The frequency of radiation emiited when the electron falls n =4 to n=1 in a hydrogen atom will be ( given ionization energy of `H= 2.18 xx 10 ^(-18)J "atom "^(-1) 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^(2)eV`
= `[(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`
`becauseh=6.625xx10^(-34)`
`v=(13.6-0.85)/(6.625xx10^(-34))xx1.6xx10^(-19)=3.08xx10^(15)s^(-1)`.