The shortest wavelength in the Lyman series of hydrogen spectrum is `912 Å` correcponding to a photon energy of `13.6eV`. The shortest wavelength in the Balmer series is about

Options

A. `912 Å//2`
B. `912 Å`
C. `912 xx 2 Å`
D. `912 xx 4 Å`

Answer

D

Explanation

For Lyman series, the series limit wavelength is given by
`(1)/(lambda) = R [(1)/(1^(2)) - (1)/(oo^(2))] = R` or `lambda = (1)/(R)`
For Balmer series, the series limit wavelength is given by
`(1)/(lambda) = R [(1)/(2^(2)) - (1)/(oo^(2))] = (R)/(4)` or `lambda = (4)/(R)`
Clearly `lambda = 4 [(1)/(R)] or lambda' = 4lambda`