If the wave length of 1st line of Balmer series of hydrogen is 6561Å, the wavelength of the 2nd line of series will be:

4860 Å

The correct answer is Option (3) → 4860 Å

Using Rydberg's formula,

$\frac{1}{λ}=R_H\left(\frac{1}{{n_f}^2}-\frac{1}{{n_i}^2}\right)$

$∴\frac{1}{λ_1}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$  $[n_f=2,n_i=3]$

$\frac{1}{λ_1}=R_H\left(\frac{9-4}{36}\right)=\frac{5R_H}{36}$

$λ_1=\frac{36}{5R_H}$

$\frac{1}{λ_2}=R_H\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$   $[n_f=2,n_i=4]$

$\frac{1}{λ_2}=R_H\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3R_H}{16}$

$λ_2=\frac{16}{3R_H}$

$\frac{λ_1}{λ_2}=\frac{36}{5R_H}×\frac{3R_H}{16}=\frac{27}{20}$

$⇒λ_2=\frac{20}{27}λ_1=\frac{20}{27}×6561Å$

$=4860Å$