1.

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)

Answer»

`1.03xx10^(3)s^(-1)`
`3008xx10^(15)s^(-1)`
`2.00xx10^(15)s^(-1)`
`1.54xx10^(-15)s^(-1)`

Solution :`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)`


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