In hydrogen atom, ratio of the shortest wavelength in the Balmer series to that in the Paschen series is

In hydrogen atom, ratio of the shortest wavelength in the Balmer series to that in the Paschen series is
  1. $9: 4$
  2. $3: 1$
  3. $4: 9$
  4. $1: 3$

Solution

The shortest wavelength in Balmer series is given by $\frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{\mathrm{R}}{4}...(i)$
The shortest wavelength in the Paschen series given by $\frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{\infty}\right)=\frac{R}{9}...(ii)$
Dividing equation (ii) by equation (i), $\frac{\lambda_1}{\lambda_2}=\frac{R}{9} \times \frac{4}{R}=\frac{4}{9}$

Asked in: MHT CET 2024 (09 May Shift 1)

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