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 6.10 MeV .

A)What is the speed of the deuterons when they exit? v=?

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? d=?

C)If the beam current is 390 A how many deuterons strike the target each second? (Q)= ?

Explanation / Answer

A)
1/2 mv2 = 6.1 x 106 eV
1 eV = 1.6 x 10-19 J
1/2 mv2 = (6.1 x 106 ) x (1.6 x 10-19)

0.5 x 3.34 x 10-27 x v2 = (6.1 x 106 ) x (1.6 x 10-19)
v2 = 5.844 x 1014
v = sqrt(5.844 x 1014)
= 2.42 x 107 m/s

B)
Equating the centripetal force and the magnetic force,
mv2/r = Bqv
r = mv/Bq
q is the charge of the deuteron, q = 1.6 x 10-19 C
= [(3.34 x 10-27) x (2.42 x 107) / 1.25 x (1.6 x 10-19)]
= 0.404 m

C)
I = nq/t
I = 390 x 10-6 A
q = 1.6 x 10-19 C
t = 1 s

390 x 10-6 = n x (1.6 x 10-19)
n = 243.75 x 1013
= 2.4375 x 1015

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