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📘 Atomic Physics — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Atomic Physics for JEE Main with previous year questions on Bohr's model, energy levels, and atomic spectra. Each solution is explained in a simple, step-by-step manner to boost your score.
Atomic Physics — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (4 April Shift 1)
Considering the Bohr model of hydrogen like atoms, the ratio of the radius of the \(5^{th}\) orbit of the electron in \(Li^{2+}\) and \(He^+\) is:
(1) \(\frac{3}{2}\) (2) \(\frac{4}{9}\) (3) \(\frac{9}{4}\) (4) \(\frac{2}{3}\)
(1) \(\frac{3}{2}\) (2) \(\frac{4}{9}\) (3) \(\frac{9}{4}\) (4) \(\frac{2}{3}\)
✅ Answer & Explanation:
Correct Option: (4)
- Bohr radius formula: \(r_n \propto \frac{n^2}{Z}\).
- Since \(n=5\) for both, \(\frac{r_{Li}}{r_{He}} = \frac{Z_{He}}{Z_{Li}}\).
- Atomic numbers: \(Z_{He} = 2\), \(Z_{Li} = 3\).
- Ratio \(= \mathbf{2 : 3}\).
Q2. JEE Main 2025 (3 April Shift 2)
Assertion A: The Bohr model is applicable to hydrogen and hydrogen-like atoms only.
Reason R: The formulation of Bohr model does not include repulsive force between electrons.
(1) Both A and R are true but R is NOT the correct explanation of A.
(2) A is false but R is true.
(3) Both A and R are true and R is the correct explanation of A.
(4) A is true but R is false.
Reason R: The formulation of Bohr model does not include repulsive force between electrons.
(1) Both A and R are true but R is NOT the correct explanation of A.
(2) A is false but R is true.
(3) Both A and R are true and R is the correct explanation of A.
(4) A is true but R is false.
✅ Answer & Explanation:
Correct Option: (3)
The Bohr model only considers the attractive force between a single electron and the nucleus. It cannot account for the complex inter-electronic repulsions found in multi-electron atoms. Thus, it is restricted to hydrogen-like systems (single electron).
The Bohr model only considers the attractive force between a single electron and the nucleus. It cannot account for the complex inter-electronic repulsions found in multi-electron atoms. Thus, it is restricted to hydrogen-like systems (single electron).
Q3. JEE Main 2025 (2 April Shift 2)
Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of \(Li^{++}\) ion in its ground state is given by \(\frac{1}{X} a_0\), where \(X\) is _______. (Where \(a_0\) is the first Bohr's radius.)
✅ Answer & Explanation:
Correct Answer: 3
- Radius \(r = \frac{n^2}{Z} a_0\).
- For \(Li^{++}\) ground state: \(n = 1\), \(Z = 3\).
- \(r = \frac{1^2}{3} a_0 = \frac{1}{3} a_0\).
- Thus, X = 3.
Q4. JEE Main 2025 (2 April Shift 1)
Considering Bohr's atomic model for hydrogen atom :
(A) the energy of H atom in ground state is same as energy of \(He^+\) ion in its first excited state.
(B) the energy of H atom in ground state is same as that for \(Li^{++}\) ion in its second excited state.
(C) the energy of H atom in its ground state is same as that of \(He^+\) ion for its ground state.
(D) the energy of \(He^+\) ion in its first excited state is same as that for \(Li^{++}\) ion in its ground state.
Choose the correct answer: (1) (B), (D) only (2) (A), (B) only (3) (A), (D) only (4) (A), (C) only
(A) the energy of H atom in ground state is same as energy of \(He^+\) ion in its first excited state.
(B) the energy of H atom in ground state is same as that for \(Li^{++}\) ion in its second excited state.
(C) the energy of H atom in its ground state is same as that of \(He^+\) ion for its ground state.
(D) the energy of \(He^+\) ion in its first excited state is same as that for \(Li^{++}\) ion in its ground state.
Choose the correct answer: (1) (B), (D) only (2) (A), (B) only (3) (A), (D) only (4) (A), (C) only
✅ Answer & Explanation:
Correct Option: (2)
- Energy \(E = -13.6 \frac{Z^2}{n^2} \, \text{eV}\).
- H ground: \(-13.6(\frac{1}{1}) = -13.6\).
