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 .

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 380 A how many deuterons strike the target each second?

Problem 24.30 Enhanced with Feedback 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 x 10 27 kg. The deuterons exit the cyclotron with a kinetic energy of 3.60 MeV You may want to review Part A What is the speed of the deuterons when they exit? Express your answer with the appropriate units. Value Units Subm My Answers Give Up Incorrect, Try Again; 5 attempts remaining Part B lf he magnetic field inside the cyclotron is 1.25 T what is the dia of the deuterons largest orbit, just before they exit? meter Express your answer with the appropriate units. d Value Units Submit My Answers Give Up

Explanation / Answer

1) Ek= (1/2) mv2,

=>: v=sqrt(2Ek/m)
where Ek= 3.6 MeV * 1.60217657*10-19 J =5.767835652*10-19 J
=> v=1.85*104 m/s

2) 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, qv x B, therefore
mv^2/r=qvB, where q is the charge of one proton
rearrange for r: r=mv/qB
multiply final answer by 2 to find diameter (this is the radius)
=> d=2*mv/qB = 6.21 *10-4 meter

3)I=q/t
since t=1s, I=q. Therefore the total charge that strikes the target each second is 380 C.
Thus the NUMBER of the deuterons = total charge by each charge of a deuteron
= (380*10-6)/(1.6*10-19) = 237.5*1010

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