1.

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 atom^(-1) and h = 6.625 xx 10^(-34) Js)

Answer»

`1.54xx10^(15)s^(-1)`
`1.03xx10^(15)JS^(-1)`
`3.08xx10^(15)s^(-1)`
`2.0xx10^(15)s^(-1)`

Solution :`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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