MHT CET · Physics · Atomic Physics
In hydrogen atom, ratio of the shortest wavelength in the Balmer series to that in the Paschen series is
- A \(9: 4\)
- B \(3: 1\)
- C \(4: 9\)
- D \(1: 3\)
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),
\(\frac{\lambda_1}{\lambda_2}=\frac{R}{9} \times \frac{4}{R}=\frac{4}{9}\)
\(\frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{\mathrm{R}}{4}...(i)\)
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),
\(\frac{\lambda_1}{\lambda_2}=\frac{R}{9} \times \frac{4}{R}=\frac{4}{9}\)
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