Question

An electron of kinetic energy 13.0 keV moves in a circular orbit perpendicular to a magnetic field of 0.360 T. Find the radius of the orbit. Tries 0/6

Problem 26-32b: Find the period of the motion.

Answer 1

A)

Velocity, v = sqrt(2K/m)

v = sqrt(2 x 13 x 1.6 x 10^-16/9.1 x 10^-31) = 6.76 x 10^7 m/s

Radius, r = mv/qB

r = (9.1 x 10^-31 x 6.76 x 10^7)/(1.6 x 10^-19 x 0.36)

r = 1.07 x 10^-3 m

B)

Time, t = 2 pi m/qB

t = (2 pi x 9.1 x 10^-31)/(1.6 x 10^-19 x 0.36)

t = 9.93 x 10^-11 sec

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