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

Explanation

For the longest wavelength transition in Balmer series
`n_(1)=2 and n_(2)=3`
`barv=R (1/n_(1)^2-1/n_2^2)`
`barv=1.09679 xx 10^(7) xx (1/2^2-1/3^2) (therefore R=1.09679 xx 10^(7) m^(-1))`
`=1.09679 xx 10^7 xx 0.139`
`=1.515 xx 10^6 m^(-1)`