Question

In hydrogen spectral series, the wave numbers of the first Lyman line and the first Balmer line are in the ratio

• A. 1 : 2
• B. 2 : 1
• C. 27 : 5
• D. 5 : 27

Hint:

Lyman is UV, Balmer is Visible. UV has higher energy and higher frequency/wave number than visible light, so the ratio \(\bar{\nu}_L / \bar{\nu}_B\) must be greater than 1.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The wave number (\(\bar{\nu}\)) of a spectral line in the hydrogen atom represents the reciprocal of the wavelength of the emitted radiation.

Step 2: Key Formula or Approach:
Rydberg Formula: \(\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\).

Step 3: Detailed Explanation:
1. First line of Lyman series (transition from \(n=2 \to n=1\)):
\[ \bar{\nu}_L = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \]
2. First line of Balmer series (transition from \(n=3 \to n=2\)):
\[ \bar{\nu}_B = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) = \frac{5R}{36} \]
3. Ratio:
\[ \frac{\bar{\nu}_L}{\bar{\nu}_B} = \frac{3R/4}{5R/36} = \frac{3}{4} \times \frac{36}{5} \]
\[ \text{Ratio} = 3 \times \frac{9}{5} = \frac{27}{5} \]

Step 4: Final Answer:
The wave numbers are in the ratio 27 : 5.