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 of finding the transition in the \( \text{He}^+ \) ion that will have the same wave number as the first line of the Lyman series of hydrogen, we can follow these steps:

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

### Step 2: Calculate the Wave Number for Hydrogen
The wave number (\( \bar{\nu} \)) for a transition can be calculated using the Rydberg formula:
\[ \bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant for hydrogen, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

For the first line of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

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

### Step 3: Set Up the Equation for \( \text{He}^+ \)
For the \( \text{He}^+ \) ion, the Rydberg formula is modified to include the atomic number \( Z \) (which is 2 for helium):
\[ \bar{\nu} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting \( Z = 2 \):
\[ \bar{\nu} = R_H \cdot 2^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 4 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 4: Equate the Wave Numbers
Since we want the wave number for \( \text{He}^+ \) to be equal to that of hydrogen:
\[ 4 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3}{4} R_H \]
Dividing both sides by \( R_H \):
\[ 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3}{4} \]
Multiplying through by 4:
\[ 16 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 3 \]
Thus:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{16} \]

### Step 5: Check Possible Transitions
We need to find \( n_1 \) and \( n_2 \) such that:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{16} \]

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

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

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

4. **Option 4: Transition from \( n_2 = 6 \) to \( n_1 = 4 \)**
\[ \frac{1}{4^2} - \frac{1}{6^2} = \frac{1}{16} - \frac{1}{36} = \frac{36 - 16}{576} = \frac{20}{576} \quad \text{(not valid)} \]

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