Question

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.00 cm and 2.60 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0560 T. Determine the energy (in keV) of the incident electron.

Answer 1

given

r1 =1cm= 0.010 m

r2 =2.6cm= 0.026 m;

charge of electron( q) = 1.60 x 10^-19 C;

B = 0.056 T;

mass of electron(m) = 9.11 x 10^-31 kg

First calculate the velocity of each electron:

v1 = r₁qB / m =.01*1.6*10^-19*0.056/ 9.11 x 10^-31

= 9.8355 x 10^7 m/s

v2 = r2qB / m=.026*1.6*10^-19*0.056/ 9.11 x 10^-31

= 2.557 x 10^8 m/s

now kinetic energy of each electron:

K1 = .5mv² = 4.41 x 10^-15 J

K2 = .5mv² = 29.78 x 10^-15 J

Third, calculate the total kinetic energy:

K = K1 + K2 = 3.419 x 10^-14 J

now convert in electron volts:

(3.4194 x 10^-14 J) / (1.60 x 10^-19 J) = 213687.5 eV

= 213.69 keV