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

Explanation

`barv = R (1/(n_1^2) - 1/(n_2^2))`
For Balmer series `n_1 = 2`, for longest wavelength, `barv` should be minimum so that `n_2 = 3`
`barv = (1.097 xx 10^(7)m^(-1)) (1/(2^2) - 1/(3^2))`
`=1.097 xx 10^7 xx 5/36`
`= 1.523 xx 10^6 m^(-1)`.