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If $a 0$ be the radius of first Bohr’s orbit of H-atom ,the de-Broglie’s wave

length of an electron revolving in the second Bohr’s orbit will be

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π a 0

π a 0

π a 0

π a 0

answer is B.

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Detailed Solution

∵ m v r n = n h 2 π

and

p = h λ

2 π r = n λ

∵ r = 4 a 0

2 π ⋅ 4 a 0 = 2 λ

λ = 4 a 0 π

h λ ⋅ r 2 = h π ⇒ λ = π r 2

∴ r 2 = 4 a 0

∴ λ = 4 a 0 π

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