The de-Broglie wavelength of electgron in `3^(rd)` orbit of `He^(+1)` ion is approximately

Options

A. `2A^(@)`
B. `3A^(@)`
C. `4A^(@)`
D. `5A^(@)`

Explanation

To find the de-Broglie wavelength of the electron in the third orbit of the He\(^+\) ion, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (\(\lambda\)) is related to the radius of the orbit (\(r\)) and the principal quantum number (\(n\)) by the formula:
\[ \lambda = \frac{2\pi r}{n} \]

### Step 2: Determine the radius of the orbit
For a hydrogen-like atom, the radius of the \(n^{th}\) orbit is given by:
\[ r_n = \frac{0.529 \, \text{Å} \cdot n^2}{Z} \]
where \(Z\) is the atomic number. For the He\(^+\) ion, \(Z = 2\).

### Step 3: Substitute \(n\) and \(Z\) into the radius formula
For the third orbit (\(n = 3\)):
\[ r_3 = \frac{0.529 \, \text{Å} \cdot 3^2}{2} = \frac{0.529 \, \text{Å} \cdot 9}{2} = \frac{4.761 \, \text{Å}}{2} = 2.3805 \, \text{Å} \]

### Step 4: Substitute \(r\) and \(n\) into the de-Broglie wavelength formula
Now, substitute \(r\) and \(n\) into the de-Broglie wavelength formula:
\[ \lambda = \frac{2\pi (2.3805 \, \text{Å})}{3} \]

### Step 5: Calculate \(\lambda\)
Calculating the above expression:
\[ \lambda = \frac{2 \cdot 3.14159 \cdot 2.3805}{3} \approx \frac{14.943}{3} \approx 4.981 \, \text{Å} \]

### Step 6: Round the answer
Rounding the answer gives us approximately:
\[ \lambda \approx 5 \, \text{Å} \]

### Conclusion
Therefore, the de-Broglie wavelength of the electron in the third orbit of the He\(^+\) ion is approximately \(5 \, \text{Å}\).