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

Answer

`1.523 xx 10^(6) m^(-1)`

Explanation

`barv = 109678 (1/(n_1^2) - 1/(n_2^2))`
For longest wavelength, `n_2 = 3, n_1 = 2`
`barv = 109678 (1/(2^2) - 1/(3^2))`
`= 109678 xx 5/36 = 1.523 xx 10^4 cm^(-1)`
`= 1.523 xx 10^6 m^(-1)`