Question

If the average life time of an excited state of hydrogen is of the order of 10−8 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 8×106Time period for nth energy level electron is, T=2πrnvn=4π2mh×rn2n rn=n2a0 T=4π2mh×n3a02 Required number of revolutions, N=10−8T After substituting n=2, m=9.1×10−31 kg, and h=6.63×10−34 J−s,we get N=8×106

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