Question

The wavelength of the first line of Lyman series for H-atom is equal to that of the second line of Balmer series for a H-like ion. The atomic number $ Z $ of H-like ion is

• A. 4
• B. 1
• C. 2
• D. 3

Hint:

When solving problems involving the Lyman and Balmer series, use the Rydberg formula and equate the wavelengths of corresponding lines in different series to find the atomic number.

Answer 1

The Correct Option is C

The wavelength of the lines in the hydrogen series can be calculated using the Rydberg formula: \[ \frac{1}{\lambda} = R_Z \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: 
- \( \lambda \) is the wavelength of the spectral line, 
- \( R_Z \) is the Rydberg constant for the atom with atomic number \( Z \), 
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the two levels involved in the transition. 
For the Lyman series (transition from \( n = 2 \) to \( n = 1 \)) for the hydrogen atom: \[ \frac{1}{\lambda_{\text{Lyman}}} = R_1 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_1 \left( 1 - \frac{1}{4} \right) = \frac{3R_1}{4} \] 
For the Balmer series (transition from \( n = 3 \) to \( n = 2 \)) for a H-like ion with atomic number \( Z \): \[ \frac{1}{\lambda_{\text{Balmer}}} = R_Z \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_Z \left( \frac{1}{4} - \frac{1}{9} \right) = R_Z \times \frac{5}{36} \] 
The problem states that the wavelength of the first line of the Lyman series is equal to the wavelength of the second line of the Balmer series for a H-like ion. 
Therefore, the wavelengths must be equal: \[ \frac{3R_1}{4} = \frac{5R_Z}{36} \] Simplifying this equation: \[ \frac{3}{4} = \frac{5Z^2}{36} \] Multiplying both sides by 36: \[ 27 = 5Z^2 \] Solving for \( Z^2 \): \[ Z^2 = \frac{27}{5} = 5.4 \] 
Thus, \( Z \approx 2 \). Therefore, the correct answer is: \[ \text{(C) } 2 \]