JEE Advanced 2026 Paper 1, Physics Section 2 Q1: The Line Spectra of the Hydrogen Atom

JEE Advanced2026Paper 1Physics Section 2

Consider a hydrogen atom with $v_{k}$ , $r_{k}$ and $K_{k}$ denoting the velocity, orbital radius and kinetic energy of the electron in the $k^{th}$ orbit, respectively. The electron undergoes a transition from the $n^{th}$ orbit, emitting radiation corresponding to the Lyman series. Considering $h$ to be the Planck's constant and $ϵ_{0}$ the permittivity of the free space, the correct statement(s) is/are:

A
Magnitude of change in kinetic energy of electron can be expressed as $\frac{h}{4 π} \frac{n v _{n}}{r _{n}} - \frac{v _{1}}{r _{1}}$ .
B
Magnitude of change in de Broglie wavelength of the electron can be expressed as $\frac{e ^{2}}{4 ϵ _{0}} \frac{1}{K _{n}} - \frac{1}{K _{1}}$ .
C
Frequency of the radiation emitted can be expressed as $\frac{e ^{2}}{8 π ϵ _{0} h} (\frac{1}{r _{1}} - \frac{1}{r _{n}})$ .
D
Magnitude of change in total energy of the electron can be expressed as $\frac{h}{2 π} \frac{v _{1}}{r _{1}} - \frac{n v _{n}}{r _{n}}$ .

Solution

Core Bohr relations. Coulomb force as centripetal force: $\frac{m v _{k}^{2}}{r _{k}} = \frac{e ^{2}}{4 π ϵ _{0} r _{k}^{2}}$ , so $m v_{k}^{2} r_{k} = \frac{e ^{2}}{4 π ϵ _{0}}$ . Quantization: $m v_{k} r_{k} = \frac{k h}{2 π}$ . Kinetic energy: $K_{k} = \frac{1}{2} m v_{k}^{2} = \frac{e ^{2}}{8 π ϵ _{0} r _{k}}$ .

Option (A). $\frac{h}{4 π} \frac{n v _{n}}{r _{n}} - \frac{v _{1}}{r _{1}} = \frac{1}{2} m v_{n}^{2} - m v_{1}^{2} \cdot \frac{1}{1}$ , after using $m v_{k} r_{k} = k h / (2 π)$ . This equals $∣ K_{n} - K_{1} ∣ = Δ K$ . Correct.

Option (B). The given expression evaluates to $λ_{n}^{2} - λ_{1}^{2}$ -type combination rather than $∣ λ_{n} - λ_{1} ∣$ , so it does not match the change in de Broglie wavelength. Incorrect.

Option (C). Lyman emission frequency: $h f = K_{1} - K_{n} = \frac{e ^{2}}{8 π ϵ _{0}} (\frac{1}{r _{1}} - \frac{1}{r _{n}})$ , so $f = \frac{e ^{2}}{8 π ϵ _{0} h} (\frac{1}{r _{1}} - \frac{1}{r _{n}})$ . Correct.

Option (D). The expression evaluates to $∣ m v_{1}^{2} - m v_{n}^{2} ∣ = 2∣ K_{1} - K_{n} ∣$ , which equals $2Δ E$ rather than $Δ E$ . Incorrect.

Correct answers: (A) and (C).

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JEE Advanced 2026 Paper 1, Physics Section 2 Q1: The Line Spectra of the Hydrogen Atom

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