Which of the following transitions in `He^(+)` ion will give rise to a spectral line which has the same wavelength as some spectral line in the hydrogen atom ?

Options

A. n=4 to n=2
B. n=6 to n=5
C. n=6 to n=3
D. none of these

Answer

To determine which transition in the \( He^+ \) ion gives rise to a spectral line with the same wavelength as a spectral line in the hydrogen atom, we can use the Rydberg formula for both ions.

### Step-by-Step Solution:

1. **Understand the Rydberg Formula**:
The Rydberg formula for the wavelength of spectral lines is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Apply the Formula for \( He^+ \)**:
For the \( He^+ \) ion, the atomic number \( Z = 2 \):
\[ \frac{1}{\lambda_{He^+}} = 4R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

3. **Apply the Formula for Hydrogen**:
For hydrogen, the atomic number \( Z = 1 \):
\[ \frac{1}{\lambda_{H}} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

4. **Set the Wavelengths Equal**:
To find transitions that yield the same wavelength, we set the two equations equal:
\[ 4R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

5. **Simplify the Equation**:
Dividing both sides by \( R \) (assuming \( R \neq 0 \)):
\[ 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
This implies:
\[ 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 0 \]

6. **Identify Possible Transitions**:
We need to check the options given in the question:
- For \( n_2 = 4 \) and \( n_1 = 2 \):
\[ \frac{1}{\lambda_{He^+}} = 4R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{4-1}{16} \right) = \frac{12R}{16} = \frac{3R}{4} \]
- For hydrogen, if \( n_2 = 2 \) and \( n_1 = 1 \):
\[ \frac{1}{\lambda_{H}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus, the wavelengths match.

7. **Check Other Options**:
- For other transitions, check if they yield the same result.
- After checking options B, C, and D, we find that only options A and D yield the same wavelength.

### Conclusion:
The transitions in \( He^+ \) that give rise to spectral lines with the same wavelength as some spectral lines in hydrogen are **Option A and Option D**.