Question

What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

• A. 5:1
• B. 5:0.81
• C. 5:1.75
• D. 5:2.7

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Wave number (\(\bar{\nu}\)) is calculated using the Rydberg formula: \(\bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\). The "first line" or "lowest energy line" of any series corresponds to the transition from the shell immediately above (\(n_2 = n_1 + 1\)).

Step 2: Key Formula or Approach:
1. Balmer Series: \(n_1 = 2\), first line \(n_2 = 3\). 2. Brackett Series: \(n_1 = 4\), first line \(n_2 = 5\). 3. Ratio = \(\bar{\nu}_{Balmer} / \bar{\nu}_{Brackett}\).

Step 3: Detailed Explanation:
1. Balmer first line: \(\bar{\nu}_B = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36} R_H\). 2. Brackett first line: \(\bar{\nu}_{Br} = R_H \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = R_H \left( \frac{1}{16} - \frac{1}{25} \right) = \frac{9}{400} R_H\). 3. Ratio: \(\frac{5/36}{9/400} = \frac{5}{36} \times \frac{400}{9} = \frac{2000}{324} \approx 6.17\). 4. Comparing with Option (b): \(5 / 0.81 \approx 6.17\).

Step 4: Final Answer:
The ratio is 5:0.81.