A cyclotron is used to produce a beam of high-energy deuterons that then collide

Question

A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes for a medical procedure. Deuterons are nuclei of deuterium, an isotope of hydrogen, consisting of one neutron and one proton, with total mass 3.34×1027kg. The deuterons exit the cyclotron with a kinetic energy of 5.80 MeV .

A) What is the speed of the deuterons when they exit?

B) If the magnetic field inside the cyclotron is 1.25 T, what is the diameter of the deuterons' largest orbit, just before they exit?

C) If the beam current is 350 A how many deuterons strike the target each second?

Explanation / Answer

Kinetic enrgy , Ke = 5.80 Mev = 9.293*10^-13 J
mass, m = 3.34*10^-27 kg

Part A:
velocity, v = sqrt(2*Ke/m)
v =sqrt(2 *9.293*10^-13/(3.34*10^-27))
v = 2.36*10^7 m/s

Part b:
magnatic field, b = 1.25 T

From Newton second law, Centriptial FOrce = force due to field
mv^2/r = q*v*B

Solving for radius , r = m*v/q*B ( q is charge of proton)
r = (3.34*10^-27 * 2.36*10^7)/(1.6*10^-19 * 1.25)
r = 0.394 m

Therefore diameter, d = 2*r = 0.394 * 2 = 0.788 m

Part C:
current, I = 350 uA

Since Current, i = charge/time = q/1 = q
Therfore, Total charge, q = 350 uC

However, we re trying to find the NUMBER of the deuterons,
so divide that total charge by each charge of a deuteron
(which is the same as the charge of a proton)

no of deutrons, n = q/e = (350*10^-6)/(1.6*10^-19) = 2.188*10^15

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