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A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons?If the radius of its ‘dees’ is 60 cm, what is thekinetic energy (in MeV) of the proton beam produced by the accelerator. |
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Answer» Solution : The OSCILLATOR FREQUENCY should be same as PROTON’s cyclotron frequency. Using eqs. We have `B = 2pi mv//q = 6.3 xx 1.67 xx 10^(-27) xx 10^7 // (1.6 xx 10^(-19) ) = 0.66 T` FINAL velocity of protons is `V = r xx 2pi v = 0.6 m xx 6.3 xx 10^7 = 3.78 xx 10^7 m//s` `E= 1//2 mv^2 = 1.67 xx 10^(-27) xx 14.3 xx 10^14 // (2 xx 1.6 xx 10^(-13)) = 7MeV` . |
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