Avarage lifetime of a hydrogen atom excited to `n =2` state `10^(-6)s` find the number of revolutions made by the electron on the average before it jump to the ground state

Answer

A, B

Explanation

`n=2`, `T=10^6 s`
Frequency `=(me^4)/(4in_0^2n^3h^2)`
So, Time period `=i/f`
`=(4in_0^2n^3h^2)/(me^4)`
`= (4 xx (3.85) xx (2)^2 xx (6.63)^3)/(9.3 xx (1.6)^4)`
`xx(10^-24 xx 10^-102)/(10^-76)`
`=8.2 xx 10^8` revolution.