If the series limit wavelength of the Lyman series for hydrogen atom is `912 Å`, then the series limit wavelength for the Balmer series for the hydrogen atom is

Options

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

Answer

C

Explanation

`1/ lamda = R [ (1)/( n_(1)^(2))-(1)/(n_(2)^(2))]`
for limiting wavelength of Lyman series
` n_1 =1,n_2 =oo (1)/( lamda)=R`
for limiting wavelength of Balmer series
`n_1 =2, n_2 =oo`
` (1)/(lamda_B) =R ((1)/(4)) implies lamda _B = 4/R`
` therefore lamda_B =4 lamda_L = 4 xx 9 12 Å`