Question

The first member of the Balmer series of hydrogen atom has wavelength of $6563\stackrel{0}{A}$. Calculate the wavelength and frequency of the second member of the same series . Given $C=3\times {10}^{8}m{s}^{-1}$

Answer 1

The wavelength of different lines of the Balmer series are given by
$\frac{1}{\lambda }=R_H(\frac{1}{2^2}-\frac{1}{n_i^2}),where{n}_i=3,4,\mathrm{5...}ForfirstmemberofBalmerseries,n_2=3\frac{1}{{\lambda }_{\alpha }}=R_H(\frac{1}{4}-\frac{1}{9})=\frac{5R_H}{36}{\lambda }_{\alpha }=\frac{36}{5R_H}=6563\underset{0}{A}\frac{1}{\lambda }=R_H(1-\frac{1}{9})=\frac{8R_H}{9}\therefore {\lambda }_{\beta }=\frac{9}{8R_4}\Rightarrow \frac{{\lambda }_{\beta }}{{\lambda }_{\alpha }}=\frac{\frac{9}{8R_H}}{\frac{36}{5R_H}}=\frac{5}{32}{\lambda }_{\beta }=\frac{5}{32}\lambda \alpha =\frac{5}{32}\times 6563=1025.5\stackrel{0}{A}$