JEE Mains · Physics · STD 12 - 12. atoms
If the wavelength of the first member of Lyman series of hydrogen is \(\lambda\). The wavelength of the second member will be _______.
- A \(\frac{27}{32} \lambda\)
- B \(\frac{32}{27} \lambda\)
- C \(\frac{27}{5} \lambda\)
- D \(\frac{5}{27} \lambda\)
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 \((i)\) & \((ii)\)…
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