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The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ a 0 is Bohr radius of first shell of hydrogen atom]

Options

  1. Ah 2 64 π 2 m a 0 2
  2. Bh 2 32 π 2 m a 0 2
  3. Ch 2 16 π 2 m a 0 2
  4. Dh 2 4 π 2 m a 0 2

Correct answer

B. h 2 32 π 2 m a 0 2

Step-by-step solution

K . E . = 1 2 m v 2 ...(i) mvr = n h 2 π v = n h 2 π m r ...(ii) Putting (ii) and (i) K . E . = 1 2 m n 2 h 2 4 π 2 m 2 r 2 = n 2 h 2 8 π 2 r 2 m Now r = 4 a 0 (since n = 2 ) and a 0 = Bohr's radius n 2 h 2 8 π 2 × 16 a 0 2 m = n 2 h 2 128 π 2 a 0 2 m Now n = 2 K . E . = 4 h 2 128 π 2 m a 0 2 = h 2 32 π 2 m a 0 2

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