For a hydrogen atom, the ratio of the largest wavelength of the Lyman series to that of the Balmer series is:
Show Hint
- \( 5 : 36 \)
- \( 5 : 27 \)
- \( 3 : 4 \)
- \( 27 : 5 \)
The Correct Option is B
Solution and Explanation
The wavelength of light emitted in any spectral series for a hydrogen atom can be calculated using the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: \( \lambda \) is the wavelength of the emitted radiation, \( R_H \) is the Rydberg constant (\( 1.097 \times 10^7 \, \text{m}^{-1} \)), \( n_1 \) and \( n_2 \) are the principal quantum numbers of the two energy levels involved.
For the Lyman series, the transition is from \( n_2 \to 1 \), and for the Balmer series, the transition is from \( n_2 \to 2 \).
Step 1: Largest wavelength in Lyman series
The largest wavelength in the Lyman series corresponds to the transition from \( n_2 = 2 \) to \( n_1 = 1 \): \[ \frac{1}{\lambda_{\text{Lyman}}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \times \frac{3}{4} \] Thus: \[ \lambda_{\text{Lyman}} = \frac{4}{3R_H} \]
Step 2: Largest wavelength in Balmer series
The largest wavelength in the Balmer series corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \): \[ \frac{1}{\lambda_{\text{Balmer}}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \times \frac{5}{36} \] Thus: \[ \lambda_{\text{Balmer}} = \frac{36}{5R_H} \]
Step 3: Ratio of the wavelengths
The ratio of the largest wavelength of the Lyman series to the largest wavelength of the Balmer series is: \[ \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} = \frac{\frac{4}{3R_H}}{\frac{36}{5R_H}} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27} \]
Thus, the correct answer is option (2): \( \frac{5}{27} \).
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