calculate the wavenumber for the longest wavelength transition in the balmer series of atomic hydrogen .

Explanation

According to Balmer formula
`bar(v)=1/lambda=R(1/n_(1)^(2)-1/n_(2)^(2))`
If wavelength `(lambda)` is to be longest, then energy and wave number `(bar(v))` must be minimum. For this `n_(2)` must be lowest.
`:. N_(1)=2, n_(2)=3` (Balmer Series)
`bar(v)=R_(H)(1/2^(2)-1/3^(2))`
`=R_(H)xx5/36`
`=109678xx5/36`
`=15233 cm^(-1)=15233xx10^(2) m^(-1)`
`=1.5233xx10^(6) m^(-1)`