An `alpha-`particle is describing a circle of radius 0.45 m in a field of magnetic induction of 1.2 T. Find its speed, frequency of rotation and kinetic energy. What potential difference will be required to accelerate the particle so as to give this much of the energy to it? The mass of `alpha-`particle is `6.8 xx 10^(-27) kg` and its charge is twice the charge of proton, i.e.,`3.2 xx 10^(-19) C`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the speed of the alpha particle
The magnetic force provides the centripetal force required for circular motion. The equation relating these forces is:

\[ BQV = \frac{MV^2}{R} \]

Where:
- \( B \) = magnetic field induction = 1.2 T
- \( Q \) = charge of the alpha particle = \( 3.2 \times 10^{-19} \, C \)
- \( R \) = radius of the circular path = 0.45 m
- \( M \) = mass of the alpha particle = \( 6.8 \times 10^{-27} \, kg \)

Rearranging the equation to find \( V \):

\[ V = \frac{BQR}{M} \]

Substituting the values:

\[ V = \frac{(1.2 \, T)(3.2 \times 10^{-19} \, C)(0.45 \, m)}{6.8 \times 10^{-27} \, kg} \]

Calculating:

\[ V = \frac{(1.2)(3.2)(0.45)}{6.8 \times 10^{-27}} = \frac{0.1728 \times 10^{-19}}{6.8 \times 10^{-27}} \approx 2.56 \times 10^{7} \, m/s \]

### Step 2: Calculate the frequency of rotation
The frequency \( f \) can be calculated using the formula:

\[ f = \frac{V}{2\pi R} \]

Substituting the values:

\[ f = \frac{2.56 \times 10^{7} \, m/s}{2\pi(0.45 \, m)} \]

Calculating:

\[ f \approx \frac{2.56 \times 10^{7}}{2 \times 3.14 \times 0.45} \approx \frac{2.56 \times 10^{7}}{2.827} \approx 9.05 \times 10^{6} \, Hz \]

### Step 3: Calculate the kinetic energy
The kinetic energy \( E \) of the alpha particle can be calculated using:

\[ E = \frac{1}{2} MV^2 \]

Substituting the values:

\[ E = \frac{1}{2} (6.8 \times 10^{-27} \, kg)(2.56 \times 10^{7} \, m/s)^2 \]

Calculating:

\[ E = \frac{1}{2} (6.8 \times 10^{-27})(6.5536 \times 10^{14}) \approx 2.23 \times 10^{-12} \, J \]

### Step 4: Calculate the potential difference required
The potential difference \( V \) required to accelerate the particle to give it this energy is given by:

\[ E = QV \]

Rearranging gives:

\[ V = \frac{E}{Q} \]

Substituting the values:

\[ V = \frac{2.23 \times 10^{-12} \, J}{3.2 \times 10^{-19} \, C} \]

Calculating:

\[ V \approx 6.97 \times 10^{6} \, V \approx 6.97 \, MV \]

### Summary of Results
- Speed \( V \approx 2.56 \times 10^{7} \, m/s \)
- Frequency \( f \approx 9.05 \times 10^{6} \, Hz \)
- Kinetic Energy \( E \approx 2.23 \times 10^{-12} \, J \)
- Potential Difference \( V \approx 6.97 \times 10^{6} \, V \)