Question

What is the ratio of the wavelength of the Lyman series limit to that of the Paschen series limit in the hydrogen spectrum?

• A. \( \frac{1}{9} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{9}{3} \)
• D. \( \frac{2}{2} \)

Hint:

When finding the ratio of wavelengths in spectral lines, remember that the energy level transitions determine the wavelengths through the Rydberg formula.

Answer 1

The Correct Option is A

Ratio of Wavelengths: Lyman Limit to Paschen Limit 

Step 1: Energy of Series Limit

The series limit corresponds to transitions to a fixed lower level from \( n = \infty \). 
Energy of a photon emitted during a transition in hydrogen is: \[ E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\text{ eV} \] For the **series limit**, \( n_2 \to \infty \), so: \[ E_{\text{limit}} = 13.6 \left( \frac{1}{n_1^2} \right) \]

Step 2: Lyman and Paschen Series Limits

- For **Lyman series**: \( n_1 = 1 \) \[ E_L = 13.6 \cdot \frac{1}{1^2} = 13.6\ \text{eV} \] - For **Paschen series**: \( n_1 = 3 \) \[ E_P = 13.6 \cdot \frac{1}{3^2} = 13.6 \cdot \frac{1}{9} = 1.51\ \text{eV} \]

Step 3: Relationship with Wavelength

Since \( E = \frac{hc}{\lambda} \), wavelength is inversely proportional to energy: \[ \lambda \propto \frac{1}{E} \] Therefore, the ratio of wavelengths is: \[ \frac{\lambda_L}{\lambda_P} = \frac{E_P}{E_L} = \frac{1.51}{13.6} = \frac{1}{9} \]

✅ Final Answer:

\(\boxed{\frac{1}{9}}\)