Ncert Exercises & Solutions

2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

For the Balmer series, ni = 2. Thus, the expression of wavenumber

(\vec{v})

is given by,

\vec{v} = [\frac{1}{{(2)}^{2}} - \frac{1}{n_{f}^{2}}] (1.097 x 10^{7} m^{- 1})

Wave number

(\vec{v})

is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition,

(\vec{v})

has to be the smallest.
For

(\vec{v})

to be minimum, nf should be minimum.
For the Balmer series, a transition from ni = 2 to nf = 3 is allowed.
Hence, taking nf = 3, we get:

(\vec{v}) = (1.097 x 10^{7}) [\frac{1}{2^{2}} - \frac{1}{3^{2}}]

\vec{v} = (1.097 x 10^{7}) [\frac{1}{4} - \frac{1}{9}]

= (1.097 x 10^{7}) (\frac{9 - 4}{36})

= (1.097 x 10^{7}) (\frac{5}{36})

\vec{v} = 1.5236 x 10^{6} m^{- 1}