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Question

If the average life time of an excited state of hydrogen is of the order of 108 s, then the number of revolutions an electron will make when it is in n = 2 state before coming to ground state will be [Take the radius of the first orbit of a hydrogen atom (a0)as 0.53 Å and all standard data if required]

A
109
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B
8×106
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C
2 × 103
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D
1010
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Solution

The correct option is B 8×106
Time period for nth energy level electron is,
T=2πrnvn=4π2mh×rn2n
rn=n2a0
T=4π2mh×n3a02
Required number of revolutions, N=108T
After substituting n=2, m=9.1×1031 kg,
and h=6.63×1034 Js,we get N=8×106

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