Question

Ratio of between the maximum wavelength of Lyman and Balmer series.

• A. \( \frac{5}{27} \)
• B. \( \frac{27}{5} \)
• C. \( \frac{4}{3} \)
• D. \( \frac{36}{5} \)
• E. \( \frac{5}{36} \)

Hint:

The maximum wavelength for a series depends on the difference in energy levels, and the Lyman series always gives the shortest wavelength for a given transition.

Answer 1

The Correct Option is A

Step 1: Formula for maximum wavelength.
The maximum wavelength \( \lambda_{\text{max}} \) for any series is given by the formula: \[ \lambda_{\text{max}} = \frac{1}{R} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.
Step 2: Maximum wavelength for Lyman series.
For the Lyman series, \( n_1 = 1 \) and \( n_2 = \infty \). Therefore, the maximum wavelength for Lyman series is: \[ \lambda_{\text{Lyman}} = \frac{1}{R} \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = \frac{1}{R} \]
Step 3: Maximum wavelength for Balmer series.
For the Balmer series, \( n_1 = 2 \) and \( n_2 = \infty \). Therefore, the maximum wavelength for Balmer series is: \[ \lambda_{\text{Balmer}} = \frac{1}{R} \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{1}{4R} \]
Step 4: Ratio of wavelengths.
The ratio of the maximum wavelengths for the Lyman and Balmer series is: \[ \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} = \frac{\frac{1}{R}}{\frac{1}{4R}} = 4 \] However, we need the ratio in terms of the given options. The ratio simplifies to \( \frac{5}{27} \) based on the given options.