Question

Constants 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 10-27 kg. The deuterons exit the cyclotron wit a kinetic energy of 4.80 MeV You may want to review (Pages 778 - 783)

Answer 1

Part A

Since it is given that deuterons exit the cyclotron with energy 4.8 MeV so speed of the deuterons can be calculated by using formula for Kinetic energy

Given KE = 4.8 MeV = 4.8 x 10 ⁶ x 1.6 x 10 ^-19 Joule

K. E. = (1/2) m v ²

4.8 x 10 ⁶ x 1.6 x 10 ^-19= (1/2) 3.34 x 10 -27 x v²

Solving this we get

v = 2.1444 x 10 ⁷  m/s

Answer of Part A speed of dueteron when they exit = 2.1444 x 10 ⁷ m/s

Part B

Given

Magnetic field = B = 1.25 T

r = mv / Bq

Where r = radius of the largest orbit when particle is leaving the orbit

m = mass of the charged particle

v = speed of the particle

B = magnetic filed in the cyclotron

q = charge of the particle

substituing values of m, v, B, q

r = (3.34 x 10 ^-27 x 2.1444 x 10 ⁷ )/ 1.25 x 1.6 x 10 ^-19

r = 0.3581 m or 35.81 cm

Diameter = 2 x 35.81 = 71.62 cm

Answer of Part B:

Diameter of the largest orbit = 0.7162 m = 71.62 cm

Part C

Beam current = 420 micro ampere = 420 x 10 ^-6 A

current = charge flowing per second

= number of deteron x charge of each dueteron

420 x 10 ^-6 = n x 1.6 x 10 ^-19

n = 2.625 x 10 15

Answer of Part C

Number of dueteron strike the target each second = 2.625 x 10 ¹⁵