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 3.34×10−27kg3.34×10−27kg. The deuterons exit the cyclotron with a kinetic energy of 5.80 MeVMeV .

Part A

What is the speed of the deuterons when they exit?

Express your answer with the appropriate units.

Part B

If the magnetic field inside the cyclotron is 1.25 TT, what is the diameter of the deuterons' largest orbit, just before they exit?

Express your answer with the appropriate units.

Part C

If the beam current is 360 μAμA how many deuterons strike the target each second?

Express your answer as the number of the deuetrons.

Answer 1

given
mass of deutron, m = 3.34*10^-27 kg
kinetic energy of the deutron, KE = 5.80 MeV

= 5.80*10^6*1.6*10^-19 J

we know, charge of deutron, q = 1.6*10^-19 C

part A) let v is the speed of deutron.

use, KE = (1/2)*m*v^2

==> v^2 = 2*KE/m

v = sqrt(2*KE/m)

= sqrt(2*5.8*10^6*1.6*10^-19/(3.34*10^-27))

= 2.36*10^7 m/s <<<<<<<<--------------Answer

part B) we know, radius of the charged particle in the cyclotron,

r = m*v/(B*q)

so, diameter d = 2*r

= 2*m*v/(B*q)

= 2*3.34*10^-27*2.36*10^7/(1.25*1.6*10^-19)

= 0.788 m <<<<<<<<<<<-----------------Answer

part C) now use, I = dQ/dt

==> dQ = I*dt

= 360*10^-6 A*s

= 360*10^-6 (C/s)*s

= 360*10^-6 C

the number of the deuetrons striking the target per second, N = dQ/e

= 360*10^-6/(1.6*10^-19)

= 2.25*10^15 <<<<<<<<<<<-----------------Answer