Question

An electron of kinetic energy 40 keV moves in a circular orbit perpendicular to a magnetic field of 0.375 T.

Find the radius of the orbit.

find the frequency if the motion.

Answer 1

let v is the speed of the electron

we know, KE = (1/2)*m*v^2

==> v = sqrt(2*KE/m)

= sqrt(2*40*1.6*10^-19/(9.1*10^-31))

= 3.75*10^6 m/s

we know, radius of orbit, r = m*v/(B*q)

= 9.1*10^-31*3.75*10^6/(0.375*1.6*10^-19)

= 5.69*10^-5 m <<<<<<<<<<<<-----------------Answer

frequency of motion, f = B*q/(2*pi*m)

= 0.375*1.6*10^-19/(2*pi*9.1*10^-31)

= 1.05*10^10 Hz <<<<<<<<<<<<-----------------Answer