Nuclear Physics – Class 12 Physics | JEE Main Notes, Formulas & Numericals

CLASS 12 PHYSICS JEE MAIN

⚛️ NUCLEAR PHYSICS

Complete Notes | Important Formulas | Solved Numericals | MCQs

Understand the nucleus, master the formulas, and practise for your exam!

📚 Table of Contents

1. Introduction to Nuclear Physics
2. Composition and Size of the Nucleus
3. Mass Defect and Binding Energy
4. Nuclear Force
5. Radioactivity and Decay Law
6. Alpha, Beta and Gamma Decay
7. Nuclear Fission and Fusion
8. Important Formula Sheet
9. Solved Numerical Problems
10. JEE Main Practice Questions
11. Quick Revision and Answer Key

1. Introduction to Nuclear Physics

Nuclear physics is the branch of physics that studies the atomic nucleus, its structure, stability, radioactive decay, and nuclear reactions.

What is an atomic nucleus?

The nucleus is the small, dense central part of an atom. It contains positively charged protons and electrically neutral neutrons. Almost all the mass of an atom is concentrated in its nucleus.

A = Z + N

• Z: Atomic number = number of protons
• N: Number of neutrons
• A: Mass number = total number of protons and neutrons

Example: Carbon-12

Carbon-12 has atomic number 6 and mass number 12.

N = A − Z = 12 − 6 = 6

Therefore, carbon-12 contains 6 protons and 6 neutrons.

Important nuclear terms

Term	Meaning	Example
Isotopes	Same Z, different A	¹H, ²H, ³H
Isobars	Same A, different Z	¹⁴C and ¹⁴N
Isotones	Same number of neutrons	¹⁴C and ¹⁵N

2. Nuclear Size and Density

The radius of a nucleus depends on its mass number.

R = R₀ A^1/3

Where:

• R = nuclear radius
• R₀ ≈ 1.2 × 10⁻¹⁵ m
• A = mass number

JEE Main shortcut:

R ∝ A^1/3

If the mass number increases by a factor of 8, the radius increases by a factor of 2.

Nuclear density

The nuclear density is approximately independent of the mass number because nuclear mass is proportional to A, while nuclear volume is also proportional to A.

ρ = Mass / Volume

Using nuclear mass approximately equal to A mₙ and radius R₀A^1/3:

ρ ≈ 3mₙ / (4πR₀³)

Taking mₙ ≈ 1.67 × 10⁻²⁷ kg and R₀ = 1.2 × 10⁻¹⁵ m gives a density of approximately 2.3 × 10¹⁷ kg/m³.

3. Mass Defect and Binding Energy

The total mass of separate protons and neutrons is greater than the mass of the bound nucleus. The difference is called the mass defect.

Δm = Zmₚ + Nmₙ − M_nucleus

Here:

• mₚ = mass of a proton
• mₙ = mass of a neutron
• M = actual mass of the nucleus

### Binding energy Binding energy is the energy required to separate a nucleus completely into its individual protons and neutrons. According to Einstein's mass-energy relation:

E = Δmc²

When mass defect is measured in atomic mass units (u):

BE (MeV) = Δm (u) × 931.5

### Using atomic masses When the mass of a neutral atom is provided, it is often convenient to use hydrogen-atom mass rather than proton mass:

Δm = Zm_H + Nmₙ − M_atom

This expression accounts for the electrons on both sides, assuming the electron binding-energy correction is negligible. ### Binding energy per nucleon

BE per nucleon = Total BE / A

A larger binding energy per nucleon generally indicates a more tightly bound nucleus. The curve peaks around the iron/nickel region, which helps explain the energy released in both fission of very heavy nuclei and fusion of light nuclei.

Exam tip: Do not confuse total binding energy with binding energy per nucleon. Always check whether the question asks for the energy of the whole nucleus or energy per nucleon.

4. Nuclear Force

The nuclear force binds protons and neutrons inside the nucleus. Its main properties are:

• It is very strong compared with the electromagnetic force at nuclear distances.
• It is short-ranged, acting significantly over distances of a few femtometres.
• It is approximately charge-independent for proton-proton, neutron-neutron and proton-neutron interactions after accounting for electromagnetic effects.
• It shows saturation: each nucleon interacts strongly with nearby nucleons rather than equally with every nucleon in the nucleus.

5. Radioactivity and Nuclear Decay

Radioactivity is the spontaneous transformation of an unstable nucleus into a more stable configuration, accompanied by the emission of particles or radiation. The rate of radioactive decay is proportional to the number of undecayed nuclei present.

dN/dt = −λN

The solution is:

N = N₀e^−λt

Where:

• N₀ = initial number of undecayed nuclei
• N = number remaining after time t
• λ = decay constant
• t = elapsed time

### Half-life Half-life is the time required for half of the radioactive nuclei to decay.

T½ = 0.693 / λ

After n half-lives:

N = N₀ / 2ⁿ

### Mean life

τ = 1 / λ = T½ / 0.693

### Activity Activity is the number of decays per second.

Activity = −dN/dt = λN

The SI unit of activity is the becquerel (Bq).

