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JEE Advanced Physics Atomic Physics 2026 JEE Advanced 2026 (Paper 1)

JEE Advanced Physics Question (2026) — Solution

Question

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 _0 the permittivity of the free space, the correct statement(s) is/are:

Options

  1. A. Magnitude of change in kinetic energy of electron can be expressed as h 4 | n v_n r_n - v_1 r_1 |.
  2. B. Magnitude of change in de Broglie wavelength of the electron can be expressed as e^2 4 _0 | 1 K_n - 1 K_1 |.
  3. C. Frequency of the radiation emitted can be expressed as e^2 8 _0 h ( 1 r_1 - 1 r_n ).
  4. D. Magnitude of change in total energy of the electron can be expressed as h 2 | v_1 r_1 - n v_n r_n |.

Answer

C. Frequency of the radiation emitted can be expressed as e^2 8 _0 h ( 1 r_1 - 1 r_n ).

Step-by-step solution

From Bohr's quantization condition, the angular momentum of an electron in the k^ th orbit is given by: m v_k r_k = k h 2 The kinetic energy of the electron in the k^ th orbit is: K_k = 1 2 m v_k^2 = 1 2 (m v_k r_k) v_k r_k Substituting the value of angular momentum: K_k = 1 2 ( k h 2 ) v_k r_k = k h v_k 4 r_k The magnitude of change in kinetic energy for a transition from the n^ th orbit to the 1^ st orbit is: | K| = |K_n - K_1| = h 4 | n v_n r_n - v_1 r_1 | This makes statement (A) correct. Since the total energy E_k = -K_k, the magnitude of change in total energy is equal to the magnitude of change in kinetic energy: | E| = | K| = h 4 | n v_n r_n - v_1 r_1 | This makes statement (D) incorrect. The electrostatic force provides the necessary centripetal force: m v_k^2 r_k = e^2 4 _0 r_k^2 K_k = 1 2 m v_k^2 = e^2 8 _0 r_k The total energy is E_k = -K_k = - e^2 8 _0 r_k . The energy of the emitted photon during the transition is: E = E_n - E_1 = - e^2 8 _0 r_n - ( - e^2 8 _0 r_1 ) = e^2 8 _0 ( 1 r_1 - 1 r_n ) The frequency of the emitted radiation is = E h : = e^2 8 _0 h ( 1 r_1 - 1 r_n ) This makes statement (C) correct. The de Broglie wavelength of the electron is _k = h m v_k . Using Bohr's quantization m v_k = k h 2 r_k , we get: _k = 2 r_k k From the kinetic energy relation r_k = e^2 8 _0 K_k , substituting r_k gives: _k = 2 k ( e^2 8 _0 K_k ) = e^2 4 _0 k K_k The magnitude of change in de Broglie wavelength is: | | = | _n - _1| = e^2 4 _0 | 1 n K_n - 1 K_1 | This makes statement (B) incorrect. Answer: Magnitude of change in kinetic energy of electron can be expressed as h 4 | n v_n r_n - v_1 r_1 |.; Frequency of the radiation emitted can be expressed as e^2 8 _0 h ( 1 r_1 - 1 r_n ).

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