Question

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.50 cm and 3.20 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0380 T. Determine the energy (in keV) of the incident electron.

Answer 1

r1 = 0.015 m; r2 = 0.032 m; q = 1.60 x 10^-19 C; B = 0.038 T; m = 9.11 x 10^-31 kg

calculate the velocity of each electron:

v1 = r1qB / m = 1.00109 x 10⁸ m/s

v2 = r2qB / m = 2.13567 x 10⁸ m/s

calculate the kinetic energy of each electron:

K1 = 0.5mv² = 4.5649 x 10^-15 J

K2 = 0.5mv² = 20.7757 x 10^-15 J

calculate the total kinetic energy:

K = K1 + K2 = 25.3406 x 10^-15 J

convert into electron volts:

(25.3406 x 10^-15 J) / (1.60 x 10^-19 J) = 158.38 keV

energy (in keV) of the incident electron = 158.38 keV