Problem 24.30 A cyclotron is used to produce a beam of high-energy deuterons tha

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Problem 24.30

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 4.60 MeV .

Part A

What is the speed of the deuterons when they exit?

Express your answer with the appropriate units.

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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?

Express your answer with the appropriate units.

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Part C

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

Problem 24.30

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 4.60 MeV .

Part A

What is the speed of the deuterons when they exit?

Express your answer with the appropriate units.

v =

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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?

Express your answer with the appropriate units.

d =

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Part C

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

Q = deuetrons

Explanation / Answer

1) Ek= (1/2) mv^2, isolate for velocity

v=sqrt(2Ek/m)

where Ek= 4.6MeV * 1.6*10^(-19) J

v = sqrt(2*4.6*10^6*1.6*10^(-19)/(3.34*10^(-27)))

v = 2.1*10^7 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

r = 3.34*10^(-27)*2.1*10^7/(1.6*10^(-19)*1.25)

r = 0.35 m

diameter = 0.7 m

3)I=q/t

since t=1s, I=q. Therefore the total charge that strikes the target each second is 350 microcoulomb.

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) and get

deuterons= 350*10^(-6)/(1.6*10^(-19)) = 2.19*10^15

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