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

Answer

`10^(-8)xx(2.19xx10^(6))/(2pi(0.529xx10^(-10)))xx((1)^(2))/((2)^(3))=8.2xx10^(6)`

Explanation

Orbital frequency
`f=("Velocity")/(2pir)=(v_(n))/(2pir_(h))`
velocity of the electron in `n^(th)` orbital
`v_(n)=(2.2xx10^(6)Z)/(n)m//s(2.2xx10^(6)(1))/(2) " "n=2 " "z=1`
`1.1xx10^(6) ms^(-1)`
Now radius
`r_(n)=0.53xx10^(-10)(n^(2))/(z) =4xx0.53xx10^(-10)m=2.12xx10^(-10)m`
`f=` number of revolution in one `sec=(v_(n))/(2pir_(n))`
`:.` number of revolution
`N=fxxt=(1.1xx10^(6))/(2pixx2.12xx10^(-10))xx10^(-8)=8.2xx10^(6)` revolution