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An electron is moving in a circular path...

An electron is moving in a circular path under the influence fo a transerve magnetic field of `3.57xx10^(-2)T`. If the value of `e//m` is `1.76xx10^(141) C//kg`. The frequency of revolution of the electron is

A

6.28 MHz

B

6.28 MHz

C

1 GHz

D

100 MHz

Text Solution

Verified by Experts

The correct Answer is:
C
We know that
`f = (qB)/(2 pi m)`
`f = (eB)/(2 pi m)`
`f = (1.76 xx 10^(11) xx 3.57 xx 10^(-2))/(2 xx 3.14) Hz`
`f = 10^(9)` Hz or 1 GHz
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