If the wavelength of the first line of the Balmer series of hydrogen is 6561 Å, the wavelength of the second line of the series should be

4860 Å

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

For first line of Balmer series, n = 3

$\Rightarrow \frac{1}{\lambda_1}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5 R}{36}$; For second line n = 4

$\Rightarrow \frac{1}{\lambda_2}=R\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3 R}{16}$

∴  $\frac{\lambda_2}{\lambda_1}=\frac{20}{27} \Rightarrow \lambda_1=\frac{20}{27} \times 6561$ = 4860 Å