The wavelength of radiation emitted when in`He^(+)` electron falls infinity to stationary state would be ` (R =1.098 xx 10 ^7 m^(-1))`

Options

A. `2.2xx10^(-8) m`
B. `2.2xx10^(-9) m`
C. `120 m`
D. `22xx10^(7) m`

Answer

To solve the problem of finding the wavelength of radiation emitted when an electron in a He\(^+\) ion falls from infinity to a stationary state, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Variables**:
- We know that the electron is falling from an initial state \( n_2 = \infty \) to a final state \( n_1 = 1 \) (the stationary state).
- The atomic number \( z \) for helium (He) is 2.
- The Rydberg constant \( R \) is given as \( 1.098 \times 10^7 \, \text{m}^{-1} \).

2. **Use the Rydberg Formula**:
The Rydberg formula for the wavelength of emitted radiation is given by:
\[ \frac{1}{\lambda} = R \cdot z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Here, \( \lambda \) is the wavelength, \( R \) is the Rydberg constant, \( z \) is the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states respectively.

3. **Substitute the Values**:
Substitute \( n_1 = 1 \), \( n_2 = \infty \), and \( z = 2 \) into the formula:
\[ \frac{1}{\lambda} = 1.098 \times 10^7 \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:
\[ \frac{1}{\lambda} = 1.098 \times 10^7 \cdot 4 \cdot 1 \]

4. **Calculate \( \frac{1}{\lambda} \)**:
\[ \frac{1}{\lambda} = 1.098 \times 10^7 \cdot 4 = 4.392 \times 10^7 \, \text{m}^{-1} \]

5. **Find \( \lambda \)**:
To find \( \lambda \), take the reciprocal:
\[ \lambda = \frac{1}{4.392 \times 10^7} \approx 2.27 \times 10^{-8} \, \text{m} \]

### Final Answer:
The wavelength of the radiation emitted when an electron in He\(^+\) falls from infinity to the stationary state is approximately:
\[ \lambda \approx 2.27 \times 10^{-8} \, \text{m} \]