In a certain cyclotron a proton moves in a circle of radius 0.650m. The magnitud
ID: 1669623 • Letter: I
Question
In a certain cyclotron a proton moves in a circle of radius 0.650m. The magnitude of the magnetic field is 1.50 T.(a) What is the oscillator frequency (in Hz)?(b) What is the kinetic energy (in eV) of theproton?Explanation / Answer
(a) The oscillator frequency f = qB / 2m Where q = charge of proton = 1.6 x10-19 C B = 1.50 T m = mass of proton = 1.67 x 10-27 kg r = 0.650 m ==> f = 1.6 x10-19 * 1.50 / ( 2 * 3.14 * 1.67 x10-27 ) = 22.88 MHz (b) Kinetic energy of proton KE = 0.5 mv2 = 0.5 * m * (qBr / m)2 = 0.5(qBr)2 / m = 0.5* ( 1.6 x 10-19 * 1.50 * 0.65 )2 / 1.67x 10-27 = 0.7286 x 10-11 J = 4.547 x 107 eV