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A cyclotron's oscillator frequeney is 10 MHz. What should be the operating magnetic field for accelerating protons ? If the radius of its dees is 60 cm, what is the kinetic energy (in MeV)of the proton beam produced by the accelerator. (e= 1.60 xx 10^(-19) C, m_p = 1.67 xx 10^(-13)kg) |
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Answer» Solution :The oscillar frequency should same as proton.s cylotron frequency . We have `B= (2PI m upsilon)/(q) = (6.3 xx 1.67 xx 10^(-27)xx 10^(7))/((1.6 xx 10^(-19))) = 066 T` Final velocity of proton is `v= r xx 2 pi upsilon = 0.6 m xx 6.3 xx 10^7 = 3.78 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)))=7 MeV` |
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