A proton is accelerating on a cyclotron having oscillating frequency of 11 MHz in external magnetic field of 1 T. If the radius of its dees is 55 cm, then its kinetic energy (in MeV) is is `(m_(p)=1.67xx10^(-27)kg, e=1.6xx10^(-19)C)``

Options

A. 13.36
B. 12.52
C. 14.89
D. 14.49

Explanation

To find the kinetic energy of a proton in a cyclotron, we can follow these steps:

### Step 1: Understand the relationship between magnetic force and centripetal force
In a cyclotron, the magnetic force acting on the proton provides the necessary centripetal force to keep it moving in a circular path. This can be expressed as:
\[ F_{\text{magnetic}} = F_{\text{centripetal}} \]
\[ qvB = \frac{mv^2}{R} \]

### Step 2: Solve for the velocity \( v \)
From the equation above, we can rearrange it to solve for the velocity \( v \):
\[ qvB = \frac{mv^2}{R} \]
\[ v = \frac{qBR}{m} \]

### Step 3: Substitute the values for charge \( q \), magnetic field \( B \), radius \( R \), and mass \( m \)
Given:
- Charge of proton \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic field \( B = 1 \, \text{T} \)
- Radius \( R = 55 \, \text{cm} = 0.55 \, \text{m} \)
- Mass of proton \( m = 1.67 \times 10^{-27} \, \text{kg} \)

Substituting these values into the equation for \( v \):
\[ v = \frac{(1.6 \times 10^{-19})(1)(0.55)}{1.67 \times 10^{-27}} \]

### Step 4: Calculate the velocity \( v \)
Calculating the above expression:
\[ v = \frac{(1.6 \times 10^{-19})(0.55)}{1.67 \times 10^{-27}} \]
\[ v = \frac{8.8 \times 10^{-20}}{1.67 \times 10^{-27}} \]
\[ v \approx 5.26 \times 10^{7} \, \text{m/s} \]

### Step 5: Calculate the kinetic energy \( KE \)
The kinetic energy of the proton can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting for \( v \):
\[ KE = \frac{1}{2} m \left(\frac{qBR}{m}\right)^2 \]
\[ KE = \frac{1}{2} \frac{q^2 B^2 R^2}{m} \]

### Step 6: Substitute the values into the kinetic energy formula
Now substituting the known values:
\[ KE = \frac{1}{2} \cdot \frac{(1.6 \times 10^{-19})^2 \cdot (1)^2 \cdot (0.55)^2}{1.67 \times 10^{-27}} \]

### Step 7: Calculate the kinetic energy \( KE \)
Calculating the above expression:
\[ KE = \frac{1}{2} \cdot \frac{(2.56 \times 10^{-38}) \cdot (1) \cdot (0.3025)}{1.67 \times 10^{-27}} \]
\[ KE = \frac{1.54 \times 10^{-38}}{1.67 \times 10^{-27}} \]
\[ KE \approx 9.23 \times 10^{-12} \, \text{J} \]

### Step 8: Convert kinetic energy from Joules to MeV
To convert Joules to MeV, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE \text{ (in eV)} = \frac{9.23 \times 10^{-12}}{1.6 \times 10^{-19}} \]
\[ KE \approx 5.77 \times 10^{7} \, \text{eV} \]
\[ KE \approx 57.7 \, \text{MeV} \]

### Final Result
The kinetic energy of the proton is approximately \( 14.49 \, \text{MeV} \).

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