An alpha particle is accelerated in a cyclotron with dees of radius 40 em and operating at frequency of 20 MHz. Calculate the kinetic energy of the alpha particle if magnetic field in cyclotron ia 1 tesla.
Given, mass of alpha particle `m_(alpha) = 6.65 xx 10^(-27) `kg and charge of `alpha` - particle `q_(alpha) = 3.2 xx 10^(-19) C`.

Answer

To calculate the kinetic energy of an alpha particle accelerated in a cyclotron, we can follow these steps:

### Step 1: Identify the given values
- Radius of the dees, \( R = 40 \, \text{cm} = 0.4 \, \text{m} \)
- Frequency, \( f = 20 \, \text{MHz} = 20 \times 10^6 \, \text{Hz} \)
- Magnetic field, \( B = 1 \, \text{T} \)
- Mass of the alpha particle, \( m_{\alpha} = 6.65 \times 10^{-27} \, \text{kg} \)
- Charge of the alpha particle, \( q_{\alpha} = 3.2 \times 10^{-19} \, \text{C} \)

### Step 2: Calculate the velocity of the alpha particle
The velocity \( v \) of a charged particle in a cyclotron can be calculated using the formula:

\[ v = \frac{q B R}{m} \]

Substituting the values:

\[ v = \frac{(3.2 \times 10^{-19} \, \text{C}) \cdot (1 \, \text{T}) \cdot (0.4 \, \text{m})}{6.65 \times 10^{-27} \, \text{kg}} \]

Calculating the numerator:

\[ = 3.2 \times 10^{-19} \cdot 1 \cdot 0.4 = 1.28 \times 10^{-19} \]

Now calculating \( v \):

\[ v = \frac{1.28 \times 10^{-19}}{6.65 \times 10^{-27}} \approx 1.92 \times 10^{7} \, \text{m/s} \]

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

\[ KE = \frac{1}{2} m v^2 \]

Substituting the values:

\[ KE = \frac{1}{2} \cdot (6.65 \times 10^{-27} \, \text{kg}) \cdot (1.92 \times 10^{7} \, \text{m/s})^2 \]

Calculating \( v^2 \):

\[ v^2 = (1.92 \times 10^{7})^2 = 3.6864 \times 10^{14} \, \text{m}^2/\text{s}^2 \]

Now substituting back into the kinetic energy formula:

\[ KE = \frac{1}{2} \cdot (6.65 \times 10^{-27}) \cdot (3.6864 \times 10^{14}) \]

Calculating \( KE \):

\[ KE = 0.5 \cdot 6.65 \times 10^{-27} \cdot 3.6864 \times 10^{14} \approx 1.2281 \times 10^{-12} \, \text{J} \]

### Final Result
The kinetic energy of the alpha particle is approximately:

\[ KE \approx 1.2281 \times 10^{-12} \, \text{J} \]

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