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📘 Nuclear Physics — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Nuclear Physics for JEE Main with previous year questions on binding energy, nuclear fission, and fusion. Each solution is explained in a simple, step-by-step manner to boost your score.
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Nuclear Physics — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (8 April Shift 2)
For a nucleus of mass number \(A\) and radius \(R\), the mass density of nucleus can be represented as :
(1) \(A^3\) (2) \(A^{1/3}\) (3) \(A^{2/3}\) (4) Independent of A
(1) \(A^3\) (2) \(A^{1/3}\) (3) \(A^{2/3}\) (4) Independent of A
✅ Answer & Explanation:
Correct Option: (4)
- Nuclear density \(\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{A \cdot m_u}{\frac{4}{3}\pi R^3}\).
- Since \(R = R_0 A^{\frac{1}{3}}\), then \(R^3 = R_0^3 A\).
- Substituting this: \(\rho = \frac{A \cdot m_u}{\frac{4}{3}\pi R_0^3 A} = \frac{m_u}{\frac{4}{3}\pi R_0^3}\).
- The \(A\) terms cancel out, meaning nuclear density is independent of the mass number A.
Q2. JEE Main 2025 (28 Jan Shift 1)
Choose the correct nuclear process from the below options [p: proton, n: neutron, \(e^-\): electron, \(e^+\): positron, \(\nu\): neutrino, \(\bar{\nu}\): antineutrino]:
(1) \(n \to p + e^+ + \bar{\nu}\) (2) \(n \to p + e^+ + \nu\)
(3) \(n \to p + e^- + \nu\) (4) \(n \to p + e^- + \bar{\nu}\)
(1) \(n \to p + e^+ + \bar{\nu}\) (2) \(n \to p + e^+ + \nu\)
(3) \(n \to p + e^- + \nu\) (4) \(n \to p + e^- + \bar{\nu}\)
✅ Answer & Explanation:
Correct Option: (4)
In beta-minus (\(\beta^-\)) decay, a neutron transforms into a proton. To conserve charge and lepton number, an electron and an antineutrino are emitted.
In beta-minus (\(\beta^-\)) decay, a neutron transforms into a proton. To conserve charge and lepton number, an electron and an antineutrino are emitted.
Q3. JEE Main 2025 (23 Jan Shift 2)
Assertion (A): The binding energy per nucleon is found to be practically independent of the mass number A, for nuclei with mass numbers between 30 and 170.
Reason (R): Nuclear force is long range.
(1) (A) is true but (R) is false (2) (A) is false but (R) is true
(3) Both (A) and (R) are true and (R) is the correct explanation of (A)
Reason (R): Nuclear force is long range.
(1) (A) is true but (R) is false (2) (A) is false but (R) is true
(3) Both (A) and (R) are true and (R) is the correct explanation of (A)
✅ Answer & Explanation:
Correct Option: (1)
- Assertion is true: The BE curve is relatively flat between \(A=30\) and \(170\) due to the saturation of nuclear forces.
- Reason is false: The nuclear force is short-range (acting over a few femtometers). Its short-range nature is actually what leads to the saturation property mentioned in the assertion.
Q4. JEE Main 2020 (03 Sep Shift 2)
Using \(R = (1.3 \times 10^{-15}) A^{1/3} \, m\), the mass density of a nucleus is of the order of :
(1) \(10^3 \, kg/m^3\) (2) \(10^{10} \, kg/m^3\)
(3) \(10^{24} \, kg/m^3\) (4) \(10^{17} \, kg/m^3\)
(1) \(10^3 \, kg/m^3\) (2) \(10^{10} \, kg/m^3\)
(3) \(10^{24} \, kg/m^3\) (4) \(10^{17} \, kg/m^3\)
✅ Answer & Explanation:
Correct Option: (4)
Substituting the given radius formula and typical nucleon mass into the density equation \(\rho = \frac{m_u}{\frac{4}{3}\pi R_0^3}\) yields an order of magnitude of \(10^{17} \, kg/m^3\).
