Question

For a hydrogen atom, the ratio of the largest wavelength of the Lyman series to that of the Balmer series is:

• A. \( 5 : 36 \)
• B. \( 5 : 27 \)
• C. \( 3 : 4 \)
• D. \( 27 : 5 \)

Hint:

To find the ratio of wavelengths in different series, use the Rydberg formula for both series, calculate their wavelengths, and find the ratio.

Answer 1

The Correct Option is B

The Rydberg formula, \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), calculates the wavelength \( \lambda \) of light emitted by a hydrogen atom, where \( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \) is the Rydberg constant, and \( n_1, n_2 \) are principal quantum numbers. For the Lyman series, \( n_1 = 1 \); for the Balmer series, \( n_1 = 2 \).

Step 1: Largest wavelength in Lyman series
The largest wavelength in the Lyman series arises from the \( n_2 = 2 \to n_1 = 1 \) transition: \( \frac{1}{\lambda_{\text{Lyman}}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \times \frac{3}{4} \). Therefore, \( \lambda_{\text{Lyman}} = \frac{4}{3R_H} \).

Step 2: Largest wavelength in Balmer series
The largest wavelength in the Balmer series corresponds to the \( n_2 = 3 \to n_1 = 2 \) transition: \( \frac{1}{\lambda_{\text{Balmer}}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \times \frac{5}{36} \). Thus, \( \lambda_{\text{Balmer}} = \frac{36}{5R_H} \).

Step 3: Ratio of the wavelengths
The ratio of the largest Lyman wavelength to the largest Balmer wavelength is: \( \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} = \frac{\frac{4}{3R_H}}{\frac{36}{5R_H}} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27} \).

The correct answer is option (2): \( \frac{5}{27} \).