A cyclotron of radius 0.25 m accelerates nonrelativistic protons in a 1.7-T magnetic field. What is the kinetic energy of the protons when they leave the cyclotron? (1 eV = 1.60 × 10^-19 J, e = 1.60 × 10^-19 C, ^mproton = 1.67 × 10^-27 kg) A) 3.7 MeV B) 7.8 keV C) 8.7 keV D) 7.8 MeV E) 8.7 MeV

Question

Quizplus · Accepted Answer

The formula for the kinetic energy of a particle in a cyclotron is given by: 
KE = (q*B*R^2)/2, where q is the charge of the particle, B is the magnetic field strength, and R is the radius of the cyclotron.

Here, we have nonrelativistic protons with a charge of e and a mass of m = 1.67 × 10^-27 kg. Thus, we can calculate their kinetic energy as follows:

KE = (e*1.7 T*(0.25 m)^2)/2 
   = 8.81*10^-14 J

Converting this to electron volts (eV), we get:

KE = (8.81*10^-14 J)/(1.60*10^-19 J/eV) 
   = 550.6 keV

Therefore, the best choice is E, which is 8.7 MeV (mega electron volts) since this is the closest answer choice to 550.6 keV.

A Cyclotron of Radius 0