If the wavelength of the first line of the Balmer series of hydrogen is 6561 $\stackrel{0}{\mathrm{A}}$, the wavelength of the second line of the series should be

The wavelength of spectral line in Balmer series is given by $\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{\mathrm{nh}2}\right]$

For first line of Balmer series, n = 3

$\begin{array}{l}\Rightarrow \frac{1}{{\mathrm{\lambda }}_1}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5\mathrm{R}}{36};\mathrm{For}\mathrm{second}\mathrm{line}\mathrm{n}=4.\\ \Rightarrow \frac{1}{{\mathrm{\lambda }}_2}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3\mathrm{R}}{16}\\ \therefore \frac{{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1}=\frac{20}{27}\Rightarrow {\mathrm{\lambda }}_1=\frac{20}{27}\times 6561=4860\stackrel{0}{\mathrm{A}}\end{array}$