A cyclotron is used to produce a beam of high-energy deuterons that then collide

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 3.34×1027kg. The deuterons exit the cyclotron with a kinetic energy of 5.90 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 380 A how many deuterons strike the target each second? 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 hydrogen, consisting of one neutron and one 27 proton, with total mass 3.34 x 10 kg. The deuterons exit the cyclotron with a kinetic energy of 5.90 MeV You may want to review (ua pages 778-783) Part A What is the speed of the deuterons when they exit? Express your answer with the appropriate units Value Submit My Answers Give up Incorrect; Try Again: 2 attempts remaining Part 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? Express your answer with the appropriate units. Value Units Submit My Answers Give Up Part C If the beam current is 380 pA how many deuterons strike the target each Second? Submit My Answers Give Up

Explanation / Answer

(A) KE = m v^2 / r

5.90 x 10^6 x 1.6 x 10^-19 = (3.34 x 10^-27) v^2 / 2

v = 2.38 x 10^7 m/s

(B) q v B = m v^2 / r

  
r = m v / q B

= (3.34 x 10^-27) (2.38 x 10^7) / (1.6 x 10^-19)(1.25)

= 0.397 m

D = 2 r = 0.794 m

(C) I = Q / t

I = n e / t

(380 x 10^-6) = n (1.6 x 10^-19) / 1

n = 2.375 x 10^15 electrons

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