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