If $λ_1$ and $λ_2$ are the longest wavelengths of the Lyman and Paschen series respectively, then $λ_1:λ_2$ is:

7 : 108

The correct answer is Option (3) → 7 : 108

Using Rydberg formula,

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

for Lyman series: $n_f=1,n_i=2$

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

$⇒λ_1=\frac{4}{3R}$

for Pashen series: $n_f=3,n_i=4$

$\frac{1}{λ_2}=R_H\left(\frac{1}{9}-\frac{1}{16}\right)$

$⇒λ_2=\frac{144}{7R}$

$∴\frac{λ_1}{λ_2}=\frac{7}{108}$