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

Explanation

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)`)
`vecv=109677cm^(-1)((1)/(2^(2))-(1)/(3^(2)))`
`vecv=109677xx(5)/(36)=15232g cm^(-1)`
`vecv=(1)/(lambda)`
or `lambda=(1)/(v)=(1)/(15232g)=6.564xx10^(-5)cm`
Wavelength, `lambda=6.564xx10^(-7)m` ltbr. Energy `E=(hc)/(lambda)`
`=(6.626xx10^(-34)Jsxx30xx10^(8)ms^(-1))/(6.564xx10^(-7)m)`
`=3.028xx10^(-19)J`
Frequency `v=(c)/(lambda)=(c)/(lambda)=(3.0xx10^(8)ms^(-1))/(6.564xx10^(-7)m)`
`=0.457xx10^(15)s^(-1)=4.57xx10^(14)s^(-1)`