The frequency of light emitted for the transition n =4 to n =2 of `He^+` is equal to the transition in H atom corresponding to which of the following?

Options

A. n=3 to n=1
B. n=2 to n=1
C. n=3 to n=2
D. n=4 to n=3

Answer

To solve the problem, we need to find the transition in the hydrogen atom that has the same frequency as the transition from n=4 to n=2 in the He+ ion. We will use the formula for the frequency of emitted light during electronic transitions in hydrogen-like atoms.

### Step-by-Step Solution:

1. **Identify the Formula**:
The frequency of light emitted during a transition in a hydrogen-like atom is given by the formula:
\[ \nu = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \nu \) is the frequency,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

2. **Calculate Frequency for He+ Transition (n=4 to n=2)**:
For the He+ ion, \( Z = 2 \), \( n_1 = 2 \), and \( n_2 = 4 \):
\[ \nu_{He^+} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \cdot 4 \left( \frac{1}{4} - \frac{1}{16} \right) \]
Simplifying this:
\[ \nu_{He^+} = R \cdot 4 \left( \frac{4 - 1}{16} \right) = R \cdot 4 \cdot \frac{3}{16} = \frac{3R}{4} \]

3. **Set Up for Hydrogen Atom**:
Now, we need to find a transition in the hydrogen atom (where \( Z = 1 \)) that gives the same frequency \( \nu_{H} = \frac{3R}{4} \).

4. **Evaluate Each Option**:
- **Option A (n=3 to n=1)**:
\[ \nu_{H} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \left( \frac{8}{9} \right) \]
- **Option B (n=2 to n=1)**:
\[ \nu_{H} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
- **Option C (n=3 to n=2)**:
\[ \nu_{H} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \]
- **Option D (n=4 to n=3)**:
\[ \nu_{H} = R \cdot 1^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{7}{144} \right) \]

5. **Compare Frequencies**:
- From the calculations:
- Option A: \( \frac{8R}{9} \)
- Option B: \( \frac{3R}{4} \) (matches)
- Option C: \( \frac{5R}{36} \)
- Option D: \( \frac{7R}{144} \)

6. **Conclusion**:
The transition in the hydrogen atom that corresponds to the transition from n=4 to n=2 in He+ is **Option B: n=2 to n=1**.