A proton of mass 1.6 times 10^{-27} , kg move | vedclass

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Question

(A) The centripetal force $F$ is given by $F = m \omega^2 r$,where $\omega = 2 \pi n$ and $n$ is the frequency of revolution.Substituting $\omega = 2

Vedclass · Accepted Answer

$0.08 	imes 10^8 \, 	ext{cycles per sec}$

Vedclass · Answer

(A) The centripetal force $F$ is given by $F = m \omega^2 r$,where $\omega = 2 \pi n$ and $n$ is the frequency of revolution.Substituting $\omega = 2 \pi n$ into the force equation: $F = m (2 \pi n)^2 r = 4 \pi^2 m n^2 r$.Given values: $m = 1.6 	imes 10^{-27} \, kg$,$r = 0.10 \, m$,and $F = 4 	imes 10^{-13} \, N$.Rearranging for $n^2$: $n^2 = \frac{F}{4 \pi^2 m r}$.$n^2 = \frac{4 	imes 10^{-13}}{4 	imes (3.14)^2 	imes 1.6 	imes 10^{-27} 	imes 0.10}$.$n^2 = \frac{10^{-13}}{9.86 	imes 1.6 	imes 10^{-28}} = \frac{10^{15}}{15.776} \approx 0.0634 	imes 10^{15} = 6.34 	imes 10^{13}$.Taking the square root: $n \approx \sqrt{63.4} 	imes 10^6 \approx 7.96 	imes 10^6 \, Hz \approx 0.08 	imes 10^8 \, Hz$.

$A$ proton of mass $1.6 \times 10^{-27} \, kg$ moves in a circular orbit of radius $0.10 \, m$ under a centripetal force of $4 \times 10^{-13} \, N$. The frequency of revolution of the proton is about:

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