Question:medium
The minimum wavelength of Lyman series lines is \( P \), then the maximum wavelength of the Lyman series lines is:
The minimum wavelength of Lyman series lines is \( P \), then the maximum wavelength of the Lyman series lines is:
Show 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 \( n = \infty \to n = 1 \).
Updated On: Nov 28, 2025
- \( \frac{4P}{3} \)
- \( 2P \)
- \( \frac{2P}{3} \)
- \( \infty \)
Hide Solution
The Correct Option is A
Solution and Explanation
Step 1: Overview of the Lyman Series
The Lyman series describes electron transitions in a hydrogen atom from higher energy levels (\( n = 2, 3, 4, \dots \)) to the \( n = 1 \) level. The wavelengths are calculated using the
Rydberg formula: \[\n\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\n\] Where: - \( \lambda \) is the wavelength, - \( R_H \) is the Rydberg constant, - \( n_1 = 1 \) (Lyman series), - \( n_2 \) is the higher energy level (2, 3, 4, ...). Step 2: Determining Maximum and Minimum Wavelengths 1. Minimum Wavelength: This occurs when \( n_2 \to \infty \) (largest energy difference): \[\n \frac{1}{\lambda_{\text{min}}} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \left( 1 \right)\n \] Therefore: \[\n \lambda_{\text{min}} = \frac{1}{R_H}\n \] 2. Maximum Wavelength: This occurs when \( n_2 = 2 \to n_1 = 1 \): \[\n \frac{1}{\lambda_{\text{max}}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right)\n \] Therefore: \[\n \lambda_{\text{max}} = \frac{4}{3R_H}\n \] Step 3: Relating Maximum and Minimum Wavelengths Given \( \lambda_{\text{min}} = P \): \[\n\lambda_{\text{min}} = P = \frac{1}{R_H}\n\] Then, \( \lambda_{\text{max}} \) is: \[\n\lambda_{\text{max}} = \frac{4}{3} \times \lambda_{\text{min}} = \frac{4}{3} \times P\n\] Step 4: Conclusion The maximum wavelength is \( \frac{4P}{3} \). Thus, the answer is: \[\n\boxed{(A)} \, \frac{4P}{3}\n\]
Rydberg formula: \[\n\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\n\] Where: - \( \lambda \) is the wavelength, - \( R_H \) is the Rydberg constant, - \( n_1 = 1 \) (Lyman series), - \( n_2 \) is the higher energy level (2, 3, 4, ...). Step 2: Determining Maximum and Minimum Wavelengths 1. Minimum Wavelength: This occurs when \( n_2 \to \infty \) (largest energy difference): \[\n \frac{1}{\lambda_{\text{min}}} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \left( 1 \right)\n \] Therefore: \[\n \lambda_{\text{min}} = \frac{1}{R_H}\n \] 2. Maximum Wavelength: This occurs when \( n_2 = 2 \to n_1 = 1 \): \[\n \frac{1}{\lambda_{\text{max}}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right)\n \] Therefore: \[\n \lambda_{\text{max}} = \frac{4}{3R_H}\n \] Step 3: Relating Maximum and Minimum Wavelengths Given \( \lambda_{\text{min}} = P \): \[\n\lambda_{\text{min}} = P = \frac{1}{R_H}\n\] Then, \( \lambda_{\text{max}} \) is: \[\n\lambda_{\text{max}} = \frac{4}{3} \times \lambda_{\text{min}} = \frac{4}{3} \times P\n\] Step 4: Conclusion The maximum wavelength is \( \frac{4P}{3} \). Thus, the answer is: \[\n\boxed{(A)} \, \frac{4P}{3}\n\]
Was this answer helpful?
0