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

Explanation

For the Balmer series, ni = 2. Thus, the expression of wavenumber`(barv)` is given by,
` barv[(1)/(2)^(2)-(1)/(n_(f)^(2))](1.097 xx 10^(7) m^(-1))`
Wave number `(barv)` is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition,`(barv)` has to be the smallest.
For `(barv)` to be minimum, `n_(f)` should be minimum. For the Balmer series, a transition from `n_(i) = 2 "to" n_(f) = 3` is allowed. Hence, taking `n_(f) = 3`, we get:
` barv = (1.097xx 10^(7))[1/(2^(2))-1/3^(2)]`
`barv= (1.097xx10^(7))[1/4-1/9]`
`(1.097xx10^(7))((9-4)/36)`
` (1.097 xx 10 ^(7))(5/36)`
`barv= 1.5236xx10^(6)m^(-1)`