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 3.60 MeV .

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

B) If the beam current is 450 A how many deuterons strike the target each second?

Explanation / Answer

A)

B=1.25T , q= 1.6*10-19C, m=3.34*10-27kg , I=450A

Let us first calculate velocity of deuteron,

KE=1/2mv2

3.60MeV=1/2*3.34*10-27*v2

(3.60*106)*(1.6*10-19) J = 1/2*3.34*10-27*v2           => v= 1.86*107m/s

Radius of the orbit = r= mv/qB = (3.34*10-27*1.86*107)/(1.6*10-19*1.25) =0.31062m

diameter = d= 2r= 2*0.31062m = 0.62124m = 62.124cm

B) I= dq/dt = n*e/dt

n=(dq*dt)/e = (450A*1s)/(1.6*10-19) = 2.8*1021 electrons

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