Question

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is

• A. $\frac{5}{27}$
• B. $\frac{3}{23}$
• C. $\frac{7}{29}$
• D. $\frac{9}{31}$

Answer 1

The Correct Option is A

To solve the problem of finding the ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum, we need to follow these steps:

1. Understand the concept of spectral series in the hydrogen atom. The Lyman series corresponds to electronic transitions where the final energy level n_f = 1, and the Balmer series corresponds to transitions where n_f = 2.
2. The formula for the wavelength \lambda in a hydrogen spectral series is given by the Rydberg formula: \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 final energy level, and n_i is the initial energy level.
3. The longest wavelength for a series corresponds to the smallest energy transition, which occurs when the transition is from the adjacent highest energy level.
   • For the Lyman series, the longest wavelength occurs when n_i = 2 and n_f = 1.
   • For the Balmer series, the longest wavelength occurs at n_i = 3 and n_f = 2.
4. Calculate the longest wavelength for the Lyman series: \frac{1}{\lambda_{\text{Lyman}}} = 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)
5. Calculate the longest wavelength for the Balmer series: \frac{1}{\lambda_{\text{Balmer}}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{5}{36} \right)
6. Find the ratio of the longest wavelengths: Since \frac{1}{\lambda_{\text{Lyman}}} = \frac{3}{4} R_H and \frac{1}{\lambda_{\text{Balmer}}} = \frac{5}{36} R_H, we rearrange to find the wavelengths: \lambda_{\text{Lyman}} = \frac{4}{3 R_H} and \lambda_{\text{Balmer}} = \frac{36}{5 R_H}
7. Thus, the ratio of the Lyman to Balmer wavelength is: \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} = \frac{\frac{4}{3 R_H}}{\frac{36}{5 R_H}} = \frac{4}{3} \cdot \frac{5}{36} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} = \frac{5}{27}
8. Therefore, the correct answer is \frac{5}{27}.

Based on these calculations, the correct option is indeed \frac{5}{27}.