- \(He^+\) 1st excited (\(n=2, Z=2\)): \(-13.6(\frac{4}{4}) = -13.6\). (A is true).
- \(Li^{++}\) 2nd excited (\(n=3, Z=3\)): \(-13.6(\frac{9}{9}) = -13.6\). (B is true).
- Thus, (A) and (B) are correct.
Q5. JEE Main 2023 (01 Feb Shift 1)
A light of energy 12.75 eV is incident on a hydrogen atom in its ground state. The angular momentum of the atom in the resulting excited state is \(\frac{x}{\pi} \times 10^{-17} \, \text{eV}\cdot\text{s}\). The value of \(x\) is ______ . (use \(h = 4.14 \times 10^{-15} \, \text{eV}\cdot\text{s}\))
✅ Answer & Explanation:
Correct Answer: 828
- Energy 12.75 eV excites electron from \(n=1\) to \(n=4\) (\(\Delta E = 13.6(1 - \frac{1}{16}) = 12.75\)).
- Angular momentum \(L = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}\).
- \(L = \frac{2 \times 4.14 \times 10^{-15}}{\pi} = \frac{8.28 \times 10^{-15}}{\pi} = \frac{828 \times 10^{-17}}{\pi}\).
- Thus, x = 828.
Q6. JEE Main 2022 (27 Jun Shift 2)
Statement I: In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (\(E_1\)) to higher energy orbit (\(E_2\)), is given as \(hf = E_1 - E_2\).
Statement II: The jumping of electron from higher energy orbit (\(E_2\)) to lower energy orbit (\(E_1\)) is associated with frequency \(f = (E_2 - E_1)/h\). This is Bohr's frequency condition.
(1) Both statement I and II are true. (2) Both false. (3) I is true, II is false. (4) I is incorrect, II is true.
Statement II: The jumping of electron from higher energy orbit (\(E_2\)) to lower energy orbit (\(E_1\)) is associated with frequency \(f = (E_2 - E_1)/h\). This is Bohr's frequency condition.
(1) Both statement I and II are true. (2) Both false. (3) I is true, II is false. (4) I is incorrect, II is true.
✅ Answer & Explanation:
Correct Option: (4)
Statement I is false because jumping from lower to higher energy requires absorption of energy, and the formula given results in a negative frequency. Statement II is the correct definition of the Bohr frequency condition for emission.
Statement I is false because jumping from lower to higher energy requires absorption of energy, and the formula given results in a negative frequency. Statement II is the correct definition of the Bohr frequency condition for emission.
Q7. JEE Main 2022 (24 Jun Shift 2)
In Bohr's model, choose the correct option when the electron undergoes transitions to a higher level :
(1) All K, P and E increase. (2) K decreases, P and E increase.
(3) P decreases, K and E increase. (4) K increases, P and E decrease.
(1) All K, P and E increase. (2) K decreases, P and E increase.
(3) P decreases, K and E increase. (4) K increases, P and E decrease.
✅ Answer & Explanation:
Correct Option: (2)
- Kinetic Energy \(K \propto 1/n^2\) → decreases.
- Potential Energy \(P = -2K\) → becomes less negative → increases.
- Total Energy \(E = -K\) → becomes less negative → increases.
Q8. JEE Main 2025 (7 April Shift 1)
For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is:
(1) 5 : 36 (2) 5 : 27 (3) 3 : 4 (4) 27 : 5
(1) 5 : 36 (2) 5 : 27 (3) 3 : 4 (4) 27 : 5
✅ Answer & Explanation:
Correct Option: (2)
- Largest wavelength in Lyman series (2 → 1): \(\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R}\).
- Largest wavelength in Balmer series (3 → 2): \(\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{5R}{36} \implies \lambda_B = \frac{36}{5R}\).
- Ratio \(\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}\).
Q9. JEE Main 2025 (24 Jan Shift 1)
During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is \(2000 \, \text{\AA}\) and it becomes \(6000 \, \text{\AA}\) when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is:
(1) \(4000 \, \text{\AA}\) (2) \(2000 \, \text{\AA}\) (3) \(3000 \, \text{\AA}\) (4) \(6000 \, \text{\AA}\)
(1) \(4000 \, \text{\AA}\) (2) \(2000 \, \text{\AA}\) (3) \(3000 \, \text{\AA}\) (4) \(6000 \, \text{\AA}\)
✅ Answer & Explanation:
Correct Option: (3)
- Energy relation: \(E_{AC} = E_{AB} + E_{BC}\).
