Atomic Physics JEE Main previous year questions with solutions

5 solved JEE Main questions on Atomic Physics, free to read — no sign-in needed. The full chapter has 154 questions; sign in to attempt the remaining 149 in the exam simulator.

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
   Show answer & solution

   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 (29 Jan, Shift 1)Atomic Models
   Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance. Reason (R) : A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below :
   1. A.(A) is false but (R) is true
   2. B.(A) is true but (R) is false
   3. C.Both $(\mathbf{A})$ and $(\mathbf{R})$ are true and $(\mathbf{R})$ is the correct explanation of $(\mathbf{A})$
   4. D.Both $(\mathbf{A})$ and $(\mathbf{R})$ are true but $(\mathbf{R})$ is not the correct explanation of $(\mathbf{A})$
   Show answer & solution

   Answer: (D)

   Assertion (A) is true because a sufficiently negative potential can repel and stop the emitted electrons from leaving the surface. Reason (R) is also true because the stopping potential $V_{\text {stop }}$ varies linearly with the frequency $\nu$ of the incident light $\left(V_{\text {stop }}=\frac{h}{e} \nu-\frac{\phi}{e}\right)$. However, (R) does not directly explain why a negative potential suppresses electron emission; it only shows how much potential is needed for different frequencies. Therefore, both (A) and (R) are true, but (R) is not the correct explanation of (A).
3. Q3JEE Main 2024 (08 Apr, Shift 1)Atomic Structure
   In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number 80 , then maximum velocity of $\alpha$ - particle is _______ $\times 10^5 \mathrm{~m} / \mathrm{s}$ approximately. $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9\right.$ SI unit, mass of $\alpha$ particle $\left.=6.72 \times 10^{-27} \mathrm{~kg}\right)$
   Show answer & solution

   Answer: 156

   $\begin{aligned} v & =\sqrt{\frac{4 \mathrm{KZe}^2}{\mathrm{mr}_{\text {min }}}} \\ & =\sqrt{\frac{4 \times 9 \times 10^9 \times 80}{6.72 \times 10^{-27} \times 4.5 \times 10^{-14}}} \times 1.6 \times 10^{-19} \\ & =9.759 \times 10^{25} \times 1.6 \times 10^{-19} \\ & =156 \times 10^5 \mathrm{~m} / \mathrm{s}\end{aligned}$
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 $h=6.6\times {10}^{-34}Js$ $and c=3×108 m s-1$ ).
   Show answer & solution

   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 $\Delta E=\dfrac{hc}{{\lambda }_{1}}-\dfrac{hc}{{\lambda }_{2}}=\dfrac{hc}{{10}^{-9}}(\dfrac{1}{500}-\dfrac{1}{600})$ $=\dfrac{6.6\times {10}^{-34}\times 3\times {10}^{8}\times 100}{500\times 600\times {10}^{-9}}$ $=\dfrac{6.6\times 3}{30}\times {10}^{-19}J$ $1eV=1.6\times {10}^{-19}J$ $=\dfrac{6.6\times 3}{30\times 1.6}eV$ $=4125\times {10}^{-4}eV$
5. Q5JEE Main 2026 (22 Jan, Shift 2)Atomic Spectrum
   The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly $\_\_\_\_$ nm .
   1. A.1550
   2. B.1217
   3. C.1875
   4. D.1784
   Show answer & solution

   Answer: (B)

   Smallest wavelength of lyman $\dfrac{1}{\lambda} = R\left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right)$ $R = \dfrac{1}{\lambda} = \dfrac{1}{91} \text{ nm}^{-1}$ $\lambda_{max}$ for balmer series $n_1 = 2 \rightarrow n_2 = 3$ $\dfrac{1}{\lambda_B} = R\left(\dfrac{1}{4} - \dfrac{1}{9}\right)$ $\dfrac{1}{\lambda_B} = \dfrac{1}{91}\left(\dfrac{5}{36}\right)$ $\lambda_B = \left(\dfrac{91 \times 36}{5}\right) = 655.2 \text{ nm}$ $\lambda_{max}$ paschen $n_1 = 3 \rightarrow n_2 = 4$ $\dfrac{1}{\lambda_p} = \dfrac{1}{91}\left(\dfrac{1}{3^2} - \dfrac{1}{4^2}\right) = \dfrac{1}{91} \times \dfrac{7}{144}$ $\lambda_p = \left(\dfrac{91 \times 144}{7}\right) = 1872 \text{ nm}$ $\Delta\lambda = \lambda_P - \lambda_B = 1872 - 655.2$ $\Delta\lambda = 1216.8$ $\Delta\lambda \approx 1217$