthal quantum number ‘ l’ would be 3, and for an electron having 1 = 3, the lowest possible value for ‘ n’ would be 4.

In other words, the lowest possible energy level (‘n’) that the electron would occupy is the 4th shell (n = 4).

For an electron having magnetic quantum number m = +3, the lowest possible value of azimuthal quantum number ‘ l’ would be 3, and for an electron having 1 = 3, the lowest possible value for ‘ n’ would be 4.

In other words, the lowest possible energy level (‘n’) that the electron would occupy is the 4th shell (n = 4).