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The frequency of the radiation emitted w...

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

A

`1.54xx10^(15)s^(-1)`

B

`1.03xx10^(15)Js^(-1)`

C

`3.08xx10^(15)s^(-1)`

D

`2.0xx10^(15)s^(-1)`

Text Solution

Verified by Experts

`I.E.=E_(oo)-E_(1)=0-E_(1)`
`=2.18xx10^(18)J` ato`m^(-1)`
Thus, `E_(n)=-(2.18xx10^(-18))/(n^(2))Jmol^(-1)`
`DeltaE=E_(4)-E_(1)=-2.18xx10^(-18)(1/(4^(2))-1/(1^(2)))`
`=2.044xx10^(-18)J"ato"m^(-1)`.
`DeltaE=hv` or `v=(DeltaE)/h=(2.044xx10^(-18)J)`
`(6.625xx10^(-34)Js)`
`-3.085xx10^(15)s^(-1)`
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