Question

A proton of mass \[1.6 \times {10^{ - 27}}kg\] goes round in a circular orbit of radius 0.12 m under a centripetal force of \[6 \times {10^{ - 14}}N\]. Then find the frequency of revolution of the proton.
A. \[1.25 \times {10^6}\,cycles/\sec \]
B. \[2.50 \times {10^6}\,cycles/\sec \]
C. \[3.75 \times {10^6}\,cycles/\sec \]
D. \[5 \times {10^6}\,cycles/\sec \]

Answer 1

Hint:Whenever a body passes through the curve centripetal force acts on the body and the direction of the force is at the body and is always acting away towards the center of the curve. The centripetal force is necessary to act on the body as the curve arises to balance the forces on the body.

Formula Used:
To find the centripetal force the formula is,
\[F = \dfrac{{m{v^2}}}{r}\]
Where, m is the mass of the body, v is the speed of the body and r is radius.

Complete step by step solution:
Consider a proton of mass \[1.6 \times {10^{ - 27}}kg\] goes round in a circular orbit of radius 0.12 m under a centripetal force of \[6 \times {10^{ - 14}}N\]. Here, we need to find the frequency of revolution of the proton. We know that the centripetal force is,
\[F = \dfrac{{m{v^2}}}{r}\]
\[\Rightarrow F = mr{\omega ^2}\]………. (1)
We know that the angular frequency,
\[\omega = \dfrac{{2\pi }}{T}\]
\[ \Rightarrow \omega = 2\pi \upsilon \]
Here, \[\upsilon \] is the frequency of revolution of the proton.

Then, \[{\omega ^2} = 4{\pi ^2}{\upsilon ^2}\]
Substitute the value of \[{\omega ^2}\]in equation (1) we get,
\[F = mr4{\pi ^2}{\upsilon ^2}\]
\[{\upsilon ^2} = \dfrac{F}{{mr4{\pi ^2}}}\]
Substitute the value of \[F = 6 \times {10^{ - 14}}N\], \[m = 1.6 \times {10^{ - 27}}kg\], \[r = 0.12m\] then, we obtain,
\[{\upsilon ^2} = \dfrac{{6 \times {{10}^{ - 14}}}}{{1.6 \times {{10}^{ - 27}} \times 0.12 \times 4 \times {{\left( {3.142} \right)}^2}}}\]
\[{\upsilon ^2} = 7.9136 \times {10^{12}}\]
\[\upsilon = 2.81 \times {10^6}\,cycles/\sec \]
The nearest value to the given option is,
\[\upsilon = 2.5 \times {10^6}\,cycles/\sec \]
Therefore, the frequency of revolution of the proton is \[2.5 \times {10^6}\,cycles/\sec \].

Hence, option B is the correct answer.

Note:In this problem it is important to remember that the equation of centripetal force and thereby using the relation between the linear velocity and angular velocity, the frequency of revolution of the proton can be determined.