Question

What is the ratio of the longest wavelength in Lyman and Balmer series?

Hint:

To find the ratio of wavelengths in different series, use the Rydberg formula and compare the terms for each series.

Answer 1

Step 1: Use the formula for the longest wavelength.
The longest wavelength in the Lyman and Balmer series corresponds to the transition from \( n = 2 \) to \( n = 1 \) for the Lyman series, and from \( n = 3 \) to \( n = 2 \) for the Balmer series. The wavelength for a transition in a hydrogen atom is given by the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \) is the Rydberg constant, and \( n_1 \) and \( n_2 \) are the initial and final principal quantum numbers.
Step 2: Calculate the longest wavelength for the Lyman series.
For the Lyman series, the transition is from \( n_2 = 2 \) to \( n_1 = 1 \). Using the Rydberg formula: \[ \frac{1}{\lambda_{\text{Lyman}}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_H \] Thus, the longest wavelength in the Lyman series is: \[ \lambda_{\text{Lyman}} = \frac{4}{3 R_H} \]
Step 3: Calculate the longest wavelength for the Balmer series.
For the Balmer series, the transition is from \( n_2 = 3 \) to \( n_1 = 2 \). Using the Rydberg formula: \[ \frac{1}{\lambda_{\text{Balmer}}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36} R_H \] Thus, the longest wavelength in the Balmer series is: \[ \lambda_{\text{Balmer}} = \frac{36}{5 R_H} \]
Step 4: Find the ratio of the longest wavelengths.
Now, find the ratio \( \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} \): \[ \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} = \frac{\frac{4}{3 R_H}}{\frac{36}{5 R_H}} = \frac{4}{3} \times \frac{5}{36} = \frac{20}{108} = \frac{5}{27} \] Thus, the ratio of the longest wavelengths in the Lyman and Balmer series is: \[ \boxed{\frac{5}{27}} \]