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JEE Main201912 Apr 2019Evening ShiftPhysicsDual Nature of MatterActual

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å ). The de-Broglie wavelength of this electron is:

Options

  1. A12.9 Å
  2. B6.6 Å
  3. C9.7 Å
  4. D3.5 Å

Correct answer

C. 9.7 Å

Step-by-step solution

According to Bohr’s atomic model m v r = n h 2 π And the de-Broglie wavelength λ = h m v From the above two equations we can write, λ = 2 π r n In the 2 nd excited state λ = 2 π r 3 ∴ λ = 2 3 π × 4.65 Å = 9.7 Å

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