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. `912xx2` Å
C. `912xx4` Å
D. `(912)/(2)` Å

Answer

C

Explanation

For Series limit of Balmer series, `n_(2)=2,n_(1)=oo`
`(1)/(lamda)=R[(1)/(n_(2)^(2))-(1)/(n_(1)^(2))]=R[(1)/((2)^(2))-(1)/((oo)^(2))]=(R)/(4)`
`therefore lamda=(4)/(R)=(4)/(10967800)=4xx921xx10^(-10)`
`=4xx912` Å