NEET Physics Class 12 Chapter 4 Nuclear Physics Notes

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:

\({ }_z X^A \rightarrow{ }_{Z-1} Y^A+{ }_{+1} e^0+v+Q\)

In B+ decay, inside a nucleus, a proton is converted into a neutron, positron, and neutrino.

\(\mathrm{p} \rightarrow \mathrm{n}+{ }_{+1} \mathrm{e}^0+\mathrm{v}\)

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 \(\mathrm{A}=-\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}\)

\(-\int_{N_0}^N \frac{d N}{N}=\int_0^t \lambda d t \Rightarrow \quad N=N_0 e^{-\lambda t} \quad \text { (it is exponential decay) }\)

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

\(=\frac{1.43 \times 10^{11}}{3.7 \times 10^{10}} \mathrm{Ci}=3.86 \mathrm{Ci} \text {. }\)

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 \(\mathrm{T}=\frac{5.7 \times 10^{-14} \mathrm{~J}}{1.5 \times 1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}}=2.8 \times 10^9 \mathrm{~K} .\)