`alpha` particle is revolving in radius r with frequency f then find value of magnetic dipole moment .

Options

A. 2evr
B. evr
C. 3evr
D. 4evr

Explanation

To find the magnetic dipole moment of an alpha particle revolving in a circular path of radius \( r \) with frequency \( f \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charge of the Alpha Particle:**
An alpha particle consists of 2 protons and 2 neutrons, hence it has a charge of \( 2e \), where \( e \) is the elementary charge (\( e \approx 1.6 \times 10^{-19} \) C).

2. **Determine the Current:**
The current \( I \) due to the revolving charge can be defined as the charge passing a point per unit time. The time period \( T \) for one complete revolution is given by:
\[ T = \frac{1}{f} \]
Therefore, the current \( I \) can be expressed as:
\[ I = \frac{\text{Charge}}{\text{Time period}} = \frac{2e}{T} = 2e f \]

3. **Calculate the Area of the Circular Path:**
The area \( A \) of the circular path with radius \( r \) is given by:
\[ A = \pi r^2 \]

4. **Calculate the Magnetic Dipole Moment:**
The magnetic dipole moment \( \mu \) is given by the product of the current and the area:
\[ \mu = I \cdot A \]
Substituting the expressions for \( I \) and \( A \):
\[ \mu = (2e f) \cdot (\pi r^2) \]
Thus, we have:
\[ \mu = 2e f \pi r^2 \]

5. **Final Expression:**
We can simplify the expression to focus on the terms relevant to the magnetic dipole moment:
\[ \mu = e f \cdot (2 \pi r^2) \]

### Conclusion:
The magnetic dipole moment \( \mu \) of the alpha particle revolving in a circular path of radius \( r \) with frequency \( f \) is given by:
\[ \mu = e f \cdot (2 \pi r^2) \]