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 times 10^-27kg. The deuterons exit the cyclotron with a kinetic energy of 5.00 MeV. What is the speed of the deuterons when they exit? If the magnetic field inside the cyclotron is 1.25 T, what is the diameter of the deuterons' largest orbit, just before they exit? If the beam current is 400 /mewA, how many deuterons strike the target each second?

Explanation / Answer

a) we know that

E = (1/2)mv^2

isolate for velocity v = sqrt(2E/m)

where E= 5.00MeV = 5*10^6*1.6*10^(-19) = 8*10^(-13) J

v = sqrt (2*8*10^(-13)/(3.34*10^(-27))) = 2.19*10^7 m/s

b) We know that the equation for centripital force is mv^2/r, we also know that the only force present in the system is the force caused by the magnetic field, q*(V x B), therefore

mV^2/r = qVB, where q is the charge of one proton

so r = mv/qB

r = 3.34*10^(-27)*2.19*10^7/(1.6*10^(-19)*1.25) = 0.3657 m

diameter = 2*r = 2*3657 = 0.731 m

c) current I = q/t

since t = 1s,

I = q

Therefore the total charge that strikes the target each second is 400microcoulomb.

However, we are 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)

number of deuterons = 400*10^(-6)/(1.6*10^(-19)) = 2.5*10^(15)

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