The time period of revolution of an electron in its ground state orbit in a hydrogen atom is `1.6xx10^(-16)` s. The frequency of the electron in its first excited state (in `s^(-1)` ) is :

Options

A. `7.8xx10^(14)`
B. `7.8xx10^(16)`
C. `3.7xx10^(14)`
D. `3.7xx10^(16)`

Answer

A

Explanation

Time period of electron in orbit of atom `Tprop(n^(3))/(z^(2))`
Suppose, time period is `T_(1)` in ground state and `T_(2)` in first excited state.
`:.n_(1)=1` and `T_(1)=1.6xx10^(-16)s`
and `n_(2)=2` and `T_(2)=?`
`:(T_(2))/(T_(1))=((n_(2))/(n_(1)))^(3)" "` (z=1 same )
`:.T_(2)=T_(1)xx((2)/(1))^(3)`
`=1.6xx10^(16)xx8`
`=12.8xx10^(-16)`
`:.` Frequency `v_(2)=(1)/(T_(2))=(1)/(12.8xx10^(-16))`
`:.v_(2)=0.0781xx10^(16)`
`:.v_(2)~~7.8xx10^(14)s^(-1)`