The shortest wavelength in hydrogen spectrum of Lyman series when `R_(H)=109678 cm^(-1)` is :-

Options

A. `1002.7 Å`
B. `1215.67 Å`
C. `1127.30 Å`
D. `911.7 Å`

Answer

D

Explanation

`bar(v) = (1)/(lamda) = R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
For Lyman series, `n_(1) = 1`
`lamda` is shortest of `1//lamda` is largest when `n_(1) = oo` Hence,
`(1)/(lamda) = 109678 ((1)/(1^(2))) = 109678 cm^(-1)`
or `lamda = (1)/(109678) cm = 911.7 xx 10^(-8) cm = 911.7 Å`