Class 12 Physics Notes – Chapter 13: Nuclei

1. Atomic Masses and Composition of Nucleus

The nucleus is situated at the centre of the atom and contains protons and neutrons. The mass of an atom is very small, so it is expressed in atomic mass unit (u).

1 u = 1/12^th of the mass of a Carbon-12 (¹²C) atom.

1 u = 1.660539 × 10^-27 kg

Composition: A nucleus of an element X is represented as ^A_ZX.

• Z (Atomic Number): Number of protons.
• N (Neutron Number): Number of neutrons.
• A (Mass Number): Total number of nucleons (A = Z + N).

Isotopes: Nuclei with the same atomic number (Z) but different mass numbers (A). Example: ¹H, ²H, ³H.
Isobars: Nuclei with the same mass number (A) but different atomic numbers (Z). Example: ³H and ³He.
Isotones: Nuclei with the same number of neutrons (N). Example: ¹⁹⁸₈₀Hg and ¹⁹⁷₇₉Au.

2. Size of the Nucleus

By performing scattering experiments, it is found that the volume of a nucleus is directly proportional to its mass number (A). The radius (R) of a nucleus is given by:

R = R₀ A^1/3

Where R₀ is a constant ≈ 1.2 × 10^-15 m (or 1.2 fm).
Note: The density of nuclear matter is constant and independent of the mass number A. It is approximately 2.3 × 10¹⁷ kg/m³.

3. Mass-Energy and Nuclear Binding Energy

Einstein showed that mass and energy are equivalent and convertible into each other. E = mc².
The energy equivalent of 1 atomic mass unit (1 u) is 931.5 MeV.

Mass Defect (ΔM)

The total mass of a nucleus is always less than the sum of the masses of its constituent protons and neutrons. This difference in mass is called the mass defect.

ΔM = [Z m_p + (A – Z) m_n] – M

Binding Energy (E_b)

The energy required to separate a nucleus completely into its constituent protons and neutrons is called nuclear binding energy.

E_b = ΔM c²

Binding Energy per Nucleon (E_bn)

It is the average energy per nucleon needed to separate a nucleus into its individual nucleons. E_bn = E_b / A.
The value of E_bn is maximum (≈ 8.75 MeV) for Iron (⁵⁶Fe), meaning it is the most stable nucleus.

4. Nuclear Force

The strong attractive force between nucleons (protons and neutrons) inside the nucleus is called the nuclear force.

• It is the strongest force in nature.
• It is a short-range force (effective only up to a few femtometres).
• It is charge-independent (force between p-p, n-n, and p-n is approximately the same).
• It becomes repulsive if the distance between nucleons is less than 0.7 fm.

5. Radioactivity

Discovered by A.H. Becquerel, radioactivity is the spontaneous emission of radiation from an unstable nucleus. There are three main types of radioactive decay:

1. α-decay: A helium nucleus (⁴₂He) is emitted.
2. β-decay: Electrons or positrons are emitted.
3. γ-decay: High energy photons (hundreds of keV) are emitted.

6. Nuclear Energy (Fission and Fusion)

• Nuclear Fission: A heavy nucleus (like ²³⁵U) breaks into two smaller, intermediate-mass nuclei when bombarded by a neutron, releasing a large amount of energy.
• Nuclear Fusion: Two light nuclei (like Hydrogen) combine to form a heavier, more stable nucleus (like Helium), releasing an enormous amount of energy. Fusion is the source of energy in the Sun and stars.

Multiple Choice Questions (MCQs)

1. 1 atomic mass unit (1 u) is equivalent to an energy of:
a) 931.5 J
b) 931.5 MeV
c) 1.6 × 10^-19 J
d) 1 MeV
Answer: (b) 931.5 MeV

2. Nuclei having the same mass number (A) but different atomic numbers (Z) are called:
a) Isotopes
b) Isobars
c) Isotones
d) Isomers
Answer: (b) Isobars

3. The radius (R) of a nucleus is related to its mass number (A) by the formula:
a) R = R₀ A
b) R = R₀ A²
c) R = R₀ A^1/3
d) R = R₀ A^1/2
Answer: (c) R = R₀ A^1/3

4. The density of nuclear matter is:
a) Directly proportional to A
b) Inversely proportional to A
c) Independent of mass number A
d) Proportional to A^1/3
Answer: (c) Independent of mass number A

5. The binding energy per nucleon (E_bn) is maximum for which of the following nuclei?
a) ⁴He
b) ²³⁵U
c) ¹⁶O
d) ⁵⁶Fe
Answer: (d) ⁵⁶Fe

6. The nuclear force is:
a) Long-ranged and charge-dependent
b) Short-ranged and charge-dependent
c) Short-ranged and charge-independent
d) Long-ranged and charge-independent
Answer: (c) Short-ranged and charge-independent

7. The neutron was discovered by:
a) J.J. Thomson
b) James Chadwick
c) Ernest Rutherford
d) Niels Bohr
Answer: (b) James Chadwick

8. In nuclear fission, a heavy nucleus splits into two lighter nuclei. The energy is released because the binding energy per nucleon of the products is:
a) Less than that of the original nucleus
b) Greater than that of the original nucleus
c) Equal to that of the original nucleus
d) Zero
Answer: (b) Greater than that of the original nucleus

