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

Explanation

For Lyman series `n_i =1`
For shortest wavelength in Lyman 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)/(3 xx 109678) = 9.117 xx 10^(-6) cm = 911.7 A^(0)`
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`
`(1)/(lambda) = R_H [ (1)/(1^(2) ) - (1)/((2)^2) ] = (3)/(4) R_H`
`lambda = (4)/(3) (1)/(R_H) = (4)/(3 xx 109678)`
`= 1215.7 xx 10^(-8) cm= 1215.7 A^0`