Question

The frequency of light emitted during the \(n = 4 \to n = 2\) transition in hydrogen is \( \frac{3}{7} \) times that of which transition in Li atom?

• A. \( 4 \to 3 \)
• B. \( 4 \to 1 \)
• C. \( 3 \to 2 \)
• D. \( 5 \to 3 \)

Hint:

Frequency of radiation is proportional to \( Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \). Apply it for different atoms and transitions.

Answer 1

The Correct Option is A

Frequency of transition: \[ f \propto Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] For H: \( Z = 1 \), transition \(4 \to 2\): \[ f_H \propto \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3}{16} \] Let’s try Li (\( Z = 3 \)), transition \(4 \to 3\): \[ f_{Li} \propto 9 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 9 \left( \frac{1}{9} - \frac{1}{16} \right) = 9 \cdot \frac{7}{144} = \frac{63}{144} \] Ratio: \[ \frac{f_H}{f_{Li}} = \frac{3/16}{63/144} = \frac{3}{16} \cdot \frac{144}{63} = \frac{3 \cdot 9}{63} = \frac{27}{63} = \frac{3}{7} \]