Nuclear Physics JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Nuclear Physics, free to read — no sign-in needed. The full chapter has 36 questions; sign in to attempt the remaining 31 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   A nuclear reactor starts producing a radioactive nuclide $X$ from $t = 0$, at a constant rate of $\alpha$ per second. Each decay of $X$ produces energy $E_0$, which is utilized to heat a liquid of mass $m$ and specific heat $s$. Assuming no heat loss from the liquid and taking $\lambda$ as the decay constant of $X$, the rate of increase in the temperature of the liquid is:
   1. A.$\dfrac{\alpha E_0}{m s}(1 - e^{-\lambda t})$
   2. B.$\dfrac{\alpha E_0}{m s}(e^{\lambda t} - 1)$
   3. C.$\dfrac{\lambda E_0}{m s}(1 - e^{-\lambda t})$
   4. D.$\dfrac{E_0}{m s}(\alpha - \lambda e^{-\lambda t})$
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   Answer: (A)

   Let $N$ be the number of active nuclei at time $t$. The rate of change of the number of nuclei is given by $\dfrac{dN}{dt} = \alpha - \lambda N$ Integrating with the initial condition $N = 0$ at $t = 0$: $\int_{0}^{N} \dfrac{dN}{\alpha - \lambda N} = \int_{0}^{t} dt$ $-\dfrac{1}{\lambda} \ln\left(\dfrac{\alpha - \lambda N}{\alpha}\right) = t$ $N = \dfrac{\alpha}{\lambda} (1 - e^{-\lambda t})$ The rate of decay of the nuclide at time $t$ is $A = \lambda N = \alpha (1 - e^{-\lambda t})$ Since each decay produces energy $E_0$, the rate of energy production is $\dfrac{dE}{dt} = A E_0 = \alpha E_0 (1 - e^{-\lambda t})$ This energy is used to heat the liquid. The rate of heat absorption is $\dfrac{dQ}{dt} = m s \dfrac{dT}{dt}$ Equating the rate of energy production to the rate of heat absorption: $m s \dfrac{dT}{dt} = \alpha E_0 (1 - e^{-\lambda t})$ $\dfrac{dT}{dt} = \dfrac{\alpha E_0}{m s} (1 - e^{-\lambda t})$ Answer: $\dfrac{\alpha E_0}{m s}(1 - e^{-\lambda t})$
2. Q2JEE Advanced Adv 2023 (Paper 2)
   In a radioactive decay process, the activity is defined as $A=-\dfrac{dN}{dt}$, where $N(t)$ is the number of radioactive nuclei at time $\left(t\right)$. Two radioactive sources, ${S}_{1}$ and ${S}_{2}$ have same activity at time $t=0$. At a later time, the activities of ${S}_{1}$ and ${S}_{2}$ are ${A}_{1}$ and ${A}_{2}$, respectively. When ${S}_{1}$ and ${S}_{2}$ have just completed their ${3}^{rd}$ and ${7}^{th}$ half-lives, respectively, the ratio $\dfrac{{A}_{1}}{{A}_{2}}$ is _____.
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   Answer: 16

   Activity for both the cases can be written as, ${A}_{1}={A}_{0}{e}^{-{\lambda }_{1}{t}_{1}}$ also ${A}_{2}={A}_{0}{e}^{-{\lambda }_{2}{t}_{2}}$. Half-life for the reaction can be written as, ${\left({t}_{\dfrac{1}{2}}\right)}_{1}=\dfrac{\ln 2}{{\lambda }_{1}}$ and hence at ${t}_{1}=\dfrac{3\ln 2}{{\lambda }_{1}}$. ${A}_{1}={A}_{0}{e}^{-{\lambda }_{1}\left(\dfrac{3\ln 2}{{\lambda }_{1}}\right)}$ $\Rightarrow {A}_{1}={A}_{0}{e}^{-3\ln 2}...\left(i\right)$ Similarly, for the second case, we can write ${t}_{2}=7\left(\dfrac{\ln 2}{{\lambda }_{2}}\right)$ and therefore, ${A}_{2}={A}_{0}{e}^{-{\lambda }_{2}\left(\dfrac{7\ln 2}{{\lambda }_{2}}\right)}$ ${A}_{2}={A}_{0}{e}^{-7\ln 2}...\left(ii\right)$ From $\left(i\right)$ and $\left(ii\right)$, we get $\dfrac{{A}_{1}}{{A}_{2}}=\dfrac{{A}_{0}{e}^{-3\ln 2}}{{A}_{0}{e}^{-7\ln 2}} =\dfrac{{A}_{0}{e}^{\ln {2}^{-3}}}{{A}_{0}{e}^{\ln {2}^{-7}}}=\dfrac{{2}^{-3}}{{2}^{-7}}=16$ Therefore, required ration would be$∴\dfrac{{A}_{1}}{{A}_{2}}=16$
3. Q3JEE Advanced Adv 2022 (Paper 2)
   In a radioactive decay chain reaction, ${}_{90}^{230}Th$ nucleus decays into $Po84214$ nucleus. The ratio of the number of $\alpha$ to number of ${\beta }^{-}$ particles emitted in this process is _______.
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   Answer: 2

