If $λ_1$ and $λ_2$ are the longest wavelengths of the Lyman and Paschen series respectively, then $λ_1:λ_2$ is:
Answer & explanation
Correct answer: option 3
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: $n_f=3,n_i=4$
$\frac{1}{λ_2}=R_H\left(\frac{1}{9}-\frac{1}{16}\right)$
$⇒λ_2=\frac{144}{7R}$
$∴\frac{λ_1}{λ_2}=\frac{7}{108}$