Question

The ratio of the minimum wavelength of the Balmer series to the maximum wavelength in the Brackett series in the hydrogen spectrum is:

• A. \( 25:16 \)
• B. \( 4:36 \)
• C. \( 9:100 \)
• D. \( 100:9 \)

Hint:

For hydrogen spectral series, remember that shorter wavelengths correspond to larger energy transitions. The Rydberg formula helps determine the ratio of different series' wavelengths.

Answer 1

The Correct Option is C

Step 1: Understanding the Rydberg Formula The wavelength \( \lambda \) of spectral lines in hydrogen is given by the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H \) is the Rydberg constant, \( n_1 \) represents the lower energy level, and \( n_2 \) represents the upper energy level.
Step 2: Calculating the Minimum Wavelength of the Balmer Series The Balmer series corresponds to transitions to \( n_1 = 2 \). The shortest wavelength occurs when \( n_2 = \infty \): \[ \frac{1}{\lambda_{\text{min, Balmer}}} = R_H \left( \frac{1}{2^2} - 0 \right) = R_H \times \frac{1}{4} \] \[ \lambda_{\text{min, Balmer}} = \frac{4}{R_H} \]
Step 3: Calculating the Maximum Wavelength of the Brackett Series The Brackett series corresponds to transitions to \( n_1 = 4 \). The longest wavelength occurs when \( n_2 = 5 \): \[ \frac{1}{\lambda_{\text{max, Brackett}}} = R_H \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \] \[ = R_H \left( \frac{1}{16} - \frac{1}{25} \right) = R_H \left( \frac{25 - 16}{400} \right) = R_H \times \frac{9}{400} \] \[ \lambda_{\text{max, Brackett}} = \frac{400}{9R_H} \]
Step 4: Calculating the Ratio \[ \frac{\lambda_{\text{min, Balmer}}}{\lambda_{\text{max, Brackett}}} = \frac{\frac{4}{R_H}}{\frac{400}{9R_H}} = \frac{4 \times 9}{400} = \frac{36}{400} = \frac{9}{100} \] Thus, the correct answer is: \[ \boxed{9:100} \]