A cyclotron is used to produce a beam of high energy deuterons that then collide
ID: 2307909 • Letter: A
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 x 10^-27kg. The deuterons exit the cyclotron with a kinetic energy of 5.00 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 400 microAmps how many deuterons strike the target each second.
Explanation / Answer
Mass of deuteron m = 3.34 x10 -27 kg
Kinetic energy at exit point K = 5 MeV
= 5 x1.6 x10 -13 J
We know K = (1/2) mv 2
From this the speed of the deuterons when they exit v = (2K/m) 1/2
Substitute values you get v = [(2x5x1.6 x10 -13 ) /(3.34x10 -27 )] 1/2
= (4.79x10 14) 1/2
= 21.88 x10 6 m/s
(B).Magnetic field B = 1.25 T
In magnetic field ,Bvq = mv 2 / r
Bq = mv / r
From this radius of the orbit when they exit r = mv /Bq
Here v = speed of the deutrron when they exit
q = charge of deuteron = 1.6 x10 -19 C
Substitute vales you get r = (3.34x10 -27)(21.88 x10 6) /(1.25 x1.6 x10 -19)
= 0.3655 m
Therefore the diameter of the deuterons largest orbit, just before they exit D = 2r
D = 2(0.3655 m)
= 0.731 m
(C). Current i = 400 x10 -6 A
time t = 1 s
Charge Q = i t
= 400 x10 -6 C
So, Number of deuterons strike the target each second N = Q / q
N = (400x10 -6) /(1.6x10 -19)
= 2.5x10 15
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