Calculate the wave number for the longest wavelength transition in the Balmer series fo atomic hydrogen . `( R_(H) = 109677 cm^(-1)).`

Explanation

For Balmer series ` n_1 =2`
If this line possesses longest wavelength
( i.e., lowest energy ) then ` n_2 =3`
` overline (v) = 1/ (lambda) = 109677 [1/2^2 - 1/3^2]`
` = 1. 523 xx 10^4 cm^(-1) = 1. 523 xx 10^6 m^(-1)`.