A cyclotron (see figure below) designed to accelerate protons has an outer radiu

Question

A cyclotron (see figure below) designed to accelerate protons has an outer radius of 0.315 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 552 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.736 T.

(a) How many revolutions does a proton make in the cyclotron?
____________________________revolutions

(b) For what time interval does the proton accelerate?
___________________________s

(c) Find their maximum kinetic energy.

__________________________eV

( The cyclotron frequency for the protons in this cyclotron = 70510000 rad/s, the speed at which protons exit the cyclotron = 22210000 m/s)

The black, dashed, curved lines represent the path of the Alternating particles. D2 After being accelerated, the particles exit here North pole of magnet

Explanation / Answer

The speed at which protons exit the cyclotron = 2.221e7 m/s

(a) Number of revolutions = 2.57e6 / 2 * 552 (because it passes twice between the dees)

Number of revolutions = 2331 revolutions

(b) time interval, t = 2331 * 2*pi / 70510000 = 2.077e-4 seconds

(c) maximum kinetic energy

K.E = 1/2mv2

K.E = 1/2*1.67e-27*2.221e72

K.E = 4.118e-13 J

Now, to convert it into eV

K.E = 4.118e-13 J / 1.6e19

K.E = 2.57e6 m/s

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