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

To solve the problem of finding the ratio of the minimum wavelength of the Balmer series to the maximum wavelength in the Brackett series in the hydrogen spectrum, we employ the Rydberg formula for hydrogen:

\( \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \)

where \( R \) is the Rydberg constant, \( n_1 \) and \( n_2 \) are integers representing the final and initial energy levels, respectively, with \( n_2 > n_1 \).

For the Balmer series:

The Balmer series has \( n_1 = 2 \). The minimum wavelength \((\lambda_{\text{min}})\) occurs at the transition from the \( n_2 = \infty \) to \( n_1 = 2 \):

\( \frac{1}{\lambda_{\text{min\, Balmer}}} = R \left( \frac{1}{2^2} \right) = \frac{R}{4} \)

For the Brackett series:

The Brackett series has \( n_1 = 4 \). The maximum wavelength \((\lambda_{\text{max}})\) occurs at the transition from \( n_2 = 5 \) to \( n_1 = 4 \):

\( \frac{1}{\lambda_{\text{max\, Brackett}}} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{16} - \frac{1}{25} \right) = R \left( \frac{25-16}{400} \right) = \frac{9R}{400} \)

Ratio Calculation:

We need the ratio \( \frac{\lambda_{\text{min\, Balmer}}}{\lambda_{\text{max\, Brackett}}} \):

\[\frac{\lambda_{\text{min\, Balmer}}}{\lambda_{\text{max\, Brackett}}} = \frac{\frac{4}{R}}{\frac{400}{9R}} = \frac{4 \times 9}{400} = \frac{36}{400} = \frac{9}{100}\]

Thus, the ratio of the minimum wavelength of the Balmer series to the maximum wavelength in the Brackett series is \( 9:100 \).

Answer 2

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: \[ \boxed{9:100} \]