Question

The wavelength of the second line of the Balmer series is 486.4 nm. What is the wavelength of the first line of Lyman series?

• A. 78.8 nm
• B. 121.6 nm
• C. 418.2 nm
• D. 610.5 nm

Hint:

Use the Rydberg formula to calculate wavelengths in hydrogen spectra. For Lyman series, the transition starts from \( n_1 = 1 \).

Answer 1

The Correct Option is B

The wavelength of the lines in the hydrogen spectrum can be calculated using the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( \lambda \) is the wavelength, \( n_1 \) and \( n_2 \) are the principal quantum numbers, and \( R_H \) is the Rydberg constant. - The second line of the Balmer series corresponds to the transition from \( n_2 = 4 \) to \( n_1 = 2 \), with a wavelength of 486.4 nm. - The first line of the Lyman series corresponds to the transition from \( n_2 = 2 \) to \( n_1 = 1 \). Using the Rydberg formula for the first line of Lyman series: \[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \] which simplifies to: \[ \frac{1}{\lambda_1} = R_H \left( 1 - \frac{1}{4} \right) = R_H \times \frac{3}{4} \] Thus, the wavelength is: \[ \lambda_1 = \frac{4}{3} \times \lambda_2 = 121.6 \, \text{nm} \] So, the correct answer is (B).