If n=2 for `He^(+)` ion than find out the wave length

Options

A. 3.33 Å
B. 6.42 Å
C. 1.47 Å
D. 2.37 Å

Explanation

To find the wavelength of the `He^(+)` ion when \( n = 2 \), we can use Bohr's model of the hydrogen-like atom. Here’s a step-by-step solution:

### Step 1: Understand the formula
According to Bohr's theory, the relationship between the radius of the orbit, the principal quantum number (\( n \)), and the atomic number (\( z \)) is given by:
\[ r = \frac{n^2}{z} \cdot a_0 \]
where \( a_0 \) is the Bohr radius, approximately \( 0.529 \) Å.

### Step 2: Identify the values
For the `He^(+)` ion:
- \( n = 2 \)
- \( z = 2 \) (since helium has an atomic number of 2)

### Step 3: Calculate the radius
Substituting the values into the formula for the radius:
\[ r = \frac{n^2}{z} \cdot a_0 = \frac{2^2}{2} \cdot 0.529 \, \text{Å} = \frac{4}{2} \cdot 0.529 \, \text{Å} = 2 \cdot 0.529 \, \text{Å} = 1.058 \, \text{Å} \]

### Step 4: Use the wavelength formula
The wavelength (\( \lambda \)) can be derived from the relationship:
\[ \lambda = \frac{2 \pi r}{n} \]
Substituting the value of \( r \):
\[ \lambda = \frac{2 \pi (1.058 \, \text{Å})}{2} \]
This simplifies to:
\[ \lambda = \pi (1.058 \, \text{Å}) \approx 3.33 \, \text{Å} \]

### Step 5: Final answer
Thus, the wavelength for the `He^(+)` ion when \( n = 2 \) is approximately:
\[ \lambda \approx 3.33 \, \text{Å} \]