Atomic Physics JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Atomic Physics, free to read — no sign-in needed. The full chapter has 25 questions; sign in to attempt the remaining 20 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 1)
   Consider an electron in the $n=3$ orbit of a hydrogen-like atom with atomic number $Z$. At absolute temperature $T$, a neutron having thermal energy $k_{\mathrm{B}} T$ has the same de Broglie wavelength as that of this electron. If this temperature is given by $T=\frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$, (where $h$ is the Planck's constant, $k_B$ is the Boltzmann constant, $m_{\mathrm{N}}$ is the mass of the neutron and $a_0$ is the first Bohr radius of hydrogen atom) then the value of $\alpha$ is ________
   Show answer & solution

   Answer: 72

   $\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{KZe}^2}{\mathrm{r}^2}$ $\mathrm{mv}^2 \mathrm{r}=\frac{1}{4 \pi \epsilon_0} \mathrm{Ze}^2$ $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$ (1)/(2) gives $\mathrm{v}=\frac{\frac{\mathrm{Ze}^2}{4 \pi \epsilon_0}}{\frac{\mathrm{nh}}{2 \pi}}=\frac{\mathrm{Ze}^2}{2 \epsilon_0 \mathrm{nh}}$ $\begin{aligned} & \frac{h}{m v}=\frac{h}{\sqrt{2 m_N \cdot K_B T}} \\ & T=\frac{m^2 Z^2 e^4}{8 \epsilon_0^2 n^2 h^2 m_N K_B} \\ & n=3 \Rightarrow T=\frac{m^2 Z^2 e^4}{72 \epsilon_0^2 h^2 m_N K_B} \\ & \frac{(1)}{(2)^2} \Rightarrow \frac{1}{m r}=\frac{\frac{Z e^2}{4 \pi \epsilon_0}}{\frac{n^2 h^2}{4 \pi^2}}\end{aligned}$ $\begin{aligned} & \mathrm{r}=\frac{\mathrm{n}^2 \mathrm{~h}^2 \epsilon_0}{\pi \mathrm{Ze}^2 \cdot \mathrm{~m}} \Rightarrow \mathrm{a}_0=\frac{\mathrm{h}^2 \epsilon_0}{\pi \mathrm{e}^2 \mathrm{~m}} \\ & \mathrm{a}_0^2=\frac{\mathrm{h}^4 \epsilon_0^2}{\pi^2 \mathrm{e}^4 \mathrm{~m}^2} \\ & \mathrm{Ta}_0^2=\frac{\mathrm{m}^2 \mathrm{Z}^2 \mathrm{e}^4}{72 \epsilon_0 \mathrm{~h}^2 \mathrm{~m}_{\mathrm{N}} \mathrm{k}_{\mathrm{B}}} \cdot \frac{\mathrm{h}^4 \epsilon_0^2}{\pi^2 \mathrm{e}^4 \mathrm{~m}^2} \\ & \mathrm{~T}=\frac{\mathrm{h}^2 \mathrm{Z}^2}{72 \pi^2 \mathrm{a}_0^2 \mathrm{~m}_{\mathrm{N}} \mathrm{k}_{\mathrm{B}}} \Rightarrow \alpha=72\end{aligned}$
2. Q2JEE Advanced Adv 2024 (Paper 1)
   A particle of mass $m$ is moving in a circular orbit under the influence of the central force $F(r)=-k r$, corresponding to the potential energy $V(r)=k r^2 / 2$, where $k$ is a positive force constant and $r$ is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by $L=n \hbar$, where $\hbar=h /(2 \pi), h$ is the Planck's constant, and $n$ a positive integer. If $v$ and $E$ are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?
   1. A.$r^2=n \hbar \sqrt{\frac{1}{m k}}$
   2. B.$v^2=n \hbar \sqrt{\frac{k}{m^3}}$
   3. C.$\frac{L}{m r^2}=\sqrt{\frac{k}{m}}$
   4. D.$E=\frac{n \hbar}{2} \sqrt{\frac{k}{m}}$
   Show answer & solution

