Visible spectrum — UPSC Physics Optional IFoS 2011 Paper II

The wavelengths (\lambda) in the visible spectrum of H atom can be expressed by the empirical formula \lambda = \left( \frac{n_1^2}{n_1^2 - 4} \right) \cdot G where n_1 is an integer and n_1 = 3, 4 etc. and G is an empirical constant. Prove from the above, the wave number \bar{\nu} = R_H \left( \frac{1}{2^2} - \frac{1}{n_1^2} \right), \text{ where } R_H = \frac{4}{G}.