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A free electron of 2 . 6 eV energy collides with a H + ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. h = 6 . 6 × 10 - 34 J s

Options

  1. A9 . 0 × 10 27   MHz
  2. B1 . 45 × 10 9   MHz
  3. C0 . 19 × 10 15   MHz
  4. D1 . 45 × 10 16   MHz

Correct answer

B. 1 . 45 × 10 9   MHz

Step-by-step solution

By Energy conservation K . E . e = T . E . H +   + E P h o t o n 2 . 6 = - 13 . 6 4 + h f h f = 6 eV f = 6 × 1 . 6 × 10 - 19 6 . 626 × 10 - 34 f = 1 . 45 × 10 15   Hz = 1 . 45 × 10 9   MHz

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