Calculate frequency of the radiation emitted when electron jumps from n=3 to n=2 in a hydrogen atom

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Calculate frequency of the radiation emitted when electron jumps from n=3 to n=2 in a hydrogen atom

Text Solution

Solution : In hydrogen spectrum the spectral lines are expressed are expressed in term of wave number `vecv` obey the following formula ltbr. Wave number `vecv=R_(H)((1)/( n_(i)^(2))-(1)/(n_(f)^(2)))` (where `R_(H)=` Rydberg constant `109677cm^(-1)`) <br> `vecv=109677cm^(-1)((1)/(2^(2))-(1)/(3^(2)))` <br> `vecv=109677xx(5)/(36)=15232g cm^(-1)` <br> `vecv=(1)/(lambda)` <br> or `lambda=(1)/(v)=(1)/(15232g)=6.564xx10^(-5)cm` <br> Wavelength, `lambda=6.564xx10^(-7)m` ltbr. Energy `E=(hc)/(lambda)` <br> `=(6.626xx10^(-34)Jsxx30xx10^(8)ms^(-1))/(6.564xx10^(-7)m)` <br> `=3.028xx10^(-19)J` <br> Frequency `v=(c)/(lambda)=(c)/(lambda)=(3.0xx10^(8)ms^(-1))/(6.564xx10^(-7)m)` <br> `=0.457xx10^(15)s^(-1)=4.57xx10^(14)s^(-1)`

Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.

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