An electron of kinetic energy 460 keV moves in a circular or

An electron of kinetic energy 46.0 keV moves in a circular orbit perpendicular to a magnetic field of 0.810 T. Find the radius of the orbit. Find the period of the motion.

Solution

radius of the orbit is calculated as follows:

r = mv/qB

= (m/qB){sqrt(2K/m)}

   = (9.11*10^-31)/(1.6x10^-19)(0.81)*{sqrt(2*46*10^3*1.6x10^-19 / (9.11*10^-31))}

   = 8.935e-4 m

= 0.894x10^-3 m

= 0.894 mm

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frequency of the motion:

f = qB/2*pi*m

= 1.6*10^-19*0.81 /2*pi*9.11x10^-31

= 2.265e+10 Hz

period of the motion:

T = 1/f = 4.41e-11 s