Atomic StructureHard

Question

The average lifetime of an electron in an excited state of hydrogen atom is about 10–8 s. How many revolutions does an electron in the n = 2 state make before dropping to the n = 1 state?

Options

A.108
B.8.33 × 106
C.6.67 × 107
D.1.04 × 106

Solution

$N = \frac{10^{- 8}\text{ sec}}{T_{n,z}} = \frac{10^{- 8}}{1.5 \times 10^{- 16} \times \frac{2^{3}}{1^{2}}} = 8.33 \times 10^{6}$

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