4.) One electron collides elastically with a second electroninitially at rest. A

Question

4.) One electron collides elastically with a second electroninitially at rest. After the collision, the radii of theirtrajectories are 1.00 cm and 3.10 cm. Thetrajectories are perpendicular to a uniform magnetic field ofmagnitude 0.0580 T. Determine the energy(in keV) of the incident electron.
1 keV

Explanation / Answer

we have the velocity of a charge in the magnetic field(trajectories are perpendicular to a uniform magnetic field ) mv^2/R =Bvq   => v =Bq/(mR) => the velocity of two electron : v1 = Be/(mR1)       v2 = Be/(mR2) conservation of energy => energy of incident : E = (1/2)mv1^2 +(1/2)mv2^2 =(1/2)(Be)^2/m (R1^2+R2^2) =5.02x10^-14 J =313.75keV

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