Remember: After 1 half-life, 50% remains; after 2, 25%; after 3, 12.5%; after 4, 6.25%.

6. Alpha, Beta and Gamma Decay

Radiation	Nature	Change in A	Change in Z
Alpha (α)	Helium nucleus, ⁴₂He	−4	−2
Beta minus (β⁻)	Electron emission	0	+1
Beta plus (β⁺)	Positron emission	0	−1
Gamma (γ)	High-energy photon	0	0

### Alpha decay Example:

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He

The mass number decreases by 4 and atomic number decreases by 2. ### Beta-minus decay A neutron changes into a proton, an electron and an electron antineutrino.

n → p + e⁻ + ν̄ₑ

In the nucleus, A remains unchanged and Z increases by 1. ### Gamma decay An excited nucleus emits a gamma photon and moves to a lower energy state. Its mass number and atomic number remain unchanged.

7. Nuclear Fission and Fusion

☢️ Nuclear Fission

A heavy nucleus splits into lighter nuclei, releasing energy and usually some neutrons. Example (one possible channel):

²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n + Energy

Fission is used in nuclear reactors. The exact products can vary between reaction channels.

☀️ Nuclear Fusion

Two light nuclei combine to form a heavier nucleus, releasing energy when the reaction products are more tightly bound. Example:

²H + ³H → ⁴He + n + 17.6 MeV

Fusion powers the Sun and other stars through nuclear reactions.

Fission	Fusion
Heavy nucleus splits	Light nuclei combine
Can sustain a chain reaction	Requires suitable high-temperature and confinement conditions
Used in nuclear power reactors	Occurs naturally in stars

8. Important Formula Sheet for JEE Main

NUCLEAR RADIUS

R = R₀A^1/3

MASS DEFECT

Δm = Zmₚ + Nmₙ − M_nucleus

BINDING ENERGY

BE = Δmc²

1 u × c² ≈ 931.5 MeV

DECAY LAW

N = N₀e^−λt

HALF-LIFE

T½ = 0.693/λ

MEAN LIFE

τ = 1/λ

ACTIVITY

A = λN

REMAINING MASS

m = m₀(1/2)^t/T½

REACTION ENERGY

Q = (m_initial − m_final)c²

Useful constants

Quantity	Value
Speed of light, c	3 × 10⁸ m/s
1 atomic mass unit	1.6605 × 10⁻²⁷ kg
1 u in energy units	931.5 MeV/c²
1 eV	1.602 × 10⁻¹⁹ J
1 MeV	1.602 × 10⁻¹³ J
Nuclear radius constant R₀	Approximately 1.2 fm

9. Solved Numerical Problems

Numerical 1: Calculate nuclear radius

Question: Calculate the approximate radius of a nucleus with mass number A = 64. Take R₀ = 1.2 fm.

Solution:

R = R₀A^1/3

Since 64^1/3 = 4,

R = 1.2 × 4 = 4.8 fm

Answer: 4.8 × 10⁻¹⁵ m.

Numerical 2: Binding energy from mass defect

Question: The mass defect of a nucleus is 0.20 u. Find its binding energy.

Solution:

BE = Δm × 931.5 MeV

BE = 0.20 × 931.5 = 186.3 MeV

Answer: 186.3 MeV.

Numerical 3: Radioactive half-life

Question: A radioactive sample initially has 8000 undecayed nuclei. How many remain after 3 half-lives?

Solution:

N = N₀ / 2ⁿ

N = 8000 / 2³ = 8000 / 8

Answer: 1000 nuclei remain.

Numerical 4: Decay constant

Question: A radioactive isotope has a half-life of 20 minutes. Find its decay constant in min⁻¹.

Solution:

λ = 0.693/T½

λ = 0.693/20 = 0.03465 min⁻¹

Answer: 0.03465 min⁻¹.

Numerical 5: Binding energy per nucleon

Question: A nucleus has mass number 12 and total binding energy 92.16 MeV. Find its binding energy per nucleon.

Solution:

BE per nucleon = 92.16/12

= 7.68 MeV

Answer: 7.68 MeV per nucleon.

Numerical 6: Energy released in a nuclear reaction

Question: The total initial rest mass in a nuclear reaction is 10.010 u, while the total final rest mass is 10.000 u. Calculate the energy released.

Solution:

Δm = 10.010 − 10.000 = 0.010 u

Q = 0.010 × 931.5 MeV

Q = 9.315 MeV

Since the initial mass is greater than the final mass, the reaction releases energy. Answer: 9.315 MeV.

10. JEE Main Practice MCQs

Attempt these questions before opening the answers. Select the correct option for each problem.

Q1. The radius of a nucleus is proportional to which quantity?

A. A   B. A²   C. A^1/3   D. A^2/3

Show answer and explanation

Answer: C. The nuclear radius is R = R₀A^1/3.

Q2. The SI unit of radioactive activity is:

A. Gray   B. Becquerel   C. Sievert   D. Tesla

Show answer and explanation

Answer: B. One becquerel corresponds to one decay per second.

Q3. A radioactive sample has a half-life of 5 hours. The fraction remaining after 15 hours is:

A. 1/2   B. 1/4   C. 1/8   D. 1/16

Show answer and explanation

Answer: C. Fifteen hours corresponds to three half-lives, so the fraction remaining is (1/2)³ = 1/8.

