The frequency of radiation emitted when the electron falls from n=4 to n=1 in a hydrogen atom will be (given ionisation energy of `H=2.18xx10^(18)J"atom"^(-1)` and `h=6.625xx10^(-25)Js`)

Options

A. `1.03xx10^(3)s^(-1)`
B. `3008xx10^(15)s^(-1)`
C. `2.00xx10^(15)s^(-1)`
D. `1.54xx10^(-15)s^(-1)`

Answer

B

Explanation

`E_(1)=2.18xx10^(-18)J"atom"^(-1)`
`E_(4)=(2.18xx10^(-18))/16=0.136xx10^(-18)J` atom
Energy released `=(2.18-0.136)xx10^(-18)J"atom"^(-1)`
`=2.044xx10^(-18)J"atom"^(-1)`
Now `hv=E`
`v=E/h=(2.044xx10^(-18))/(6.625xx10^(-34))=3.08xx10^(15)s^(-1)`