9. The source of energy in stars like the Sun is:
a) Nuclear fission
b) Nuclear fusion
c) Chemical combustion
d) Radioactivity
Answer: (b) Nuclear fusion

10. The mass defect (ΔM) of a nucleus is given by:
a) M – [Z m_p + (A – Z) m_n]
b) [Z m_p + (A – Z) m_n] – M
c) Z m_p + (A – Z) m_n + M
d) M – Z m_p
Answer: (b) [Z m_p + (A – Z) m_n] – M

Textbook Exercise Solutions

Q 13.1: Obtain the binding energy (in MeV) of a nitrogen nucleus (¹⁴₇N), given m (¹⁴₇N) = 14.00307 u.
Solution:
For ¹⁴₇N: Z = 7 protons, N = 14 – 7 = 7 neutrons.
Mass of 1 proton (m_p) = 1.007825 u (using mass of H-atom to include electrons).
Mass of 1 neutron (m_n) = 1.008665 u.
Mass defect ΔM = [7 m_p + 7 m_n] – m(¹⁴₇N)
ΔM = [7(1.007825) + 7(1.008665)] – 14.00307
ΔM = [7.054775 + 7.060655] – 14.00307 = 14.11543 – 14.00307 = 0.11236 u.
Binding Energy E_b = ΔM × 931.5 MeV
E_b = 0.11236 × 931.5 ≈ 104.7 MeV.

Q 13.2: Obtain the binding energy of the nuclei ⁵⁶₂₆Fe and ²⁰⁹₈₃Bi in units of MeV from the following data: m(⁵⁶₂₆Fe) = 55.934939 u, m(²⁰⁹₈₃Bi) = 208.980388 u.
Solution:
For ⁵⁶₂₆Fe: Z = 26, N = 30.
ΔM = [26(1.007825) + 30(1.008665)] – 55.934939
ΔM = [26.20345 + 30.25995] – 55.934939 = 56.4634 – 55.934939 = 0.528461 u.
Total BE = 0.528461 × 931.5 = 492.26 MeV.
BE per nucleon (E_bn) = 492.26 / 56 ≈ 8.79 MeV.

For ²⁰⁹₈₃Bi: Z = 83, N = 126.
ΔM = [83(1.007825) + 126(1.008665)] – 208.980388
ΔM = [83.649475 + 127.09179] – 208.980388 = 210.741265 – 208.980388 = 1.760877 u.
Total BE = 1.760877 × 931.5 = 1640.26 MeV.
BE per nucleon (E_bn) = 1640.26 / 209 ≈ 7.84 MeV.

Q 13.3: A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of ⁶³₂₉Cu atoms (of mass 62.92960 u).
Solution:
For ⁶³₂₉Cu: Z = 29, N = 34.
Mass defect of one atom Δm = [29(1.007825) + 34(1.008665)] – 62.92960
Δm = [29.226925 + 34.29461] – 62.92960 = 63.521535 – 62.92960 = 0.591935 u.
Binding energy per atom = 0.591935 × 931.5 = 551.38 MeV.
Number of atoms in 3.0 g coin = (Mass / Atomic Mass) × Avogadro’s Number
N = (3.0 / 63) × 6.023 × 10²³ ≈ 2.868 × 10²² atoms.
Total energy required = N × BE per atom = 2.868 × 10²² × 551.38 MeV ≈ 1.58 × 10²⁵ MeV.
(In Joules: 1.58 × 10²⁵ × 1.6 × 10^-13 J ≈ 2.53 × 10¹² J).

Q 13.4: Obtain approximately the ratio of the nuclear radii of the gold isotope ¹⁹⁷₇₉Au and the silver isotope ¹⁰⁷₄₇Ag.
Solution:
Nuclear radius R = R₀ A^1/3.
Ratio R_Au / R_Ag = (A_Au / A_Ag)^1/3 = (197 / 107)^1/3
R_Au / R_Ag = (1.841)^1/3 ≈ 1.23.

Q 13.5: The Q value of a nuclear reaction A + b → C + d is defined by Q = [ m_A + m_b – m_C – m_d ] c². Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i) ¹₁H + ³₁H → ²₁H + ²₁H
(ii) ¹²₆C + ¹²₆C → ²⁰₁₀Ne + ⁴₂He
(Given masses: ¹H = 1.007825 u, ²H = 2.014102 u, ³H = 3.016049 u, ¹²C = 12.000000 u, ²⁰Ne = 19.992439 u, ⁴He = 4.002603 u)
Solution:
(i) Q = [ m(¹H) + m(³H) – 2 m(²H) ] × 931.5 MeV
Q = [ 1.007825 + 3.016049 – 2(2.014102) ] × 931.5
Q = [ 4.023874 – 4.028204 ] × 931.5 = -0.00433 × 931.5 = -4.03 MeV.
Since Q is negative, the reaction is Endothermic.

(ii) Q = [ 2 m(¹²C) – m(²⁰Ne) – m(⁴He) ] × 931.5 MeV
Q = [ 2(12.000000) – (19.992439 + 4.002603) ] × 931.5
Q = [ 24.000000 – 23.995042 ] × 931.5 = 0.004958 × 931.5 = 4.62 MeV.
Since Q is positive, the reaction is Exothermic.