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 3.60 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?

Explanation / Answer

a. apply centripetal force = magnetic force

i.e mv^2/r = qvB,

where m = mass o the charged particle

v = velocity, r = radius ,

q = charge = 1.6*10^-19 C

B = magnetic field

so now , r = mv/qB

also KE = PE

0.5 mv^2 = eV

or v = sqrt(2eV/m)

v = sqrt(2 * 3.6 e 6 *1.6 e -19)/(3.34 e -27)

v = 1.85 e 7 m/s

-------------------------------------

use r = mv/qB

r = 3.34 e -27 * 1.85 e 7 /(1.6e -19 * 1.25)

r = 0.308 m

diameter D = 2 r = 0.617 m

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