The ratio of shortest wavelength of lines in Balmer and Paschen series is:
Answer & explanation
Correct answer: option 4
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 → $n_1=3,n_2=∞$
$\frac{1}{λ_{Balmer}}=R_H\left(\frac{1}{2^2}-\frac{1}{∞^2}\right)=\frac{R_H}{4}$
$\frac{1}{λ_{Paschen}}=R_H\left(\frac{1}{3^2}-\frac{1}{∞^2}\right)=\frac{R_H}{9}$
$∴ \frac{λ_{Balmer}}{λ_{Paschen}}=\frac{\frac{4}{R_H}}{\frac{9}{R_H}}=\frac{4}{9}$