A proton of mass `1.6 xx 10^(-27)kg` goes round in a circular orbit of radius 0.10 m under a centripetal force of `4 xx 10^(-13) N`. then the frequency of revolution of the proton is about

Options

A. `0.08 xx 10^(8)` cycles per sec
B. `4 xx 10^(8)` cycles per sec
C. `8 xx 10^(8)` cycles per sec
D. `12 xx 10^(8)` cycles per sec

Answer

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

### Step 1: Write down the given values
- Mass of the proton, \( m = 1.6 \times 10^{-27} \, \text{kg} \)
- Radius of the circular orbit, \( r = 0.10 \, \text{m} \)
- Centripetal force, \( F_c = 4 \times 10^{-13} \, \text{N} \)

### Step 2: Use the formula for centripetal force
The centripetal force acting on an object moving in a circular path is given by the formula:
\[ F_c = m \omega^2 r \]
where \( \omega \) is the angular velocity in radians per second.

### Step 3: Rearrange the formula to find \( \omega \)
We can rearrange the formula to solve for \( \omega \):
\[ \omega^2 = \frac{F_c}{m r} \]
Substituting the known values:
\[ \omega^2 = \frac{4 \times 10^{-13}}{(1.6 \times 10^{-27})(0.10)} \]

### Step 4: Calculate \( \omega^2 \)
Calculating the denominator:
\[ (1.6 \times 10^{-27})(0.10) = 1.6 \times 10^{-28} \]
Now substituting back into the equation:
\[ \omega^2 = \frac{4 \times 10^{-13}}{1.6 \times 10^{-28}} = 2.5 \times 10^{15} \]

### Step 5: Find \( \omega \)
Taking the square root to find \( \omega \):
\[ \omega = \sqrt{2.5 \times 10^{15}} = 5 \times 10^{7} \, \text{rad/s} \]

### Step 6: Calculate the frequency \( f \)
The frequency \( f \) in cycles per second (Hz) can be calculated from the angular velocity using the relation:
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of \( \omega \):
\[ f = \frac{5 \times 10^{7}}{2\pi} \]
Using \( \pi \approx 3.14 \):
\[ f \approx \frac{5 \times 10^{7}}{6.28} \approx 7.96 \times 10^{6} \, \text{Hz} \]

### Step 7: Express the frequency in scientific notation
To express \( 7.96 \times 10^{6} \) in terms of \( 10^{8} \):
\[ f \approx 0.0796 \times 10^{8} \, \text{Hz} \approx 0.08 \times 10^{8} \, \text{Hz} \]

### Final Answer
Thus, the frequency of revolution of the proton is approximately:
\[ \boxed{0.08 \times 10^{8} \, \text{Hz}} \]