- In terms of wavelength: \(\frac{hc}{\lambda_{AC}} = \frac{hc}{\lambda_{AB}} + \frac{hc}{\lambda_{BC}}\).
- \(\frac{1}{2000} = \frac{1}{\lambda_{AB}} + \frac{1}{6000}\).
- \(\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000} = \frac{3-1}{6000} = \frac{2}{6000} = \frac{1}{3000}\).
- \(\lambda_{AB} = \mathbf{3000 \, \text{\AA}}\).
Q10. JEE Main 2024 (29 Jan Shift 2)
The minimum potential difference needed to observe any Balmer series lines in the emission spectra of hydrogen atom is \(\frac{\alpha}{10} \, \text{V}\), where \(\alpha\) is _________. (Assume atom starts in ground state).
✅ Answer & Explanation:
Correct Answer: 121
- To emit Balmer lines, the electron must first be excited to at least the \(n=3\) level (so it can fall to \(n=2\)).
- Excitation energy \(E_{1 \to 3} = 13.6 \left( 1 - \frac{1}{9} \right) = 13.6 \times \frac{8}{9} \approx 12.088 \, \text{eV}\).
- Potential difference \(V \approx 12.1 \, \text{V}\).
- Comparing with \(\alpha/10\), we get \(\alpha = 121\).
Q11. JEE Main 2024 (06 Apr Shift 1)
The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :
(1) 4 : 1 (2) 1 : 4 (3) 2 : 1 (4) 1 : 2
(1) 4 : 1 (2) 1 : 4 (3) 2 : 1 (4) 1 : 2
✅ Answer & Explanation:
Correct Option: (1)
- Shortest wavelength occurs when \(n_2 = \infty\).
- Shortest Lyman: \(\frac{1}{\lambda_L} = R(1/1^2 - 0) = R \implies \lambda_L = \frac{1}{R}\).
- Shortest Balmer: \(\frac{1}{\lambda_B} = R(1/2^2 - 0) = \frac{R}{4} \implies \lambda_B = \frac{4}{R}\).
- Ratio \(\frac{\lambda_B}{\lambda_L} = \mathbf{4 : 1}\).
Q12. JEE Main 2022 (29 Jul Shift 1)
Find the ratio of energies of photons produced due to transition of an election of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
(1) 3 : 4 (2) 4 : 3 (3) 1 : 4 (4) 4 : 1
(1) 3 : 4 (2) 4 : 3 (3) 1 : 4 (4) 4 : 1
✅ Answer & Explanation:
Correct Option: (1)
- \(E_1\) (2 → 1): \(13.6(1 - \frac{1}{4}) = 13.6(\frac{3}{4})\).
- \(E_2\) (\(\infty\) → 1): \(13.6(1 - 0) = 13.6\).
- Ratio \(E_1 / E_2 = \mathbf{3 / 4}\).
Q13. JEE Main 2022 (25 Jul Shift 2)
\(\frac{x}{x+4}\) is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) third level to second level and (ii) highest level to second level. Value of \(x\) is:
✅ Answer & Explanation:
Correct Answer: 5
- \(E_1\) (3 → 2): \(13.6(\frac{1}{4} - \frac{1}{9}) = 13.6(\frac{5}{36})\).
- \(E_2\) (\(\infty\) → 2): \(13.6(\frac{1}{4} - 0) = 13.6(\frac{1}{4})\).
- Ratio \(\frac{E_1}{E_2} = \frac{\frac{5}{36}}{\frac{1}{4}} = \frac{5}{9}\).
- Equating: \(\frac{x}{x+4} = \frac{5}{9} \implies 9x = 5x + 20 \implies 4x = 20 \implies \mathbf{x = 5}\).
Q14. JEE Main 2021 (16 Mar Shift 1)
The wavelengths of first and third spectral lines of H-atom in Balmer series (\(\lambda_1, \lambda_3\)) are related by a factor of approximately \(x \times 10^{-1}\). The value of \(x\) is ______ .