   Answer: A,B,C

   The central force will provide necessary centripetal force $\Rightarrow \mathrm{kr}=\frac{\mathrm{mv}^2}{\mathrm{r}}$ or, $\mathrm{kr}^2=\mathrm{mv}^2$ ...(1) By quantisation rule $\mathrm{n} \hbar=\mathrm{mvr}$ or, $\frac{\mathrm{n} \hbar}{\mathrm{r}}=\mathrm{mv}$ ...(2) $\begin{aligned} & \frac{(1)}{(2)^2} \Rightarrow \frac{\mathrm{kr}^2}{\frac{\mathrm{n}^2 \hbar^2}{\mathrm{r}^2}}=\frac{\mathrm{mv}^2}{\mathrm{~m}^2 \mathrm{v}^2} \\ & \Rightarrow \frac{\mathrm{k}}{\mathrm{n}^2 \hbar^2} \mathrm{r}^4=\frac{1}{\mathrm{~m}} \\ & \Rightarrow \mathrm{r}=\left(\frac{\mathrm{n}^2 \hbar^2}{\mathrm{~km}}\right)^{\frac{1}{4}} \Rightarrow \mathrm{r}^2=\frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}\end{aligned}$ $\begin{aligned} & \text {(2) Using (1), } \mathrm{K} \cdot \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}=\mathrm{mv}^2 \\ & \Rightarrow \mathrm{v}^2=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}^3}}\end{aligned}$ (3) $\frac{\mathrm{L}}{\mathrm{mr}^2}=\frac{\mathrm{mvr}}{\mathrm{mr}^2}=\frac{\mathrm{v}}{\mathrm{r}}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ from (1) $\begin{aligned} & \text{(4) } \mathrm{E}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{kr}^2=\frac{\mathrm{n} \hbar}{2} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}+\frac{1}{2} \mathrm{k} \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}} \\ & \mathrm{E}=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\end{aligned}$
3. Q3JEE Advanced Adv 2020 (Paper 1)
   A particle of mass $m$ moves in circular orbits with potential energy $U\left(r\right)=Fr$, where $F$ is a positive constant and $r$ is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle's orbit is denoted by $R$ and its speed and energy are denoted by $v$ and $E$ respectively, then for the ${n}^{th}$ orbit (here $h$ is the Planck's constant)
   1. A.$R\propto {n}^{1/3}$ and $v\propto {n}^{2/3}$
   2. B.$R\propto {n}^{2/3}$ and $v\propto {n}^{1/3}$
   3. C.$E=\dfrac{3}{2}{\left(\dfrac{{n}^{2}{h}^{2}{F}^{2}}{4{\pi }^{2}m}\right)}^{1/3}$
   4. D.$E=2{\left(\dfrac{{n}^{2}{h}^{2}{F}^{2}}{4{\pi }^{2}m}\right)}^{1/3}$
   Show answer & solution

   Answer: B,C

   $U=F\cdot r$ ${F}_{c}=-\dfrac{dU}{dr}=-F$ $\dfrac{m{v}^{2}}{r}=F$ $mur=\dfrac{nh}{2\pi }$ $\Rightarrow \dfrac{\left(mu{r}^{2}\right)}{mu\left(4r\right)}=\dfrac{{n}^{2}{h}^{2}}{4{\pi }^{2}\cdot F}$ $\Rightarrow m{r}^{3}=\dfrac{{n}^{2}{h}^{2}}{4{\pi }^{2}F}$ $\Rightarrow r={\left(\dfrac{{n}^{2}{h}^{2}}{4{\pi }^{2}mF}\right)}^{1/3}$ $\Rightarrow r\propto {n}^{2/3}$ $E=U+K=F\cdot r+1/2m{v}^{2}$ $=F\cdot r+1/2F\cdot r$ $E=3/2F\cdot r$ Put value of $r$, $E=\dfrac{3}{2}Fr{\left(\dfrac{{n}^{2}{h}^{2}}{4{\pi }^{2}mF}\right)}^{1/3}$
4. Q4JEE Advanced Adv 2019 (Paper 2)
   A free hydrogen atom after absorbing a photon of wavelength ${\lambda }_{a}$ gets excited from the state $n=1$ to the state $n=4.$ Immediately after that the electron jumps to $n=m$ state by emitting a photon of wavelength ${\lambda }_{e}.$ Let the change in momentum of atom due to the absorption and the emission are $\Delta {p}_{a}$ and $\Delta {p}_{e},$ respectively. If $\dfrac{{\lambda }_{a}}{{\lambda }_{e}}=\dfrac{1}{5}.$ Which of the option(s) is/are correct? [Use $hc=1242 eV nm, 1 nm={10}^{-9} m,$ $h$ and $c$ are Planck's constant and speed of light, respectively]
   1. A.${\lambda }_{e}=418 nm$
   2. B.The ratio of kinetic energy of the electron in the state $n=m$ to the state $n=1$ is $\dfrac{1}{4}$
   3. C.$m=2$
   4. D.$\dfrac{\Delta {p}_{a}}{\Delta {p}_{e}}=\dfrac{1}{2}$
   Show answer & solution

