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 He\(^+\) falls from infinity to a stationary state, we can use the Rydberg formula for hydrogen-like atoms. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Transition
The electron is transitioning from an infinite energy level (n\(_2\) = ∞) to the ground state (n\(_1\) = 1).

### Step 2: Write the Rydberg Formula
The Rydberg formula for the wavelength of emitted radiation is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \(\lambda\) is the wavelength,
- \(R\) is the Rydberg constant,
- \(Z\) is the atomic number,
- \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and upper energy levels, respectively.

### Step 3: Substitute Values
For He\(^+\) (helium ion), the atomic number \(Z\) is 2. The values for \(n_1\) and \(n_2\) are:
- \(n_1 = 1\)
- \(n_2 = \infty\)

Substituting these values into the formula gives:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
Since \(\frac{1}{\infty^2} = 0\), the equation simplifies to:
\[ \frac{1}{\lambda} = R \cdot Z^2 \cdot 1 \]

### Step 4: Calculate \(\frac{1}{\lambda}\)
Substituting \(R = 1.098 \times 10^7 \, \text{m}^{-1}\) and \(Z = 2\):
\[ \frac{1}{\lambda} = 1.098 \times 10^7 \cdot 2^2 = 1.098 \times 10^7 \cdot 4 = 4.392 \times 10^7 \, \text{m}^{-1} \]

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

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