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 \( n = 4 \to n = 2 \). 
 

Answer 2

Step 1: Understanding the Question
We are asked to find a transition in a He⁺ ion that gives the same wavelength as a transition in a hydrogen atom.

Step 2: Using the Rydberg Formula
The Rydberg formula gives the wavelength for a spectral line: \[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Where:

\( R \) is the Rydberg constant

\( Z \) is the atomic number (for H, \( Z = 1 \); for He⁺, \( Z = 2 \))

\( n_1 \) is the lower energy level

\( n_2 \) is the higher energy level

 

Step 3: Matching Wavelengths
To get the same wavelength in He⁺ as in hydrogen, the expressions for \( \frac{1}{\lambda} \) must be equal. That means: \[ Z_{\text{He}}^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)_{\text{He}} = \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)_{\text{H}} \] Since \( Z_{\text{He}} = 2 \), we get: \[ 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)_{\text{He}} = \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)_{\text{H}} \]

Step 4: Check Transitions
Let’s try the hydrogen atom’s transition \( n = 2 \rightarrow n = 1 \), which gives: \[ \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4} \] We want: \[ 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3}{4} \Rightarrow \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3}{16} \] Try \( n_1 = 2 \), \( n_2 = 4 \): \[ \frac{1}{2^2} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{3}{16} \]

Step 5: Final Answer
He⁺ transition \( n = 4 \rightarrow n = 2 \) gives the same wavelength as a transition in hydrogen atom.

Correct Option: Option 1 — \( n = 4 \rightarrow n = 2 \)