Question

The wavelength of spectral line corresponding to electron transition \( n = 3 \) to \( n = 2 \) for \( \text{Li}^{2+} \) ion is \( \lambda \,\text{nm} \). What is the wavelength (in nm) corresponding to electron transition \( n = 4 \) to \( n = 1 \) for \( \text{He}^+ \) ion?

• A. \( \frac{\lambda}{4} \)
• B. \( \frac{\lambda}{3} \)
• C. \( \frac{\lambda}{9} \)
• D. 5\( \lambda \)

Hint:

Use \( \lambda \propto \frac{1}{Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)} \) to compare transitions in hydrogen-like ions.

Answer 1

The Correct Option is B

Energy difference for hydrogen-like atoms is: \[ E \propto Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \Rightarrow \lambda \propto \frac{1}{Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)} \] For Li\(^{2+} \): \( Z = 3 \), transition from 3 to 2 \[ \lambda_1 \propto \frac{1}{3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)} = \frac{1}{9 \left( \frac{1}{4} - \frac{1}{9} \right)} = \frac{1}{9 \cdot \frac{5}{36}} = \frac{4}{5} \] For He\(^{+} \): \( Z = 2 \), transition from 4 to 1 \[ \lambda_2 \propto \frac{1}{2^2 \left( \frac{1}{1^2} - \frac{1}{4^2} \right)} = \frac{1}{4 \cdot \frac{15}{16}} = \frac{4}{15} \] So, \[ \frac{\lambda_2}{\lambda_1} = \frac{4/15}{4/5} = \frac{1}{3} \Rightarrow \lambda_2 = \frac{\lambda}{3} \]