Question

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is 1.9 T.
(a) What is the diameter of the largest orbit, just before the protons exit the cyclotron? Express your answer with the appropriate units. d = 57 cm Previous
(b) A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1.0 ms?

Answer 1

Given Information"

Kinectic energy = 6.5 MeV

magnetic field = B = 1.9 T

Period = T = 1 ms

Required Information"

Diameter = d = ?

Answer:

d = 38.6 cm

f = 1000

Explanation:

(a) What is the diameter of the largest orbit, just before the protons exit the cyclotron?

The radius and hence the diameter of the largest orbit can be found by using

r = mv /|q|B

Where q = 1.60x10⁻¹⁹ C and m = 1.67x10⁻²⁷ kg and v is the speed of the particle  which can be found by using

v = √2K /m

Where K is the kinetic energy in Joules.

1 eV is equal to 1.60x10⁻¹⁹ J

K = 6.5x10⁶*1.60x10⁻¹⁹ = 1.04x10⁻¹² J

v = √2*1.04x10⁻¹² /1.67x10⁻²⁷

v = 35.29x10⁶ m/s

r = 1.67x10⁻²⁷*35.29x10⁶ /1.60x10⁻¹⁹*1.9

r = 0.193 m

d = 2r

d = 2*0.193 = 0.386

or d = 38.6 cm

(b) A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1.0 ms?

The period T and frequency (number of rotations per second) are related as

f = 1/T

f = 1/0.001

f = 1000