Question

If the longest wavelength of the spectral line of the Paschen series of \( \text{Li}^{2+} \) ion spectrum is \( x \) Å, then the longest wavelength (in Å) of the Lyman series of the hydrogen spectrum is:

• A. \( \frac{12}{7} x \)
• B. \( \frac{7x}{12} \)
• C. \( \frac{20x}{27} \)
• D. \( \frac{27x}{20} \)

Hint:

For spectral series, the longest wavelength corresponds to the transition where \( n_2 = n_1 + 1 \). The wavelength is inversely proportional to \( Z^2 \), meaning higher atomic number ions emit shorter wavelengths.

Answer 1

The Correct Option is B

Step 1: Identifying the Longest Wavelength Transition The longest wavelength in any spectral series corresponds to the transition from \( n = n_{\text{min}} + 1 \) to \( n = n_{\text{min}} \). - For the Paschen series, this occurs at \( n = 4 \to n = 3 \).
- For the Lyman series, it happens at \( n = 2 \to n = 1 \). Using the Rydberg formula: \[ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: - \( R \) is the Rydberg constant.
- \( Z \) is the atomic number.
- \( n_1 \) and \( n_2 \) represent the lower and upper energy levels, respectively.
Step 2: Wavelength Ratio for Different Series For a hydrogen-like ion, the wavelength of a given transition is inversely proportional to \( Z^2 \), meaning: \[ \lambda \propto \frac{1}{Z^2} \] Given that Lithium (\( Li^{2+} \)) has \( Z = 3 \) and Hydrogen (\( H \)) has \( Z = 1 \), the ratio of their wavelengths is: \[ \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Paschen}}} = \frac{1/Z_{\text{H}}^2}{1/Z_{\text{Li}}^2} = \frac{1/1^2}{1/3^2} = \frac{9}{1} = 9 \]
Step 3: Compute the Required Wavelength For Paschen series (longest wavelength): \[ \lambda_{\text{Paschen}} = x \] For Lyman series (longest wavelength): \[ \lambda_{\text{Lyman}} = \lambda_{\text{Paschen}} \times \frac{7}{12} \] \[ = \frac{7x}{12} \] Thus, the correct answer is \( \frac{7x}{12} \).