Question

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer 1

For the Balmer series, ni = 2. Thus, the expression of wavenumber $\overline{v}$ is given by,
$\overline{v}=[\frac{1}{(2{)}^2}-\frac{1}{n_f^2}](1.097\times {10}^7{m}^{-1})$
Wave number $\overline{v}$ is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, $\overline{v}$ as to be the smallest. For $\overline{v}$ 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:
$\overline{v}=(1.097\times {10}^7)[\frac{1}{2^2}-\frac{1}{3^2}]\overline{v}=(1.097\times {10}^7)[\frac{1}{4}-\frac{1}{9}]=(1.097\times {10}^7)\left(\frac{9-4}{36}\right)=(1.097\times {10}^7)\left(\frac{5}{36}\right)\overline{v}=1.5236\times {10}^6{m}^{-1}$