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

Explanation

For Balmer series `n=2barv=R_(H)[1/(2^(2))-1/(n_(2)^(2))]barv=1/x`
For `lamda` to be longest (maximum), `barv` should be minimum. The can be sw when `n_(2)^(1)` is minimum.
`barv=1.097xx10^(7)m^(-1)(1/(2^(2))-1/(3^(2)))=1.523xx10^(6)m^(-1)`