A cyclotron is used to produce a beam of high energy deutero

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.

Solution

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