The frequency of radiaiton emitted when the electron falls form `n =4` to `n = 1` in a hydrogen atom will be (Given ionization enegry of `H =2.18 xx 10^(-18)J `and `h= 6.625 xx 10^(-34)Js)`

Options

A. `3.08xx10^15 s^(-1)`
B. `2.00xx10^15 s^(-1)`
C. `1.54xx10^15 s^(-1)`
D. `1.03xx10^15 s^(-1)`

Answer

A

Explanation

`E_"Ionisation"=E_oo-E_n=(13.6Z_"eff"^2)/n^2eV`
`=[(13.6Z^2)/n_2^2-(13.6Z^2)/n_1^2]`
`E=hv=(13.6xx1^2)/(1)^2-(13.6xx1)^2 /(4)^2 , hv =13.6-0.85`
`because h=6.625xx10^(-34)`
`v=(13.6-0.85)/(6.625xx10^(-34))xx1.6xx10^(-19)=3.08xx10^15 s^(-1)`