COMEDK · Physics · 28. Atomic Physics
The shortest wavelengths of Paschen, Balmer and Lyman series are in the ratio
- A \(9: 1: 4\)
- B \(1: 4: 9\)
- C \(9: 4: 1\)
- D \(1: 9: 4\)
Answer & Solution
Correct Answer
(C) \(9: 4: 1\)
Step-by-step Solution
Detailed explanation
Wavelength \(\lambda\) of spectral lines in \(\mathrm{H}\)-atom is given as \[ \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \quad \text{...(i)} \] For shortest wavelength of Paschen series, \[ n_{1}=3 \text { and } n_{2}=\infty \] \(\therefore\) From…
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