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 83 questions; sign in to attempt the remaining 78 in the exam simulator.

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.$5$
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   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>
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   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 (09 Apr, Shift 2)Nuclear reactions
   The energy released in the fusion of $2 \mathrm{~kg}$ of hydrogen deep in the sun is $E_H$ and the energy released in the fission of $2 \mathrm{~kg}$ of ${ }^{235} \mathrm{U}$ is $E_U$. The ratio $\frac{E_H}{E_U}$ is approximately: (Consider the fusion reaction as $4 \mid H+2 \mathrm{e}^{-} \rightarrow{ }_2^4 \mathrm{He}+2 v+6 \gamma+26.7 \mathrm{MeV}$, energy released in the fission reaction of ${ }^{235} \mathrm{U}$ is $200 \mathrm{MeV}$ per fission nucleus and $\mathrm{N}_{\mathrm{A}}=$ $\left.6.023 \times 10^{23}\right)$
   1. A.7.62
   2. B.25.6
   3. C.15.04
   4. D.9.13
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   Answer: (A)

   In each fusion reaction, $4{ }_1^1 \mathrm{H}$ nucleus are used. Energy released per Nuclei of ${ }_1^1 \mathrm{H}=\frac{26.7}{4} \mathrm{MeV}$ $\therefore$ Energy released by $2 \mathrm{~kg}$ hydrogen $\left(\mathrm{E}_{\mathrm{H}}\right)$ $=\frac{2000}{1} \times \mathrm{N}_{\mathrm{A}} \times \frac{26.7}{4} \mathrm{MeV}$ $\qquad \qquad\quad \&$ $\begin{aligned} & \therefore \text { Energy released by } 2 \mathrm{~kg} \text { Vranium }\left(\mathrm{E}_{\mathrm{V}}\right) \\ & =\frac{2000}{235} \times \mathrm{N}_{\mathrm{A}} \times 200 \mathrm{MeV}\end{aligned}$ So, $\frac{E_H}{E_V}=235 \times \frac{26.7}{4 \times 200}=7.84$ $\therefore$ Approximately close to 7.62
4. Q4JEE Main 2026 (06 Apr, Shift 1)Nuclear energy
   The energy released when $\dfrac{7}{17.13}$ kg of $^{7}_{3}\text{Li}$ is converted into $^{4}_{2}\text{He}$ by proton bombardment is $\alpha \times 10^{32}$ eV. The value of $\alpha$ is _______. (Nearest integer) (Mass of $^{7}_{3}\text{Li} = 7.0183$ u, mass of $^{4}_{2}\text{He} = 4.004$ u, mass of proton $= 1.008$ u and $1$ u $= 931$ MeV/c$^2$ and Avogadro number $= 6.0 \times 10^{23}$)
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   Answer: 6

   The nuclear reaction for the proton bombardment of Lithium is: $^{7}_{3}\text{Li} + ^{1}_{1}\text{H} \rightarrow 2 ^{4}_{2}\text{He}$ The mass defect ($\Delta m$) for the reaction is: $\Delta m = m(^{7}_{3}\text{Li}) + m(^{1}_{1}\text{H}) - 2m(^{4}_{2}\text{He})$ $\Delta m = 7.0183 + 1.008 - 2(4.004)$ $\Delta m = 8.0263 - 8.008 = 0.0183 \text{ u}$ Energy released per reaction ($Q$) is: $Q = \Delta m \times 931 \text{ MeV} = 0.0183 \times 931 \text{ MeV} = 17.0373 \text{ MeV}$ Number of moles of $^{7}_{3}\text{Li}$ in $\dfrac{7}{17.13} \text{ kg}$: $n = \dfrac{\dfrac{7}{17.13} \times 10^3 \text{ g}}{7 \text{ g/mol}} = \dfrac{1000}{17.13} \text{ mol}$ Number of atoms of $^{7}_{3}\text{Li}$: $N = n \times N_A = \dfrac{1000}{17.13} \times 6.0 \times 10^{23} = \dfrac{6}{17.13} \times 10^{26}$ Total energy released ($E$) is: $E = N \times Q = \left(\dfrac{6}{17.13} \times 10^{26}\right) \times (17.0373 \times 10^6 \text{ eV})$ $E = \dfrac{102.2238}{17.13} \times 10^{32} \text{ eV} \approx 5.967 \times 10^{32} \text{ eV}$ Rounding to the nearest integer, we get $\alpha = 6$. Answer: $6$
5. Q5JEE Main 2026 (05 Apr, Shift 2)Nuclear energy
   Assuming the experimental mass of ${}^{12}_{6}C$ as $12\text{ u}$, the mass defect of ${}^{12}_{6}C$ atom is _______ $\text{MeV}/c^2$. (Mass of proton $= 1.00727\text{ u}$, mass of neutron $= 1.00866\text{ u}$, $1\text{ u} = 931.5\text{ MeV}/c^2$ and $c$ is the speed of the light in vacuum).
   1. A.$127.5$
   2. B.$89.03$
   3. C.$272.0$
   4. D.$92.0$
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   Answer: (B)

   The mass of $6$ protons is $6 \times 1.00727\text{ u} = 6.04362\text{ u}$. The mass of $6$ neutrons is $6 \times 1.00866\text{ u} = 6.05196\text{ u}$. Total mass of the constituent nucleons is $6.04362\text{ u} + 6.05196\text{ u} = 12.09558\text{ u}$. The mass defect $\Delta m$ is the difference between the total mass of the nucleons and the experimental mass of the atom (ignoring electron mass as per the given data): $\Delta m = 12.09558\text{ u} - 12\text{ u} = 0.09558\text{ u}$ Converting the mass defect into $\text{MeV}/c^2$: $\Delta m = 0.09558 \times 931.5\text{ MeV}/c^2 = 89.03277\text{ MeV}/c^2 \approx 89.03\text{ MeV}/c^2$ Answer: $89.03$