Q: The Lymann series of the hydrogen spectrum can be represented by the equation.

$\large \nu = 3.2881 \times 10^{15} (\frac{1}{1^2} – \frac{1}{n^2} ) $

(where n = 2, 3, ….) Calculate the maximum and minimum wavelength of lines in this series.

Solution: $\large \frac{1}{\lambda} = \frac{\nu}{c} = \frac{3.2881 \times 10^{15}}{3 \times 10^8} \times 10^{15} (\frac{1}{1^2} – \frac{1}{n^2} ) $

Wavelength is maximum , when n is minimum so that 1/n^{2} is maximum

$\large \frac{1}{\lambda_{max}} = \frac{3.2881 \times 10^{15} }{3 \times 10^8} (\frac{1}{1^2} – \frac{1}{2^2} ) $

$\large \lambda_{max} = \frac{3 \times 10^8}{3.2881 \times 10^{15}} \times \frac{4}{3}$

= 1.2165 × 10^{–7} m = 121.67 nm

Wavelength is minimum , when n is ∞ ;

$\large \frac{1}{\lambda_{min}} = \frac{3.2881 \times 10^{15} }{3 \times 10^8}

λ_{min} = 0.9124 × 10^{–7} m 91.24 nm