Atomic Physics JEE Main previous year questions with solutions

4 solved JEE Main questions on Atomic Physics, free to read — no sign-in needed. The full chapter has 145 questions; sign in to attempt the remaining 141 in the exam simulator.

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1. Q1JEE Main 2026 (23 Jan, Shift 1)Atomic Spectrum
   In hydrogen atom spectrum, $(R \rightarrow$ Rydberg's constant $)$ A. the maximum wavelength of the radiation of Lyman series is $\frac{4}{3 R}$ B. the Balmer series lies in the visible region of the spectrum C. the minimum wavelength of the radiation of Paschen series is $\frac{9}{R}$ D. the minimum wavelength of Lyman series is $\frac{5}{4 R}$ Choose the correct answer from the options given below :
   1. A.B, D Only
   2. B.A, B and C Only
   3. C.A, B and D Only
   4. D.A, B Only
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   Answer: (B)

   A. Maximum wavelength of Lyman series ($n=1$ to $n=2$): $\frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$ $\lambda = \frac{4}{3R}$ TRUE B. The Balmer series lies in the visible region of the spectrum. TRUE C. Minimum wavelength of Paschen series ($n=3$ to $n=\infty$): $\frac{1}{\lambda'} = R\left(\frac{1}{9}\right)$ $\lambda' = \frac{9}{R}$ TRUE D. Minimum wavelength of Lyman series ($n=1$ to $n=\infty$): $\frac{1}{\lambda} = R\left(1\right)$ $\lambda = \frac{1}{R}$, NOT $\frac{5}{4R}$ FALSE So A, B and C are correct.
2. Q2JEE Main 2025 (02 Apr, Shift 1)Atomic Structure
   Considering Bohr's atomic model for hydrogen atom : (A) the energy of H atom in ground state is same as energy of $\mathrm{He}^{+}$ion in its first excited state. (B) the energy of H atom in ground state is same as that for $\mathrm{Li}^{++}$ion in its second excited state. (C) the energy of H atom in its ground state is same as that of $\mathrm{He}^{+}$ion for its ground state. (D) the energy of $\mathrm{He}^{+}$ion in its first excited state is same as that for $\mathrm{Li}^{++}$ion in its ground state Choose the correct answer from the options given below :
   1. A.(B), (D) only
   2. B.(A), (B) only
   3. C.(A), (D) only
   4. D.(A), (C) only
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   Answer: (B)

   $\begin{aligned} & \mathrm{E} \propto \frac{\mathrm{Z}}{\mathrm{n}^2} \\ & \mathrm{Z}_{\mathrm{H}}=1 \quad \mathrm{Z}_{\mathrm{He}^{+}}=2 \quad \mathrm{Z}_{\mathrm{Li}^{+2}}=3\end{aligned}$ $\begin{aligned} & 1^{\text {st }} \text { excited state } \Rightarrow \mathrm{n}=2 \\ & 2^{\text {nd }} \text { excited state } \Rightarrow \mathrm{n}=3\end{aligned}$ From the given statements only A \& B are correct.
3. Q3JEE Main 2024 (29 Jan, Shift 2)Atomic Models
   Given below are two statements: Statement I: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model. Statement II: An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model. In the light of the above statements, choose the most appropriate from the options given below.
   1. A.Both statement I and statement II are false
   2. B.Statement I is false but statement II is true
   3. C.Statement I is true but statement II is false
   4. D.Both statement I and statement II are true
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   Answer: (C)

   Rutherford proposed that most of the mass of an atom is concentrated in a small, dense nucleus at the centre, which contains positively charged protons. Electrons, carrying negative charge, revolve around the nucleus in orbits.In the Bohr model, electrons do not move in random orbits but have specific energy levels or shells around the nucleus. The electrons are not in a "spherical cloud" but are found in discrete orbits at fixed distances from the nucleus. Rutherford's model, on the other hand, did not specify the fixed orbits of electrons and was later refined by Bohr to explain the stability of atoms.Hence, statement I is true but statement II is false.
4. Q4JEE Main 2023 (13 Apr, Shift 2)Bohr's Atomic Model
   An atom absorbs a photon of wavelength $500nm$ and emits another photon of wavelength $600nm$. The net energy absorbed by the atom in this process is $n\times {10}^{-4}eV$. The value of$n$is [Assume the atom to be stationary during the absorption and emission process] (Take $h=6.6\times {10}^{-34}Js$ $and c=3×108 m s-1$ ).
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   Answer: 4125

   The energy of a photon is given by $E=\dfrac{hc}{\lambda }$. It is given that ${\lambda }_{1}=500nm$, ${\lambda }_{2}=600nm$. The net energy absorbed is $\Delta E=\dfrac{hc}{{\lambda }_{1}}-\dfrac{hc}{{\lambda }_{2}}=\dfrac{hc}{{10}^{-9}}(\dfrac{1}{500}-\dfrac{1}{600})$ $=\dfrac{6.6\times {10}^{-34}\times 3\times {10}^{8}\times 100}{500\times 600\times {10}^{-9}}$ $=\dfrac{6.6\times 3}{30}\times {10}^{-19}J$ $1eV=1.6\times {10}^{-19}J$ $=\dfrac{6.6\times 3}{30\times 1.6}eV$ $=4125\times {10}^{-4}eV$