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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 _______.

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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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