✅ Answer & Explanation:
Correct Answer: 15
- \(\lambda_1\) (3 → 2): \(1/\lambda_1 = R(\frac{1}{4} - \frac{1}{9}) = 5R/36 \implies \lambda_1 = 36/5R = \frac{7.2}{R}\).
- \(\lambda_3\) (5 → 2): \(1/\lambda_3 = R(\frac{1}{4} - \frac{1}{25}) = 21R/100 \implies \lambda_3 = 100/21R \approx \frac{4.76}{R}\).
- Ratio \(\lambda_1 / \lambda_3 = 7.2 / 4.76 \approx 1.512 = 15.12 \times 10^{-1}\).
- Rounding gives x = 15.
Q15. JEE Main 2025 (24 Jan Shift 2)
Arrange the following in the ascending order of wavelength \((\lambda)\):
(A) Microwaves \((\lambda_1)\) (B) Ultraviolet rays \((\lambda_2)\)
(C) Infrared rays \((\lambda_3)\) (D) X-rays \((\lambda_4)\)
(1) \(\lambda_4 < \lambda_3 < \lambda_2 < \lambda_1\) (2) \(\lambda_3 < \lambda_4 < \lambda_2 < \lambda_1\)
(3) \(\lambda_4 < \lambda_3 < \lambda_1 < \lambda_2\) (4) \(\lambda_4 < \lambda_2 < \lambda_3 < \lambda_1\)
(A) Microwaves \((\lambda_1)\) (B) Ultraviolet rays \((\lambda_2)\)
(C) Infrared rays \((\lambda_3)\) (D) X-rays \((\lambda_4)\)
(1) \(\lambda_4 < \lambda_3 < \lambda_2 < \lambda_1\) (2) \(\lambda_3 < \lambda_4 < \lambda_2 < \lambda_1\)
(3) \(\lambda_4 < \lambda_3 < \lambda_1 < \lambda_2\) (4) \(\lambda_4 < \lambda_2 < \lambda_3 < \lambda_1\)
✅ Answer & Explanation:
Correct Option: (4)
- According to the Electromagnetic Spectrum, the order of increasing wavelength is:
- X-rays \((\lambda_4)\) < Ultraviolet \((\lambda_2)\) < Infrared \((\lambda_3)\) < Microwaves \((\lambda_1)\).
Q16. JEE Main 2022 (25 Jun Shift 2)
Which figure represents the variation of \(\ln(R/R_0)\) with \(\ln A\) (if \(R\) is nucleus radius and \(A\) is mass number)?
Choose correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (2)
- Formula: \(R = R_0 A^{\frac{1}{3}} \implies \frac{R}{R_0} = A^{\frac{1}{3}}\).
- Taking log: \(\ln(R/R_0) = \frac{1}{3} \ln A\).
- This is a straight line equation \(y = mx\) with slope 1/3. Graph (2) correctly represents this linear relationship passing through the origin.
📘 Exam Preparation Tip for Atomic Physics
- Bohr's Postulates: Electrons revolve in stable orbits without radiating energy. Angular momentum is quantized: \(L = \frac{nh}{2\pi}\). Energy is emitted or absorbed when electrons jump between orbits.
- Bohr's Formulas: - Radius: \(r_n = \frac{n^2}{Z} a_0\), where \(a_0 = 0.529 \,\text{\AA}\). - Velocity: \(v_n = \frac{Z}{n} \frac{e^2}{2\epsilon_0 h}\). - Energy: \(E_n = -13.6 \frac{Z^2}{n^2} \, \text{eV}\).
- Hydrogen Spectrum: Rydberg formula: \(\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\). Series: Lyman (\(n_1 = 1\)), Balmer (\(n_1 = 2\)), Paschen (\(n_1 = 3\)), Brackett (\(n_1 = 4\)), Pfund (\(n_1 = 5\)).
- Energy, Frequency, Wavelength: \(E = hf = \frac{hc}{\lambda}\). In electron volts: \(E(eV) = \frac{1240}{\lambda(\text{nm})}\).
- Excitation and Ionization: Excitation energy is the energy required to raise an electron to a higher level. Ionization energy is the energy required to completely remove the electron from the atom (from ground state to \(n = \infty\)).