Question

A proton goes round in a circular orbit of radius $0.01m$ under a centripetal force of $4\times {10}^{-3}$ N. What is the frequency of revolution of the proton?

• A. $2.5\times {10}^{12}$ Hz
• B. $4\times {10}^{13}$ Hz
• C. $8\times {10}^{12}{s}^{-1}$
• D. $16\times {10}^{12}{s}^{-1}$

Answer 1

The correct option is A $2.5\times {10}^{12}$ Hz

Centripetal force $\left(F_c\right)=m{\omega }^{2}R$
$\Rightarrow 4\times {10}^{-3}=(1.67\times {10}^{-27})\times {\omega }^2\times {10}^{-2}\Rightarrow {\omega }^2=\frac{(4\times {10}^{-3})}{(1.67\times {10}^{-27})\times {10}^{-2}}=2.39\times {10}^{26}\Rightarrow \omega =2.39\times {10}^{26}{s}^{-1}=1.55\times {10}^{13}{s}^{-1}$
$f=\frac{\omega }{2\pi }=\frac{1.55\times {10}^{13}}{2\pi }=2.46\times {10}^{12}Hz\approx 2.5\times {10}^{12}Hz$