Calculate the shortest and longest wavelength I hydrogen spectrum of Lyman series.
or Calculate the wavelength of the first line and the series limit for the Lyman series of hydrogen. `(R_(H)=109678cm^(-1))`.

Explanation

For Lyman series `n_(1)=1`
For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum i.e., `n_(2)=infty`
So, `(1)/(lamda)=R_(H)[(1)/(1^(2))-(1)/(infty^(2))]=R_(H)`
`lamda=(1)/(109678)=9.117xx10^(-6)cm`
=911.7Ã…
For longest wavelength in Lyman series (i.e., first line), the energy difference in two states showing transition should be minimum, i.e., `n_(2)=2`
So, `(1)/(lamda)=R_(H)[(1)/(1^(2))-(1)/(2^(2))]=(3)/(4)R_(H)`
or `lamda=(4)/(3)xx(1)/(R_(H))=(4)/(3xx109678)=1215.7xx10^(-8)cm`
=1215.7Ã…