Find the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Explanation

The equation used for explaining the line spectrum of hydrogen- `overline(v)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`, for Balmer series `n_(1)=2`
`therefore`For Balmer series, `overline(v)=R((1)/(2^(2))-(1)/(n_(2)^(2)))`
As, `overline(v)=(1)/(lamda)" "lamda` will be longest if `n_(2)` is minimum.
Thus, there `n_(2)=n_(1)+1=2+1=3` [R=`109677cm^(-1)`)
So, `overline(v)=R((1)/(2^(2))-(1)/(3^(2)))`
or, `overline(v)=109677((1)/(2^(2))-(1)/(3^(2)))cm^(-1)`
`=109677xx(5)/(36)cm^(-1)=1.5233xx10^(4)cm^(-1)`.