JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2024
- Hard
- MCQ
If the wavelength of the first member of Lyman series of hydrogen is \(\lambda\). The wavelength of the second member will be _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{27}{32} \lambda\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \) \(...(i)\)
\(\frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] \) \(...(ii)\)
On dividing \((i)\) & \((ii)\)
\(\lambda^{\prime}=\frac{27}{32} \lambda\)
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