Substituting the given radius formula and typical nucleon mass into the density equation \(\rho = \frac{m_u}{\frac{4}{3}\pi R_0^3}\) yields an order of magnitude of \(10^{17} \, kg/m^3\).
Q5. JEE Main 2025 (3 April Shift 1)
Match List-I with List-II :
Choose the correct answer: (1) A-II, B-I, C-III, D-IV (2) A-III, B-I, C-II, D-IV
| List-I (Reaction) | List-II (Process) |
|---|---|
| (A) \(_0^1n + _{92}^{235}U \to _{54}^{140}Xe + _{38}^{94}Sr + 2_0^1n\) | (I) Chemical reaction |
| (B) \(2H_2 + O_2 \to 2H_2O\) | (II) Fusion with +ve Q value |
| (C) \(_1^2H + _1^2H \to _2^3He + _0^1n\) | (III) Fission |
| (D) \(_1^1H + _1^3H \to _1^2H + _1^2H\) | (IV) Fusion with -ve Q value |
Choose the correct answer: (1) A-II, B-I, C-III, D-IV (2) A-III, B-I, C-II, D-IV
✅ Answer & Explanation:
Correct Option: (2)
- (A) Fission: A heavy nucleus splits into lighter fragments [III].
- (B) Chemical Reaction: Rearrangement of atoms into molecules [I].
- (C) Fusion (+ve Q): Deuterium fusion releases energy [II].
- (D) Fusion (-ve Q): Requires energy input to proceed [IV].
Q6. JEE Main 2025 (2 April Shift 2)
Energy released when two deuterons (\(_1H^2\)) fuse to form a helium nucleus (\(_2He^4\)) is : (Given : BE/nucleon of \(_1H^2 = 1.1 \, \text{MeV}\) and \(_2He^4 = 7.0 \, \text{MeV}\))
✅ Answer & Explanation:
Correct Option: (3)
- Total BE of reactants \(= 2 \times (2 \times 1.1) = 4.4 \, \text{MeV}\).
- Total BE of product \(= 1 \times (4 \times 7.0) = 28.0 \, \text{MeV}\).
- Energy released \(Q = BE_{product} - BE_{reactants} = 28.0 - 4.4 = \mathbf{23.6 \, \text{MeV}}\).
Q7. JEE Main 2022 (26 Jul Shift 2)
Two lighter nuclei combine: \(_1^2X + _1^2X \to _2^4Y\). BE/nucleon for X and Y are 1.1 MeV and 7.6 MeV. The energy released is ______ MeV.
✅ Answer & Explanation:
Correct Answer: 26
- BE of reactants \(= 2 \times (2 \times 1.1) = 4.4 \, \text{MeV}\).
- BE of product \(= 4 \times 7.6 = 30.4 \, \text{MeV}\).
- Energy released \(= 30.4 - 4.4 = \mathbf{26 \, \text{MeV}}\).
Q8. JEE Main 2021 (25 Jul Shift 2)
Find energy required to break aluminum nucleus \(_{13}^{27}Al\) into nucleons. (Masses: \(n=1.00866u, p=1.00726u, Al=27.18846u\)). Energy \(= x \times 10^{-3} \, J\). Find \(x\).
✅ Answer & Explanation:
Correct Answer: 27
Calculating the mass defect \(\Delta m = [13(1.00726) + 14(1.00866)] - 27.18846\) and converting to Joules using given conversion factors results in a value where x = 27.
Calculating the mass defect \(\Delta m = [13(1.00726) + 14(1.00866)] - 27.18846\) and converting to Joules using given conversion factors results in a value where x = 27.
Q9. JEE Main 2024 (29 Jan Shift 1)
The explosive in a Hydrogen bomb is a mixture of \(_1H^2, _1H^3\) and \(_3Li^6\). During the explosion the energy released is approximately :
(1) 28.12 MeV (2) 12.64 MeV (3) 16.48 MeV (4) 22.22 MeV
(1) 28.12 MeV (2) 12.64 MeV (3) 16.48 MeV (4) 22.22 MeV
✅ Answer & Explanation:
Correct Option: (4)
Using the mass data for Lithium, Deuterium, and Helium provided in the source and applying \(E = \Delta m c^2\), the total energy released in the bomb's chain reaction is approximately 22.22 MeV.
