How many revolutions does an electron in the n = 2 state of a hydrogen atom make before dropping to the n = 1 state ? The average lifetime of an excited state is `10^(-8)` s. Bohr radius `0.53 Å` and velocity` V_(1)=2.19xx10^(6)m//s`

Explanation

Velocity of electron in the nth orbit of hydrogen atom
`v_(n)=(v_(1))/(n)`
`=(2.19xx10^(6))/(n)`
`:.v_(2)=(2.19xx10^(6))/(2)=1.095xx10^(6)m//s`
and radius in `n^(th)` orbit
`r_(n)=n^(2)r_(1)`
`=(2)^(2)xx0.53Å`
`=4xx0.53Å`
`=2.12Å`
`rArr` No. of revolution in one second
`v=(v_(n))/(2pi r_(n))=(v_(2))/(2pi r_(2))`
`:.v=(1.095xx10^(6))/(2xx3.14xx2.12xx10^(-10))`
`0.0822xx10^(16)s^(-1)`
`rArr` No. of revolution in `10^(-8)` second
`N=vxx t`
`=0.0822xx10^(16)xx10^(-8)`
`:.N=8.22xx10^(6)` revolution.