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 6.40 MeV . Part A What is the speed of the deuterons when they exit? Part 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? Part C If the beam current is 360 A how many deuterons strike the target each second?

Explanation / Answer

Given that

K.E=6.4MeV=10.24*10^-13 J

mass m=3.34*10^-27 kg

now we find the speed of the deutrons

10.24*10^-13=1/2*3.34*10^-27*v^2

v^2=6.13*10^14

speed v=2.5*10^7 m/s

now we find the diameter of deutons

radius r=3.34*10^-27*2.5*10^7/1.25*3.2*10^-19=0.835/4=0.21 m

diameter D=2*0.21=0.42 m

now we find the many deutrones

I=q/t

360*10^-6=q/1

q=360*10^-6

n*3.2*10^-19=360*10^6=112.5*10^13 deutrones

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