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 3.80 MeV .

A. What is the speed of the deuterons when they exit?

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?

C. If the beam current is 370 Ahow many deuterons strike the target each second?

Explanation / Answer

isolate for velocity: v=sqrt(2Ek/m)
where Ek= 3.8 MeV * 1.60217657e-19 J
v=1.9 e7 m/s

2) We know that the equation for centripetal 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*1.9*107/(1.60217657e-19*1.25)
multiply final answer by 2 to find diameter (this is the radius)
d=0.63 m

3) I=q/t
since t=1s, I=q. Therefore the total charge that strikes the target each second is

370 microcoulomb.
to find the NUMBER of the deuterons divide that total charge by each charge of a deuteron (which is the same as the charge of a proton) and get
N/t = I/e = 370E-6/1.6E-19= 2.31E15

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.