Respuesta :
The diameter of the largest orbit just before the protons exit the cyclotron is 39 cm.
The number of orbits completed by the proton during this 1.0 ms is 14000 revolutions.
The kinetic energy for the protons can be computed by using the formula:
- K.E = 1/2mv²
mv² = 2 K.E
v = sqrt( 2*KE / M)
the kinetic energy of the medical isotopes = 6.5 MeV
substituting the values,
v = sqrt( 2* 6.5* 1.6* 10^-13 / 1.67* 10^-27 )
v = 3.53 × 10⁷ m/s
The radius of the orbit can be estimated by using the formula:
mv² / R = qvB
r = q*v / mv²
r = ( 1.67* 10^-27 * 3.53 × 10⁷ ) / ( 1.9* 1.6* 10^-19 )
r = 0.19415 m
Since diameter (D) = 2r,
D= 2(0.19415 m)
D= 0.39 m
D≅ 39 cm
The time period to complete a revolution around the spiral trajectory is:
T = 2πr / v
T = 2*3.14* 0.1941 / 3.53*10^7
T = 0.7 × 10⁻⁷ s
Finally, the number of orbits that the proton does to complete the revolution in 1 ms is:
n = t / T
n = 10^-3 / (0.7*10^-7)
n = 14285.71
n ≅ 14000 revolutions
Learn more about cyclotron here:
https://brainly.com/question/26832152
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Your question is incomplete, but most probably full question was:
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?
