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MHT CET · Physics · Atomic Physics

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

  1. A \(9: 4\)
  2. B \(3: 1\)
  3. C \(4: 9\)
  4. D \(1: 3\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4: 9\)

Step-by-step Solution

Detailed explanation

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),
\(\frac{\lambda_1}{\lambda_2}=\frac{R}{9} \times \frac{4}{R}=\frac{4}{9}\)