Calculate frequency, energy and wavelength of the radiation corresponding to the speciral line of the lowest frequency in lyman series in the spectrum of a hydrogen atom .Also calculate the energy for the coresponding line in the spectrum of `Li^(2+).(R_(H) = 109677 cm^(-1), c = 3 xx 10^(8) m s^(-1), Z = 3)`

Answer

`v=2.469xx10^(15)s^(-1) ; E= 1.64xx10^(-18)J ; lambda=1.215xx10^(-7)m`

Explanation

The intensity of spectral lines decreases as the wavelength decreases and the frequency increases. The line with lowest frequency corresponds to transition from `n_(H)=2 "to" n_(L)=1`
`bar(v)=(1)/(lambda)=R[(1)/(n_(L^(2)))-(1)/(n_(H^(2)))]`
`(1)/(lambda)bar(v)=1.097xx10^(7)[(1)/(1^(2))-(1)/(2^(2))]m^(-1)=8.23xx10^(6)m^(-1)`
`lamdba=(1)/(8.23xx10^(6))m=1.215xx10^(-7)m`
`v=(C)/(lambda)=((3.0xx10^(8)ms^(-1)))/((1.215xx10^(-7)m))=2.469xx10^(15)s^(-1)`
Energy (E)`=hv=(6.626xx10^(-34)Js)xx(2.469xx10^(15)s^(-1))=1.64xx10^(-18)J`.