An electron revolving in an orbit of radius `0.5 Å` in a hydrogen atom executes per secon. The magnetic momentum of electron due to its orbital motion will be

Options

A. `1.256 xx 10^(-23) A m^(2)`
B. `653 xx 10^(-26) A m^(2)`
C. `10^(-3) A m^(2)`
D. `256 xx 10^(-26) A m^(2)`

Explanation

To find the magnetic moment of an electron revolving in a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Radius of the orbit, \( r = 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m} \)
- Charge of the electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)
- Frequency of revolution, \( f = 10^{16} \, \text{Hz} \)

2. **Calculate the Area of the Orbit**:
The area \( A \) of the circular orbit can be calculated using the formula:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (0.5 \times 10^{-10})^2 \]
\[ A = \pi \times (0.25 \times 10^{-20}) = 0.785 \times 10^{-20} \, \text{m}^2 \]

3. **Calculate the Current \( I \)**:
The current \( I \) due to the revolving electron can be calculated using the formula:
\[ I = e \times f \]
Substituting the values of \( e \) and \( f \):
\[ I = (1.6 \times 10^{-19}) \times (10^{16}) = 1.6 \times 10^{-3} \, \text{A} \]

4. **Calculate the Magnetic Moment \( M \)**:
The magnetic moment \( M \) is given by the formula:
\[ M = I \times A \]
Substituting the values of \( I \) and \( A \):
\[ M = (1.6 \times 10^{-3}) \times (0.785 \times 10^{-20}) \]
\[ M = 1.256 \times 10^{-23} \, \text{A m}^2 \]

5. **Final Answer**:
The magnetic moment of the electron due to its orbital motion is:
\[ M = 1.256 \times 10^{-23} \, \text{A m}^2 \]