Question

The wavenumber of the first line of the Lyman series of the hydrogen spectrum is equal to the wavenumber of the second line of the Balmer series of \( X^{n+} \) ion. What is \( X^{n+} \)?

• A. \( {Li}^{2+} \)
• B. \( {Be}^{3+} \)
• C. \( {He}^{2+} \)
• D. \( {B}^{4+} \)

Hint:

This type of problem involves understanding the Rydberg formula and applying it to different series of hydrogen and ions.

Answer 1

The Correct Option is C

The wavenumber of the first line of the Lyman series of hydrogen is given by the Rydberg formula for hydrogen: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \] The second line of the Balmer series for the ion \( X^{n+} \) is also given by the Rydberg formula: \[ \frac{1}{\lambda} = R_X \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] Since the wavenumbers are equal, we can equate the Rydberg constants for both: \[ R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_X \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] After solving for \( X \), we find that the ion corresponding to this condition is \( {He}^{2+} \). 
Final Answer: \( {He}^{2+} \).