A proton of mass `1.6times10^-27`kg goes round in a circular orbit of radius 0.12 m under a certripetal force of `6 times 10^-14` N. then the frequency of revolution of the proton is about

Options

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

Answer

A

Explanation

`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.