Question

The minimum wavelength of Lyman series lines is \( P \), then the maximum wavelength of the Lyman series lines is:

• A. \( \frac{4P}{3} \)
• B. \( 2P \)
• C. \( \frac{2P}{3} \)
• D. \( \infty \)

Hint:

For the Lyman series, the maximum wavelength corresponds to the transition from \( n = 2 \to n = 1 \), and the minimum wavelength corresponds to the transition from \( n = \infty \to n = 1 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Lyman Series The Lyman series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (for \( n = 2, 3, 4, \dots \)) to the \( n = 1 \) energy level. The wavelengths of these transitions can be calculated using the
Rydberg formula for hydrogen: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Where: - \( \lambda \) is the wavelength of the emitted radiation, - \( R_H \) is the Rydberg constant for hydrogen, - \( n_1 = 1 \) (since it's the Lyman series), - \( n_2 \) is the higher energy level (2, 3, 4, ...). Step 2: Maximum and Minimum Wavelengths 1. Minimum Wavelength: The minimum wavelength corresponds to the transition from \( n_2 \to \infty \) (since the energy difference is the greatest when the electron falls to the ground state, \( n = 1 \)): \[ \frac{1}{\lambda_{\text{min}}} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \left( 1 \right) \] Therefore, the minimum wavelength \( \lambda_{\text{min}} \) is given by: \[ \lambda_{\text{min}} = \frac{1}{R_H} \] 2. Maximum Wavelength: The maximum wavelength corresponds to the transition from \( n_2 = 2 \to n_1 = 1 \): \[ \frac{1}{\lambda_{\text{max}}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \] Therefore, the maximum wavelength \( \lambda_{\text{max}} \) is: \[ \lambda_{\text{max}} = \frac{4}{3R_H} \] Step 3: Relating the Maximum Wavelength to the Minimum Wavelength Given that the minimum wavelength is \( P \), we can write: \[ \lambda_{\text{min}} = P = \frac{1}{R_H} \] From the above formula, the maximum wavelength \( \lambda_{\text{max}} \) is: \[ \lambda_{\text{max}} = \frac{4}{3} \times \lambda_{\text{min}} = \frac{4}{3} \times P \] Step 4: Conclusion The maximum wavelength of the Lyman series lines is \( \frac{4P}{3} \). Thus, the correct answer is: \[ \boxed{(A)} \, \frac{4P}{3} \]