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 3.34×10−27kg. The deuterons exit the cyclotron with a kinetic energy of 6.10 MeV .

If the magnetic field inside the cyclotron is 1.25 T, what is the diameter of the deuterons' largest orbit, just before they exit?

Answer 1

Given that deutrons exit the cyclotron with KE of 6.10 MeV, So

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

v = speed of Deutron when it exits = sqrt (2*KE/m)

KE = 6.10 MeV = 6.10*10^6*1.6*10^-19 J

m = mass of deutron

So,

v = sqrt (2*6.10*10^6*1.6*10^-19/(3.34*10^-27)) = 2.417*10^7 m/sec

Now when deutron enters into magnetic field using force balance on it

Fc = Fm

Centripetal force on ion = magnetic force

m*v^2/R = q*v*B

R = m*v/(q*B)

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

Diameter = 2*3.34*10^-27*2.417*10^7/(1.6*10^-19*1.25)

Diameter = 0.807 m = 80.7 cm

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