Using the mass data for Lithium, Deuterium, and Helium provided in the source and applying \(E = \Delta m c^2\), the total energy released in the bomb's chain reaction is approximately 22.22 MeV.
Q10. JEE Main 2024 (09 Apr Shift 1)
A star has 100% helium composition. It converts three \(_2^4He\) into one \(_{12}^6C\) via triple alpha process. Mass of star is \(2.0 \times 10^{32} \, \text{kg}\) and power is \(5.808 \times 10^{30} \, \text{W}\). The rate of converting helium to carbon is \(n \times 10^{42} \, s^{-1}\). Value of \(n\) is :
✅ Answer & Explanation:
Correct Answer: 15
Calculating the energy released per triple-alpha process from the mass defect (\(3 \times 4.0026u - 12u\)) and dividing the total power by this energy yields a conversion rate where n = 15.
Calculating the energy released per triple-alpha process from the mass defect (\(3 \times 4.0026u - 12u\)) and dividing the total power by this energy yields a conversion rate where n = 15.
Q11. JEE Main 2022 (28 Jun Shift 1)
The Q-value of a nuclear reaction and kinetic energy of the projectile particle, \(K_p\), are related as :
(1) \(Q = K_p\) (2) \((K_p + Q) < 0\) (3) \(Q < K_p\) (4) \((K_p + Q) > 0\)
(1) \(Q = K_p\) (2) \((K_p + Q) < 0\) (3) \(Q < K_p\) (4) \((K_p + Q) > 0\)
✅ Answer & Explanation:
Correct Option: (4)
For a reaction to occur, the total available energy (projectile kinetic energy + reaction Q-value) must be positive to provide for the kinetic energy of the products. Thus, \((K_p + Q) > 0\).
For a reaction to occur, the total available energy (projectile kinetic energy + reaction Q-value) must be positive to provide for the kinetic energy of the products. Thus, \((K_p + Q) > 0\).
Q12. JEE Main 2022 (24 Jun Shift 1)
Nucleus A (mass 220, BE/nucleon 5.6 MeV) splits into fragments B (mass 105) and C (mass 115) with BE/nucleon 6.4 MeV. Energy released per fission is :
(1) 0.8 MeV (2) 275 MeV (3) 220 MeV (4) 176 MeV
(1) 0.8 MeV (2) 275 MeV (3) 220 MeV (4) 176 MeV
✅ Answer & Explanation:
Correct Option: (4)
- Initial BE \(= 220 \times 5.6 = 1232 \, \text{MeV}\).
- Final BE \(= (105 + 115) \times 6.4 = 220 \times 6.4 = 1408 \, \text{MeV}\).
- Energy released \(= 1408 - 1232 = \mathbf{176 \, \text{MeV}}\).
📘 Exam Preparation Tip for Nuclear Physics
- Nuclear Size and Density: Radius \(R = R_0 A^{1/3}\) where \(R_0 = 1.2 \times 10^{-15} \, \text{m}\). Nuclear density is constant (\(\sim 10^{17} \, \text{kg/m}^3\)) and independent of A.
- Binding Energy: \(BE = \Delta m c^2\) where \(\Delta m\) is the mass defect. Binding energy per nucleon is a measure of nuclear stability. The curve peaks at \(A \approx 56\) (Iron).
- Nuclear Reactions: Q-value: \(Q = (m_{reactants} - m_{products})c^2\). For a reaction to occur spontaneously (exothermic), \(Q > 0\). For endothermic reactions, \(Q < 0\).
- Radioactivity: Decay law: \(N = N_0 e^{-\lambda t}\). Half-life: \(T_{1/2} = \frac{\ln 2}{\lambda}\). Activity: \(A = \lambda N\).
- Nuclear Fission: Splitting of a heavy nucleus into lighter fragments, releasing energy. Fission of \(^{235}U\) releases about 200 MeV per fission.
- Nuclear Fusion: Combining of light nuclei to form a heavier nucleus, releasing energy. This is the source of energy in stars.