What is the kinetic energy, in mega-electron-volts (MeV). of a proton extracted

Question

What is the kinetic energy, in mega-electron-volts (MeV). of a proton extracted from a cyclotron with radius r = 1.81 m, if the magnetic field of the cyclotron is uniform and ha; magnitude B = 0.851 T? The mass of a proton is 1.67 middot 10-27 kg. An accelerator produces alphas using an electric potential difference Vo from .50 MV to 9.00 MV as shown below: What is the kinetic energy of the alphas as they leave the accelerator if the potential difference V_0 on the accelerator is 2.50 MV and they initially start

Explanation / Answer

Answer :

Given

r = 1.81 m

B = 0.851 T

m (proton) = 1.67E-27 kg

The operation of a cyclotron is based on the fact that the period of the motion of a charged particle in a uniform magnetic field is independent of the velocity of the particle, as can be seen in the following derivation:

F = ma

qvB = mv2/r

v = qBr/m

kinetic energy of proton (KE) = 1/2*m*v^2

KE = q2B2r2/2m

KE = (1.6E-19 C)2(0.851T)2(1.81 m)2/2(1.67E-27 kg)

KE = (6.07374E-11 / 3.34 )Joules

KE = 1.8184E-11 Joules

KE = 113.49 MeV

[1 MeV = 1.6021773E-13 J

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