Q.17:- Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
For the Balmer series, n1 = 2. Hence, ṽ = R(1/22 – 1/n22)
ṽ = 1/λ (inversely proportional)
For λ to be maximum, ṽ should be minimum. This can be happened when n2 is minimum i.e. n2 = 3. Hence, ṽ = (1.097×107 m-1) (1/22 – 1/32) = 1.097×107×5/36 m-1 = 1.523×106 m-1