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 4.80 MeV .

Part A)

What is the speed of the deuterons when they exit?

Part 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?

Part C)

If the beam current is 390 μA how many deuterons strike the target each second?

Answer 1

Given,

m = 3.34 x 10^-27 kg ; KE(qV) = 4.8 MeV = 7.69 x 10^-13 J

A)We know that

qV = 1/2 m v^2 = KE

v = sqrt (2 KE/m)

v = sqrt (2 x 7.69 x 10^-13/(3.3 x 10^-27)) = 2.16 x 10^7 m/s

Hence, v = 2.16 x 10^7 m/s

b)B = 1.25 T

We know that the magnetic and centripital force balances

Fc = Fm

m v^2/R = q v B

R = m v/B q

R = 3.34 x 10^-27 x 2.16 x 10^7/(1.25 x 1.6 x 10^-19) = 0.361 m

D = 2R = 2 x 0.361 = 0.722 m

Hence, D = 0.722 m = 72.2 cm

c)I = 390 uA = 390 x 10^-6 A

I = Q/t => Q = I t

Q = 3.9 x 10^-4 x 1 = 3.9 x 10^-4 C

n = Q/e

n = 3.9 x 10^-4/(1.6 x 10^-19) = 2.44 x 10^15

Hence, n = 2.44 x 10^15