   Let number of $\alpha$ particles are ${n}_{\alpha }$ and $\beta$ particles are ${n}_{\beta }$. Given: ${}_{90}{Th}^{230}{\rightarrow }_{84}{Po}^{214}+{n}_{\alpha }\alpha +{n}_{\beta }\beta$. Change in mass number is $=230-214=4{n}_{\alpha }$ $\Rightarrow {n}_{\alpha }=4$ Number of proton, ${n}_{\beta }=84-\left(90-2{n}_{a}\right)$ $\Rightarrow {n}_{\beta }=2$ So, the ratio, $\dfrac{{n}_{\alpha }}{{n}_{\beta }}=2$
4. Q4JEE Advanced Adv 2022 (Paper 1)
   The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton be ${E}_{b}^{p}$ and the binding energy of a neutron be ${E}_{b}^{n}$ in the nucleus. Which of the following statement(s) is(are) correct?
   1. A.${E}_{b}^{p}-{E}_{b}^{n}$ is proportional to $Z\left(Z-1\right)$ where $Z$ is the atomic number of the nucleus.
   2. B.${E}_{b}^{p}-{E}_{b}^{n}$ is proportional to ${A}^{-\dfrac{1}{3}}$ where $A$ is the mass number of the nucleus.
   3. C.${E}_{b}^{p}-{E}_{b}^{n}$ is positive.
   4. D.${E}_{b}^{p}$ increases if the nucleus undergoes a beta decay emitting a positron.
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   Answer: A,B,D

   Total binding energy (without considering repulsions), ${E}_{b}=\left[Z{m}_{p}+\left(A-Z\right){m}_{n}-{m}_{x}\right]{c}^{2}$ Where, $XZA$ is the nuclei under consideration. Now, considering repulsion : Number of proton pairs $=C2Z$ $\Rightarrow$ Thus repulsion energy $\propto \dfrac{Z\left(Z-1\right)}{2}\times \dfrac{1}{4\pi {ϵ}_{0}}\dfrac{{e}^{2}}{R}$ Where $R$ is the radius of the nucleus $\Rightarrow {E}_{b}^{p}-{E}_{b}^{n}\propto Z\left(Z-1\right)∴$ there will be no repulsion term for neutrons. Also, since $R={R}_{0}{A}^{\dfrac{1}{3}}$ $\Rightarrow {E}_{b}^{p}-{E}_{b}^{n}\propto {A}^{-\dfrac{1}{3}}$ Because of repulsion among protons, ${E}_{b}^{p}\lt {E}_{b}^{n}$ Since in ${\beta }^{+}$ decay, number of protons decrease $\Rightarrow$ repulsion would decrease $\Rightarrow {E}_{b}^{p}$ increases
5. Q5JEE Advanced Adv 2022 (Paper 1)
   The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction $N716+He24\rightarrow H11+O819$ in a laboratory frame is $n$ (in $MeV$). Assume that $N716$ is at rest in the laboratory frame. The masses of $N716,He24,H11$ and $O819$ can be taken to be $16.006u,4.003u$, $1.008u$ and $19.003u$, respectively, where $1u=930MeV{c}^{-2}$. The value of $n$ is If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.
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   Answer: 2.33

   Given here: $N716+He24\rightarrow H11+O819$ Energy absorbed in the reaction is $Q=\left({m}_{N}+{m}_{He}-{m}_{H}-{m}_{O}\right)\times {c}^{2}$ $=\left(16.006+4.003-1.008-19.003\right)\times 930MeV$ $=-1.86MeV$ $=1.86MeV$ Let $v$ is velocity of alpha particle. Now, the maximum loss of kinetic energy$=$ $\dfrac{1}{2}\times \dfrac{m\times 4m}{5m}\times {v}^{2}$ $\Rightarrow \dfrac{1}{2}m{v}^{2}=\dfrac{5}{4}\times Q$ $=\dfrac{5}{4}\times \left(1.86\right)MeV$ $=2.325MeV$ $∴n=2.33$