455. Problem 34.33 (RHK) We have to estimate the total path length travelled by a deuteron during the acceleration process. We may assume an accelerating potential between the dees of 80 kV, a dee radius of 53 cm, and an oscillator frequency of 12 MHz.

 Solution:             Click For PDF Version A deuteron nucleus has the same charge as that of a proton and nearly twice its mass. Therefore, mass of deuteron, and its charge, . In a cyclotron that has uniform magnetic field , the angular velocity of circular motion of a particle of charge , mass is and the corresponding frequency is The final speed of the charged particles is determined by the radius R at which the particles leave the accelerator. The corresponding (nonrelativistic kinetic energy) of the particles is The oscillator frequency of the cyclotron is . Therefore, the final kinetic energy of the deuterons will be The accelerator potential across the dees is 80 kV. Therefore, kinetic energy of the deuterons will increase each time they cross the dee-gap by an amount Therefore, the total number of accelerations that a deuteron will receive during its acceleration in the cyclotron will be The kinetic energy of the deuteron, after undergoing acceleration each time will change by . That is it will change from to And, the speed of the deuteron will change from to to With constant speeds deuterons will travel for time Therefore, the total distance a deuteron will travel before leaving the cyclotron will be The reader is encouraged to evaluate the sum of square roots of natural numbers ranging from 1 to 208 and see the resultant value... , Therefore,