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A proton of mass 1.6 xx 10^(-27)kg goes...

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

A

`0.08 xx 10^(8)` cycles per second

B

`4 xx 10^(8)` cycles per second

C

`8 xx 10^(8)` cycles per second

D

`12 xx 10^(8)` cycles per second

Text Solution

Verified by Experts

The correct Answer is:
A
`F = momega^(2) = m 4pi^(2) f^(2)r`
`rArr 4 xx 10^(-13) = 1.6 xx 10^(-27) xx 4pi^(2) xx f^(2) xx 0.10`
`rArr f = 0.08 xx 10^(8)` cycles/sec.
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