If the average life time of an excited state of hydrogen is of the order of `10^(-8) s`, estimate how many whits an alectron makes when it is in the state `n = 2` and before it suffers a transition to state` n = 1 (Bohrredius a_(0) = 5.3 xx 10^(-11)m)`?

Options

A. `10^(7)`
B. `8 xx 10^(6)`
C. `2 xx 10^(5)`
D. none of these

Answer

B

Explanation

Time period for `n^(th)` energy level electron is,
`T = (2 pi r_(n))/(v_(n)) = (4 pi^(2) m)/(h) xx (r_(n)^(2))/(n)`
`r_(n) = n^(2) a_(0)`
`T = (4 pi^(2) m)/(h) xx n^(3) a_(0)^(2)`
Required number of revolutions, `N = (10^(-8))/(T)`
After substituting `n = 2, m = 9.1 xx 10^9-31) kg` and `h = 6.63 xx 10^(-34) J-s` , we get `N = 8 xx 10^(6)`.