   Answer: B,C

   Energy for transition of electron from one orbit to other is given by- ${E}_{2}-{E}_{1}=\left(13.6\right){Z}^{2}\left(\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right)=\dfrac{hc}{\lambda }$ $\Rightarrow \dfrac{1}{\lambda }=\dfrac{13.6}{hc}.{Z}^{2}\left(\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right)$ Now as per question $\dfrac{1}{{\lambda }_{a}}=\dfrac{13.6}{hc}.{Z}^{2}\left(\dfrac{1}{{1}^{2}}-\dfrac{1}{{4}^{2}}\right)$ and $\dfrac{1}{{\lambda }_{e}}=\dfrac{13.6}{hc}.{Z}^{2}\left(\dfrac{1}{{m}^{2}}-\dfrac{1}{{4}^{2}}\right)$ $\Rightarrow \dfrac{{\lambda }_{e}}{{\lambda }_{a}}=\dfrac{\left(1-\dfrac{1}{16}\right)}{\dfrac{1}{{m}^{2}}-\dfrac{1}{16}}=\dfrac{\left(15\right){m}^{2}}{16-{m}^{2}}$ But $\dfrac{{\lambda }_{a}}{{\lambda }_{e}}=\dfrac{1}{5}\left(given\right)$ $\Rightarrow \dfrac{16-{m}^{2}}{15{m}^{2}}=\dfrac{1}{5}\Rightarrow 16-{m}^{2}=3{m}^{2}$ $\Rightarrow {m}^{2}=4$ $\Rightarrow m=2\Rightarrow \left(C\right)$ is correct Now, $\dfrac{1}{{\lambda }_{e}}=\dfrac{13.6}{hc}.{Z}^{2}\left(\dfrac{1}{{m}^{2}}-\dfrac{1}{16}\right)$ $=\dfrac{13.6}{1242}\times 1\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\left(ashc=1242andz=1\right)$ $\Rightarrow {\lambda }_{e}=\dfrac{1242}{13.6}\times \dfrac{16}{3}=487nm\Rightarrow \left(A\right)$ is incorrect Now, kinetic energy of electron $K\propto \dfrac{1}{{n}^{2}}$ $\Rightarrow \dfrac{{K}_{2}}{{K}_{1}}=\dfrac{{1}^{2}}{{2}^{2}}=\dfrac{1}{4}\Rightarrow \left(B\right)$ is correct Also, $∆P=\dfrac{h}{\lambda }$ $\Rightarrow \dfrac{∆{P}_{a}}{∆{P}_{e}}=\dfrac{{\lambda }_{e}}{{\lambda }_{a}}=5$ $\Rightarrow \left(D\right)$ is incorrect.
5. Q5JEE Advanced Adv 2018 (Paper 2)
   Consider a hydrogen-like ionized atom with atomic number $Z$ with a single electron. In the emission spectrum of this atom, the photon emitted in the $n=2ton=1$ transition has energy $74.8eV$ higher than the photon emitted in the $n=3ton=2$ transition. Given that the ionization energy of the hydrogen atom is $13.6eV$, what is the value of $Z$?
   Show answer & solution

   Answer: 3

   $∆{E}_{2\rightarrow 1}=13.6\times {Z}^{2}\left[1-\dfrac{1}{4}\right]=13.6\times {Z}^{2}\left[\dfrac{3}{4}\right]$ $∆{E}_{3\rightarrow 2}=13.6\times {Z}^{2}\left[\dfrac{1}{4}-\dfrac{1}{9}\right]=13.6\times {Z}^{2}\left[\dfrac{5}{36}\right]$ $∆{E}_{2\rightarrow 1}=∆{E}_{3-2}+74.8$ $13.6\times {Z}^{2}\left[\dfrac{3}{4}\right]=13.6\times {Z}^{2}\left[\dfrac{5}{36}\right]+74.8$ $13.6\times {Z}^{2}\left[\dfrac{3}{4}-\dfrac{5}{36}\right]=74.8$ ${Z}^{2}=9$ $Z=+3$