Question

The ratio of the wavelength of first line of Balmer series to the first line of Lyman series is

• A. 3 : 1
• B. 17 : 5
• C. 27 : 5
• D. 27 : 5

Hint:

First Balmer: \(3 \to 2\); First Lyman: \(2 \to 1\). Use \(1/\lambda \propto (1/n_1^2 - 1/n_2^2)\) and take the inverse ratio.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Rydberg formula: \(\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)\).
Step 2: Detailed Explanation:
Balmer first line (\(n = 3 \to 2\)): \(\dfrac{1}{\lambda_B} = R\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = R \cdot \dfrac{5}{36}\)
Lyman first line (\(n = 2 \to 1\)): \(\dfrac{1}{\lambda_L} = R\left(1 - \dfrac{1}{4}\right) = R \cdot \dfrac{3}{4}\)
\[ \frac{\lambda_B}{\lambda_L} = \frac{R \cdot 3/4}{R \cdot 5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5} \]
Step 3: Final Answer:
\(\lambda_B : \lambda_L = \mathbf{27 : 5}\).