Nuclear Physics JEE Main previous year questions with solutions

5 solved JEE Main questions on Nuclear Physics, free to read — no sign-in needed. The full chapter has 88 questions; sign in to attempt the remaining 83 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (06 Apr, Shift 2)Nuclear energy
   The energy released if hydrogen atoms are combined to form $^{4}_{2}\text{He}$ is __________ MeV. (Take binding energies per nucleon of $^{2}_{1}\text{H}$ and $^{4}_{2}\text{He}$ as $1.1$ MeV and $7.2$ MeV, respectively)
   1. A.$6.1$
   2. B.$24.4$
   3. C.$26.6$
   4. D.$5$
   Show answer & solution

   Answer: (B)

   The fusion reaction is given by: $2 ^{2}_{1}\text{H} \rightarrow ^{4}_{2}\text{He}$ Binding energy of one $^{2}_{1}\text{H}$ nucleus = $2 \times 1.1 = 2.2$ MeV Total initial binding energy = $2 \times 2.2 = 4.4$ MeV Binding energy of $^{4}_{2}\text{He}$ nucleus = $4 \times 7.2 = 28.8$ MeV Energy released $\Delta E$ = Final binding energy - Initial binding energy $\Delta E = 28.8 - 4.4 = 24.4$ MeV Answer: $24.4$
2. Q2JEE Main 2025 (07 Apr, Shift 2)Nuclear structure
   <p>Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The density of the copper $\left({ }_{29}^{64} \mathrm{Cu}\right)$ nucleus is greater than that of the carbon $\left({ }_6^{12} \mathrm{C}\right)$ nucleus. Reason (R): The nucleus of mass number A has a radius proportional to $\mathrm{A}^{1 / 3}$. In the light of the above statements, choose the most appropriate answer from the options given below :</p>
   1. A.<p>$(\mathrm{A})$ is correct but $(\mathrm{R})$ is not correct</p>
   2. B.<p>$(\mathrm{A})$ is not correct but $(\mathrm{R})$ is correct</p>
   3. C.<p>Both $(\mathrm{A})$ and $(\mathrm{R})$ are correct and $(\mathrm{R})$ is the correct explanation of (A)</p>
   4. D.<p>Both (A) and (R) are correct but (R) is not the correct explanation of (A)</p>
   Show answer & solution

   Answer: (B)

   <p>$\rho=\frac{\mathrm{M}}{\mathrm{~V}}=\frac{\mathrm{m}_{\mathrm{n}} \times \mathrm{A}}{\frac{4}{3} \pi \mathrm{R}^3}=\frac{\mathrm{m}_{\mathrm{n}} \times \mathrm{A}}{\frac{4}{3} \pi \mathrm{AR}_0^3}$ So $\rho$ is almost is constant $\begin{aligned} & \mathrm{R}=\mathrm{R}_0 \mathrm{~A}^{1 / 3} \\ & \mathrm{R} \propto \mathrm{A}^{1 / 3} \end{aligned}$</p>
3. Q3JEE Main 2024 (31 Jan, Shift 1)Atomic Models
   The mass defect in a particular reaction is $0.4g$. The amount of energy liberated is $n\times {10}^{7}kWh$, where $n=$ _____. (speed of light $=3\times {10}^{8}m{s}^{-1}$)
   Show answer & solution

   Answer: 1

   Energy liberated is given by, $E=∆m{c}^{2}$ $=\left(0.4\times {10}^{-3}\right)\times {\left(3\times {10}^{8}\right)}^{2}$ $=3600\times {10}^{7}kWs$ $=\dfrac{3600\times {10}^{7}}{3600}kWh=1\times {10}^{7}kWh$
4. Q4JEE Main 2023 (01 Feb, Shift 1)Bohr's Atomic Model
   The mass of proton, neutron and helium nucleus are respectively $1.0073u,1.0087uand4.0015u$. The binding energy of helium nucleus is:
   1. A.$14.2MeV$
   2. B.$28.4MeV$
   3. C.$56.8MeV$
   4. D.$7.1MeV$
   Show answer & solution

   Answer: (B)

   Given here, ${m}_{He}=4.0015u$, ${m}_{p}=1.0073u$ and ${m}_{n}=1.0087u$. Mass defect $∆m=2{m}_{p}+2{m}_{n}-{m}_{He} =2\times 1.0073+2\times 1.0087-4.0015 =0.0305u$ Binding energy $E=∆m\times 931MeV =0.0305\times 931 =28.4MeV$
5. Q5JEE Main 2022 (29 Jul, Shift 2)Atomic Structure
   Read the following statements: (A) Volume of the nucleus is directly proportional to the mass number. (B) Volume of the nucleus is independent of mass number. (C) Density of the nucleus is directly proportional to the mass number. (D) Density of the nucleus is directly proportional to the cube root of the mass number. (E) Density of the nucleus is independent of the mass number. Choose the correct option from the following options.
   1. A.(A) and (D) only.
   2. B.(A) and (E) only.
   3. C.(B) and (E) only.
   4. D.(A) and (C) only
   Show answer & solution

   Answer: (B)

   Radius of nucleus is $R={R}_{0}{A}^{\dfrac{1}{3}}$, where, ${R}_{0}=1.2\times {10}^{-15}m$ and $A$ is mass number of nucleus. Thus, $R\propto {A}^{\dfrac{1}{3}}$ Volume of nucleus is $V=\dfrac{4}{3}\pi {R}^{3}\Rightarrow V\propto {R}^{3}\Rightarrow V\propto A$. Now, the density of nucleus is given by $\rho =\dfrac{M}{V}=\dfrac{mA}{\dfrac{4}{3}\pi {R}_{0}^{3}A}=\dfrac{m}{\dfrac{4}{3}\pi {R}_{0}^{3}}$ Thus, density is independent of $A$. Hence, only $\left(A\right)$ and $\left(E\right)$ are correct.