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Calculate the shortest and longest wavelength in hydrogen spectrum of Lyman series.

Answer» 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) = oo`
`(1)/(lambda) = R_(H) [(1)/(1^(2))-(1)/((oo)^(2))] = R_(H)`
`lambda = (1)/(3xx 109678) = 9.11 xx 10^(-6) cm = 911. 7 A^(@)`
For longest wavlength in Lyman series (i,e. first line the energy difference in two states showing transition should be minimum, i.e., `n_(2) = 2`
`(1)/(lambda) = R_(H) [(1)/(1^(2))-(1)/((2)^(2))] = (3)/(4) R_(H)`
`lambda =(4)/(3) (1)/(R_(H)) = (4)/(3 xx 109677)`
`= 1215.7 xx 10^(-8) cm = 1215.7 A^(@)`


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