The maximum velocity to which a proton can be accelerated in a cyclotron of 10 MHZ frequency and radius 50 cm is

Options

A. `6.28xx10^(8) ms^(-1)`
B. `3.14xx10^(8) ms^(-1)`
C. `6.28xx10^(7) ms^(-1)`
D. `3.14 xx10^(7) ms^(-1)`

Explanation

To find the maximum velocity to which a proton can be accelerated in a cyclotron with a frequency of 10 MHz and a radius of 50 cm, we can follow these steps:

### Step 1: Understand the relationship between velocity, radius, and frequency.
In a cyclotron, the velocity \( V \) of a charged particle is given by the formula:
\[ V = r \omega \]
where \( r \) is the radius and \( \omega \) is the angular frequency.

### Step 2: Relate angular frequency to frequency.
The angular frequency \( \omega \) can be expressed in terms of the frequency \( \nu \):
\[ \omega = 2 \pi \nu \]
where \( \nu \) is the frequency in hertz (Hz).

### Step 3: Substitute the expression for \( \omega \) into the velocity formula.
Substituting \( \omega \) into the velocity formula gives:
\[ V = r (2 \pi \nu) \]

### Step 4: Convert the radius from centimeters to meters.
Given that the radius \( r \) is 50 cm, we convert it to meters:
\[ r = 50 \, \text{cm} = \frac{50}{100} \, \text{m} = 0.5 \, \text{m} \]

### Step 5: Convert the frequency from MHz to Hz.
The frequency \( \nu \) is given as 10 MHz. We convert it to hertz:
\[ \nu = 10 \, \text{MHz} = 10 \times 10^6 \, \text{Hz} = 10^7 \, \text{Hz} \]

### Step 6: Substitute the values into the velocity formula.
Now we can substitute \( r \) and \( \nu \) into the velocity equation:
\[ V = 0.5 \, \text{m} \times (2 \pi \times 10^7 \, \text{Hz}) \]

### Step 7: Calculate the maximum velocity.
Calculating the above expression:
\[ V = 0.5 \times 2 \times 3.14 \times 10^7 \]
\[ V = 1 \times 3.14 \times 10^7 \, \text{m/s} \]
\[ V \approx 3.14 \times 10^7 \, \text{m/s} \]

### Final Answer:
The maximum velocity to which a proton can be accelerated in the cyclotron is approximately:
\[ V \approx 3.14 \times 10^7 \, \text{m/s} \]

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