The transiton in `He^(+)` ion that will have the same wave number as the first line of lyman series of hydrogen will be

Options

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

Answer

To solve the problem, we need to find the transition in the `He^(+)` ion that has the same wave number as the first line of the Lyman series of hydrogen. Let's break this down step by step.

### Step 1: Understand the Lyman Series
The Lyman series corresponds to transitions where the electron falls to the n=1 energy level from higher energy levels (n=2, 3, 4, ...). The first line of the Lyman series corresponds to the transition from n=2 to n=1.

### Step 2: Calculate the Wave Number for the First Line of Lyman Series in Hydrogen
The formula for the wave number (ṽ) is given by:
\[ \tilde{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the first line of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Substituting these values into the formula:
\[ \tilde{\nu} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \]

### Step 3: Apply the Formula for `He^(+)` Ion
For hydrogen-like ions (like `He^(+)`), the wave number is modified by the square of the atomic number (Z). For `He^(+)`, Z = 2. Therefore, the wave number formula becomes:
\[ \tilde{\nu} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting Z = 2:
\[ \tilde{\nu} = 4 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 4: Set the Wave Numbers Equal
We want the wave number for the transition in `He^(+)` to equal the wave number for the first line of the Lyman series in hydrogen:
\[ 4 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = R_H \left( \frac{3}{4} \right) \]
Dividing both sides by \( R_H \):
\[ 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3}{4} \]

### Step 5: Simplify the Equation
Dividing both sides by 4:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{16} \]

### Step 6: Solve for Possible Transitions
We need to find integers \( n_1 \) and \( n_2 \) such that:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{16} \]
Assuming \( n_1 < n_2 \), we can try different values for \( n_1 \) and \( n_2 \).

- For \( n_1 = 2 \) and \( n_2 = 4 \):
\[ \frac{1}{2^2} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16} \]
This matches our requirement.

### Conclusion
The transition in `He^(+)` that has the same wave number as the first line of the Lyman series of hydrogen is from \( n_2 = 4 \) to \( n_1 = 2 \).