An electron is moving in a circular path under the influence of a transverse magnetic field of `3.57xx10^(-2)` T.If the value of e/m is `1.76xx10^11` c/kg, the frequency of revolution of the electron is

Options

A. 6.82 MHz
B. 1 GHz
C. 100 MHz
D. 62.8 MHz

Answer

B

Explanation

`R = (mv)/(qB) , omega = (v)/( R) = (qB)/(m)`
`f = (omega)/(2pi) = (1)/(2pi ) * (qB)/(m)`
= `(1.76 xx 10^(11) xx 3.57 xx 10^(-2))/((2 xx 3.14)) = 10^(9)` Hz = 1GHz.