Q4. During alpha decay, the atomic number of the parent nucleus:

A. Increases by 2   B. Decreases by 2   C. Increases by 4   D. Remains unchanged

Show answer and explanation

Answer: B. An alpha particle contains two protons and two neutrons, so the daughter nucleus has atomic number Z − 2.

Q5. If the mass defect of a nucleus is 0.005 u, its binding energy is approximately:

A. 0.466 MeV   B. 4.66 MeV   C. 46.6 MeV   D. 931.5 MeV

Show answer and explanation

Answer: B. BE = 0.005 × 931.5 = 4.6575 MeV ≈ 4.66 MeV.

Q6. In beta-minus decay, the atomic number:

A. Decreases by 1   B. Increases by 1   C. Decreases by 2   D. Remains unchanged

Show answer and explanation

Answer: B. A neutron converts into a proton, an electron and an electron antineutrino. Thus Z increases by one while A remains constant.

Q7. The mean life τ and half-life T½ are related by:

A. τ = 0.693 T½   B. τ = T½/0.693   C. τ = 2T½   D. τ = T½²

Show answer and explanation

Answer: B. Since T½ = 0.693/λ and τ = 1/λ, τ = T½/0.693.

Q8. Which process is responsible for the Sun's energy production?

A. Chemical combustion   B. Nuclear fission only   C. Nuclear fusion   D. Radioactive decay only

Show answer and explanation

Answer: C. The Sun generates energy primarily through nuclear fusion reactions involving hydrogen nuclei.

Q9. If the radius of a nucleus doubles, its mass number becomes how many times its original value, assuming R₀ is constant?

A. 2   B. 4   C. 6   D. 8

Show answer and explanation

Answer: D. Since R ∝ A^1/3, doubling R requires A to increase by 2³ = 8.

Q10. A nuclear reaction has a positive Q-value. Under the usual rest-mass convention, this means:

A. Final rest mass is greater than initial rest mass   B. Energy is released   C. No energy is transferred   D. The atomic number must decrease

Show answer and explanation

Answer: B. A positive Q-value means the initial rest energy exceeds the final rest energy, with the difference available as released energy.

11. Numerical Practice Set

Numerical 1: Calculate the nuclear radius for A = 125. Take R₀ = 1.2 fm.

Reveal solution

R = 1.2 × 125^1/3 = 1.2 × 5 = 6 fm.

Numerical 2: A radioactive sample has a half-life of 10 days. What percentage remains after 30 days?

Reveal solution

Number of half-lives = 30/10 = 3.

Remaining fraction = (1/2)³ = 1/8.

Percentage remaining = (1/8) × 100 = 12.5%.

Numerical 3: A nucleus has mass defect 0.015 u and mass number 15. Calculate its total binding energy and binding energy per nucleon.

Reveal solution

Total BE = 0.015 × 931.5 = 13.9725 MeV.

BE per nucleon = 13.9725/15 = 0.9315 MeV per nucleon.

Numerical 4: The decay constant of a radioactive isotope is 0.02 s⁻¹. Find its mean life and half-life.

Reveal solution

Mean life: τ = 1/λ = 1/0.02 = 50 s.

Half-life: T½ = 0.693/0.02 = 34.65 s.

Numerical 5: A radioactive sample initially has a mass of 160 mg and a half-life of 4 hours. Find the remaining mass after 12 hours.

Reveal solution

Number of half-lives = 12/4 = 3.

Remaining mass = 160 × (1/2)³ = 20 mg.

Numerical 6: A nuclear reaction has a mass defect of 0.002 u. Calculate the energy released in joules. Use 1 u = 1.6605 × 10⁻²⁷ kg and c = 3 × 10⁸ m/s.

Reveal solution

Δm = 0.002 × 1.6605 × 10⁻²⁷ = 3.321 × 10⁻³⁰ kg.

Q = Δmc² = 3.321 × 10⁻³⁰ × (3 × 10⁸)².

Q = 2.989 × 10⁻¹³ J approximately.

12. Quick Revision: Important Points

• Mass number A = Z + N.
• Nuclear radius R = R₀A^1/3.
• Nuclear density is approximately constant across nuclei.
• Binding energy = mass defect × c².
• 1 u corresponds to approximately 931.5 MeV of rest energy.
• Half-life = 0.693/λ.
• Mean life = 1/λ.
• After n half-lives, the fraction remaining is 1/2ⁿ.
• Alpha decay changes A by −4 and Z by −2.
• Beta-minus decay leaves A unchanged and increases Z by 1.
• Gamma emission does not change A or Z.
• Positive Q-value indicates energy release.

JEE Main preparation strategy: Practise mass-defect calculations, binding energy per nucleon, nuclear radius ratios, half-life problems, decay constants, and energy released in nuclear reactions. Pay close attention to units and the sign of the Q-value.

🎯 Keep Practising!

Learn the concepts, revise the formula sheet, and solve numerical problems regularly.

Class 12 Physics | Nuclear Physics | JEE Main Preparation