Question:

If the wavelength of the first member of the Lyman series of hydrogen is \( \lambda \). The wavelength of the second member will be:

Updated On: Nov 20, 2024
  • \( \frac{27}{32} \lambda \)
  • \( \frac{32}{27} \lambda \)
  • \( \frac{27}{5} \lambda \)
  • \( \frac{5}{27} \lambda \)
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The Correct Option is A

Solution and Explanation

For the first member of the Lyman series:

\[ \frac{1}{\lambda} = \frac{13.6 \, z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] \quad \dots \text{(i)} \]

For the second member of the Lyman series:

\[ \frac{1}{\lambda'} = \frac{13.6 \, z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] \quad \dots \text{(ii)} \]

Dividing equation (i) by (ii):

\[ \lambda' = \frac{27}{32} \lambda \]

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