The transitionis `He^(o+)` ion that would have the same wavelength as the first Lyman line in hydtrogen spectrum is

Options

A. `2 rarr 1`
B. `5 rarr 3`
C. `4 rarr 2`
D. `6 rarr 4`

Explanation

To solve the problem of finding the transition in the \( \text{He}^+ \) ion that would have the same wavelength as the first Lyman line in the hydrogen spectrum, we can follow these steps:

### Step 1: Calculate the Wavelength of the First Lyman Line in Hydrogen
The first Lyman line corresponds to a transition from \( n_2 = 2 \) to \( n_1 = 1 \) in hydrogen. We can use the Rydberg formula:

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R_H \) is the Rydberg constant for hydrogen.
- \( n_1 = 1 \) and \( n_2 = 2 \).

Substituting these values into the formula:

\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \]

Thus, we have:

\[ \frac{1}{\lambda_1} = \frac{3R_H}{4} \]

### Step 2: Set Up the Equation for the Helium Ion
For the \( \text{He}^+ \) ion, we need to find the transition that gives the same wavelength. The Rydberg formula for \( \text{He}^+ \) (where \( Z = 2 \)) is:

\[ \frac{1}{\lambda_2} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Substituting \( Z = 2 \):

\[ \frac{1}{\lambda_2} = R_H \cdot 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 3: Equate the Two Wavelengths
Since we want \( \lambda_1 = \lambda_2 \):

\[ \frac{3R_H}{4} = 4R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Dividing both sides by \( R_H \):

\[ \frac{3}{4} = 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 4: Simplify the Equation
Rearranging gives:

\[ \frac{3}{16} = \frac{1}{n_1^2} - \frac{1}{n_2^2} \]

### Step 5: Test Possible Transitions
We can test the provided options to find valid \( n_1 \) and \( n_2 \) values that satisfy the equation \( \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{16} \).

1. **Option A**: \( n_1 = 1, n_2 = 2 \)
\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \quad \text{(not valid)} \]

2. **Option B**: \( n_1 = 3, n_2 = 5 \)
\[ \frac{1}{3^2} - \frac{1}{5^2} = \frac{1}{9} - \frac{1}{25} = \frac{25 - 9}{225} = \frac{16}{225} \quad \text{(not valid)} \]

3. **Option C**: \( n_1 = 2, n_2 = 4 \)
\[ \frac{1}{2^2} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16} \quad \text{(valid)} \]

### Conclusion
The transition in the \( \text{He}^+ \) ion that would have the same wavelength as the first Lyman line in the hydrogen spectrum is from \( n_1 = 2 \) to \( n_2 = 4 \).

### Final Answer
The correct transition is \( n_1 = 2 \) and \( n_2 = 4 \).

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