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 3.80 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 350 μA how many deuterons strike the target each second?

Answer 1

Answer:

19080667.0818 m/s

0.637294 m

$2.1875\times 10^{15}$

Explanation:

m = Mass of deuterons = $3.34\times 10^{-27}\ kg$

v = Velocity

K = Kinetic energy = 3.8 MeV

d = Diameter

B = Magnetic field = 1.25 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time = 1 s

i = Current = 350 μA

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 3.8\times 10^6\times 1.6\times 10^{-19}}{3.34\times 10^{-27}}}\\\Rightarrow v=19080667.0818\ m/s$

The speed of the deuterons when they exit is 19080667.0818 m/s

In this system the centripetal and magnetic force will balance each other

$\dfrac{mv^2}{r}=qvB\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=qvB\\\Rightarrow d=\dfrac{2mv}{qB}\\\Rightarrow d=\dfrac{2\times 3.34\times 10^{-27}\times 19080667.0818}{1.6\times 10^{-19}\times 1.25}\\\Rightarrow d=0.637294\ m$

The diameter is 0.637294 m

Current is given by

$i=\dfrac{nq}{t}\\\Rightarrow n=\dfrac{it}{q}\\\Rightarrow n=\dfrac{350\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\Rightarrow n=2.1875\times 10^{15}$

The number of deuterons is $2.1875\times 10^{15}$