A proton of mass `1.6times10^-27`kg goes round in a circular orbit of radius 0.12 m under a certripetal force of `6 times 10^-14` N. then the frequency of revolution of the proton is about

Options

A. `1.25 times 10^6` cycles per second
B. `2.50 times 10^6` cycles per second
C. `3.75 times 10^6` cycles per second
D. `5.00 times 10^6` cycles per second

Answer

To find the frequency of revolution of a proton moving in a circular orbit, we can follow these steps:

### Step 1: Write down the formula for centripetal force.
The centripetal force \( F_c \) acting on an object moving in a circular path is given by the formula:
\[ F_c = m \cdot R \cdot \omega^2 \]
where:
- \( m \) is the mass of the proton,
- \( R \) is the radius of the circular path,
- \( \omega \) is the angular velocity.

### Step 2: Relate angular velocity to frequency.
The angular velocity \( \omega \) can be expressed in terms of frequency \( \nu \) using the formula:
\[ \omega = 2 \pi \nu \]
Thus, we can substitute \( \omega \) in the centripetal force equation:
\[ F_c = m \cdot R \cdot (2 \pi \nu)^2 \]

### Step 3: Simplify the equation.
Expanding the equation gives:
\[ F_c = m \cdot R \cdot 4 \pi^2 \nu^2 \]
We can rearrange this to solve for \( \nu^2 \):
\[ \nu^2 = \frac{F_c}{m \cdot R \cdot 4 \pi^2} \]

### Step 4: Substitute the given values.
We know:
- \( F_c = 6 \times 10^{-14} \, \text{N} \)
- \( m = 1.6 \times 10^{-27} \, \text{kg} \)
- \( R = 0.12 \, \text{m} \)

Substituting these values into the equation for \( \nu^2 \):
\[ \nu^2 = \frac{6 \times 10^{-14}}{(1.6 \times 10^{-27}) \cdot (0.12) \cdot (4 \pi^2)} \]

### Step 5: Calculate \( \nu^2 \).
Calculating the denominator:
\[ 4 \pi^2 \approx 39.478 \]
Now substituting:
\[ \nu^2 = \frac{6 \times 10^{-14}}{(1.6 \times 10^{-27}) \cdot (0.12) \cdot 39.478} \]
Calculating the denominator:
\[ (1.6 \times 10^{-27}) \cdot (0.12) \cdot 39.478 \approx 7.5696 \times 10^{-28} \]
Now substituting back:
\[ \nu^2 = \frac{6 \times 10^{-14}}{7.5696 \times 10^{-28}} \approx 7.93 \times 10^{13} \]

### Step 6: Take the square root to find \( \nu \).
\[ \nu = \sqrt{7.93 \times 10^{13}} \approx 8.9 \times 10^6 \, \text{Hz} \]

### Final Answer:
The frequency of revolution of the proton is approximately \( 8.9 \times 10^6 \, \text{Hz} \).

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