Question

The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Paschen series is:

• A. \( \frac{4}{9} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{3}{4} \)

Hint:

For spectral series, use the Rydberg formula to find wavelength ratios.

Answer 1

The Correct Option is A

The minimum wavelength in a spectral series corresponds to a transition from \(n = \infty\) to a lower energy level (\(n_1\)). For the Balmer series, \(n_1 = 2\), and for the Paschen series, \(n_1 = 3\). The Rydberg formula is used: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \] where \(R_H\) is the Rydberg constant. Step 1: Wavelength for the Balmer series (\(n_1 = 2\)) For the Balmer series (\(n_1 = 2, n_2 = \infty\)): \[ \frac{1}{\lambda_{\text{Balmer}}} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R_H \cdot \frac{1}{4}. \] Therefore: \[ \lambda_{\text{Balmer}} = \frac{1}{\frac{R_H}{4}} = \frac{4}{R_H}. \] Step 2: Wavelength for the Paschen series (\(n_1 = 3\)) For the Paschen series (\(n_1 = 3, n_2 = \infty\)): \[ \frac{1}{\lambda_{\text{Paschen}}} = R_H \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R_H \cdot \frac{1}{9}. \] Therefore: \[ \lambda_{\text{Paschen}} = \frac{1}{\frac{R_H}{9}} = \frac{9}{R_H}. \] Step 3: Ratio of Wavelengths The ratio of the minimum wavelengths is: \[ \frac{\lambda_{\text{Balmer}}}{\lambda_{\text{Paschen}}} = \frac{\frac{4}{R_H}}{\frac{9}{R_H}} = \frac{4}{9}. \] Final Answer: The ratio of the minimum wavelengths is: \[ \boxed{\frac{4}{9}}. \]