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

Explanation

For Balmer series, `n_(1)=2`. Hence, `bar(v)=R(1/2^(2)-1/n_(2)^(2))`
`bar(v)=1/lambda`. For `lambda` to be longest (maximum), `bar(v)` should be minimum. This can be so when `n_(2)` is minimum, i.e., `n_(2)=3`. Hence, `bar(v)=(1.097xx10^(7) m^(-1)) (1/2^(2)-1/3^(2))=1.097xx10^(7)xx5/36 m^(-1) =1.523xx10^(6) m^(-1)`