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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. AMagnitude of change in kinetic energy of electron can be expressed as h 4 | n v_n r_n - v₁ r₁ | .
  2. BMagnitude of change in de Broglie wavelength of the electron can be expressed as e^2 4 ₀ | 1 K_n - 1 K₁ | .
  3. CFrequency of the radiation emitted can be expressed as e^2 8 ₀ h ( 1 r₁ - 1 r_n ) .
  4. DMagnitude of change in total energy of the electron can be expressed as h 2 | v₁ r₁ - n v_n r_n | .

Correct answer

A. Magnitude of change in kinetic energy of electron can be expressed as h 4 | n v_n r_n - v₁ r₁ | .

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 - K₁| = h 4 | n v_n r_n - v₁ r₁ | This makes statement (A) correct. Since the total energy E_

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