The ratio of shortest wavelength of lines in Balmer and Paschen series is:

4 : 9

The correct answer is Option (4) → 4 : 9

$\frac{1}{λ}=R_H\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$ [Rydberg formula]

where,

λ = wavelength of light

$R_H$ = Rydberg constant = $1.097×10^7m^{-1}$

$n_1$ = principle quantum number of lower energy level

$n_2$ = principle quantum number of higher energy level

In,

Balmer series → $n_1=2,n_2=∞$

Paschen series → $n_1=3,n_2=∞$

$\frac{1}{λ_{Balmer}}=R_H\left(\frac{1}{2^2}-\frac{1}{∞^2}\right)=\frac{R_H}{4}$

$\frac{1}{λ_{Paschen}}=R_H\left(\frac{1}{3^2}-\frac{1}{∞^2}\right)=\frac{R_H}{9}$

$∴ \frac{λ_{Balmer}}{λ_{Paschen}}=\frac{\frac{4}{R_H}}{\frac{9}{R_H}}=\frac{4}{9}$