Question

The ratio of the wavelengths of the first Lyman line and the second Balmer line of hydrogen atom is

• A. 3:4
• B. 1:4
• C. 2:3
• D. 1:3

Hint:

Rydberg formula: \( \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \). For Hydrogen, Z=1. - Lyman series: \( n_f = 1 \). First line is \( n_i=2 \to n_f=1 \). - Balmer series: \( n_f = 2 \). First line is \( n_i=3 \to n_f=2 \). Second line is \( n_i=4 \to n_f=2 \). Calculate \( 1/\lambda \) for each, then find the ratio of \( \lambda \)'s.

Answer 1

The Correct Option is B

The Rydberg formula for the wavelength \( \lambda \) of spectral lines of hydrogen is: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \] where \( R_H \) is the Rydberg constant, \( n_f \) is the principal quantum number of the final state, and \( n_i \) is the principal quantum number of the initial state (\(n_i>n_f\)).
First Lyman line (\( \lambda_L \)): Transition from \( n_i = 2 \) to \( n_f = 1 \).
\[ \frac{1}{\lambda_L} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \] So, \( \lambda_L = \frac{4}{3R_H} \).
Second Balmer line (\( \lambda_B \)): The Balmer series has \( n_f = 2 \).
First Balmer line is \( n_i = 3 \) to \( n_f = 2 \).
Second Balmer line is \( n_i = 4 \) to \( n_f = 2 \).
\[ \frac{1}{\lambda_B} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{4-1}{16} \right) = R_H \left( \frac{3}{16} \right) \] So, \( \lambda_B = \frac{16}{3R_H} \).
Ratio of wavelengths \( \frac{\lambda_L}{\lambda_B} \): \[ \frac{\lambda_L}{\lambda_B} = \frac{4/(3R_H)}{16/(3R_H)} = \frac{4}{3R_H} \times \frac{3R_H}{16} = \frac{4}{16} = \frac{1}{4} \] The ratio \( \lambda_L : \lambda_B = 1:4 \).
This matches option (2).