One electron collides elastically with a second electron initially at rest. Afte

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 3.20 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0540 T. Determine the energy (in keV) of the incident electron

A cyclotron designed to accelerate protons has an outer radius of 0.410 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 600 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.350 T.

(a) Find the cyclotron frequency for the protons in this cyclotron. in rad/sec
(b) Find the speed at which protons exit the cyclotron. in m/s
(c) Find their maximum kinetic energy. eV
(d) How many revolutions does a proton make in the cyclotron? in revolutions
(e) For what time interval does one proton accelerate? in seconds

Explanation / Answer

1.

Here the centripetal force is balanced by magnetic force

mv2/r =qvB

v=qBr/m

since the collision is perfectly elastic =

KE=(1/2)mV1f2 +(1/2)mV2f2=(1/2)(q2B2/m)(R12+R22)

KE=(1/2)(1.6*10-19)2*0.0542/(9.11*10-31)[0.012+0.0322]

KE=4.6*10-14 J

in eV

KE =(4.6*10-14)(1.6*10-19)

KE=287.9 KeV

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