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

For Balmer series, the longest wavelength corresponds to transition from `n = 3 ` to `n = 2`
`bar(v) = 109678 (1/(2^2) - 1/(n^2)) cm^(-1)`
`= 109678 (1/(2^2) - 1/(3^2))`
`=15233.0 cm^(-1)`
or `" " = 1.523 xx 10^(6) m^(-1)`.