Question

Which of the following transitions of He$^+$ ion will give rise to a spectral line that has the same wavelength as the spectral line in a hydrogen atom?

• A. \( n = 4 \to n = 2 \)
• B. \( n = 6 \to n = 5 \)
• C. \( n = 6 \to n = 3 \)
• D. None of these

Hint:

The He$^+$ ion behaves like a hydrogen-like atom but with a stronger nuclear attraction, leading to shorter wavelengths for similar transitions.

Answer 1

The Correct Option is A

Step 1: {Understanding the Rydberg Formula} 
The wavelength \( \lambda \) of emitted radiation during an electron transition is given by the Rydberg formula: \[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: - \( R \) is the Rydberg constant, - \( Z \) is the atomic number, - \( n_1 \) and \( n_2 \) are the principal quantum numbers. 
Step 2: {Applying to Hydrogen and Helium Ions} 
For hydrogen (\( Z = 1 \)), the wavelength of the transition \( n_2 \to n_1 \) is: \[ \frac{1}{\lambda_H} = R(1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] For the He$^+$ ion (\( Z = 2 \)), the wavelength of transition \( n_4 \to n_3 \) is: \[ \frac{1}{\lambda_{He}} = R(2)^2 \left( \frac{1}{n_3^2} - \frac{1}{n_4^2} \right) \] Equating \( \lambda_H = \lambda_{He} \), we get: \[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = 4 \left( \frac{1}{n_3^2} - \frac{1}{n_4^2} \right) \] Solving for integer values, we find \( n_3 = 2 \), \( n_4 = 4 \) satisfies the condition. Thus, the correct answer is \( n = 4 \to n = 2 \).