A proton of mass 1.6 × 10⁻²⁷ kg goes round in a circular orbit of radius 0.10 m under a centripetal force of 4 × 10⁻¹³ N. then the frequency of revolution of the proton is about

The correct option is 0.08 × 10⁸ cycles per sec.

To find the frequency of revolution of the proton, we start by analyzing the centripetal force acting on it. When a particle of mass $m$ moves in a circular orbit of radius $r$ with a constant speed $v$, the centripetal force $F$ is given by the formula:
$F=\frac{m{v}^2}{r}$

The relationship between linear velocity $v$, radius $r$, and angular velocity $\omega$ is:
$v=r\omega$
Substituting this into the centripetal force formula gives:
$F=mr{\omega }^2$

The angular velocity $\omega$ is related to the frequency of revolution $f$ by the equation:
$\omega =2\pi f$
Substituting $\omega$ in terms of $f$ into our force equation, we get:
$F=mr{\left(2\pi f\right)}^2=4{\pi }^{2}mr{f}^2$

Now, we solve for the frequency $f$:
$f^2=\frac{F}{4{\pi }^{2}mr}$
Taking the square root of both sides gives:
$f=\frac{1}{2\pi }\sqrt{\frac{F}{mr}}$

We are given the following values:
• Mass of the proton, $m=1.6\times {10}^{-27}\text{kg}$
• Radius of the orbit, $r=0.10\text{m}$
• Centripetal force, $F=4\times {10}^{-13}\text{N}$

Substituting these values into the equation:
$f=\frac{1}{2\pi }\sqrt{\frac{4\times {10}^{-13}}{1.6\times {10}^{-27}\times 0.10}}$
Simplify the term inside the square root:
$\frac{4\times {10}^{-13}}{1.6\times {10}^{-28}}=2.5\times {10}^{15}=25\times {10}^{14}$

Now, find the square root:
$\sqrt{25\times {10}^{14}}=5\times {10}^7{\text{s}}^{-1}$

Substitute this back into the frequency expression:
$f=\frac{5\times {10}^7}{2\pi }$
Using $\pi \approx 3.14$:
$f\approx \frac{5\times {10}^7}{6.28}\approx 0.796\times {10}^7\text{Hz}$
Expressing this in the format of the options:
$f\approx 0.08\times {10}^8\text{cycles per sec}$