If an electron is revolving in a circular orbit of radius `0.5A^(@)` with a velocity of `2.2xx10^(6)m//s`. The magnetic dipole moment of the revolving electron is

Options

A. `8.8xx10^(-24)Am`
B. `8.8xx10^(-23)Am`
C. `8.8xx10^(-22)Am`
D. `8.8xx10^(21)Am`

Explanation

To find the magnetic dipole moment (M) of an electron revolving in a circular orbit, we can use the formula:

\[ M = \frac{Q \cdot V \cdot R}{2} \]

where:
- \( Q \) is the charge of the electron,
- \( V \) is the velocity of the electron,
- \( R \) is the radius of the orbit.

### Step 1: Identify the values
- The charge of the electron \( Q = 1.6 \times 10^{-19} \) C (Coulombs).
- The velocity of the electron \( V = 2.2 \times 10^6 \) m/s.
- The radius of the orbit \( R = 0.5 \) Å (angstrom), which is \( 0.5 \times 10^{-10} \) m.

### Step 2: Substitute the values into the formula
Substituting the values into the formula for magnetic dipole moment:

\[ M = \frac{(1.6 \times 10^{-19}) \cdot (2.2 \times 10^6) \cdot (0.5 \times 10^{-10})}{2} \]

### Step 3: Calculate the numerator
First, calculate the product of the values in the numerator:

\[ 1.6 \times 10^{-19} \cdot 2.2 \times 10^6 \cdot 0.5 \times 10^{-10} \]

Calculating step-by-step:
1. Calculate \( 1.6 \times 2.2 = 3.52 \).
2. Now multiply by \( 0.5 = 3.52 \times 0.5 = 1.76 \).
3. Combine the powers of ten: \( 10^{-19} \cdot 10^6 \cdot 10^{-10} = 10^{-23} \).

So, we have:

\[ 1.76 \times 10^{-23} \]

### Step 4: Divide by 2
Now, divide the result by 2:

\[ M = \frac{1.76 \times 10^{-23}}{2} = 0.88 \times 10^{-23} = 8.8 \times 10^{-24} \text{ A m}^2 \]

### Final Result
Thus, the magnetic dipole moment of the revolving electron is:

\[ M = 8.8 \times 10^{-24} \text{ A m}^2 \]