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)`?

Explanation

The angular mumentum of an electctron in nth orbit is `nh//2pi`.
By Bohr's hypothesis: Again , `mvr = nh//2 pi` where `r` is the radius of the orbit.
or `nu = (nh)/(2 pi mr)` ...(i)
The time period of completing an orbit
`T = (2 pi r)/(nu) = (2 pi r(2 pi mr))/(nh)`
`T = (4 pi^(2) mr^(2))/(nh)` ...(ii)
Since the radius of the orbit is proportional to `n^(2)`, hence
`r = a_(0) n^(2)`
`:. T = (4 pi^(2) ma_(0)^(2) n^(4))/(nh) = (4 pi^(2) ma_(0)^(2) n^(3))/(h)`
Number of orbits completed in `10^(-8) s`
`= (10^(-8))/(T) = (10^(-8) xx h)/(4 pi^(2) ma_(0)^(2) n^(3))`
`= (10^(-8) xx (6.6 xx 10^(-34)))/(4 (3.14)^(2) (9.1 xx 10^(-31)) (5.3 xx 10^(-11)) ^(2) (2)^(3)) = 8 xx 10^(6)`