InterviewSolution
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InterviewSolution
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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 the kinetic energy (in MeV) of the proton beam produced by the accelerator. (e=1.60xx10^(-19)C,m_(p)=1.67xx10^(-27kg),1MeV=1.6xx10^(-13)J) |
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Answer» Solution :1. Angular frequency of a charged particle in cyclotron is, `omega=(Bq)/m` `therefore2pif=(Bq)/m` `thereforeB=(2pifm)/q` `thereforeB=((2)(3.14)(10xx10^(6))(1.67xx10^(-27)))/((1.6xx10^(-19)))` `thereforeB=6.555xx10^(-1)=0.6555T` 2. Kinetic energy of proton, `K=1/2mv^(2)` `thereforeK=1/2mr^(2)omega^(2)` `thereforeK=1/2mr^(2)(4pi^(2)f^(2))" "(becauseomega=2pif)` `thereforeK=2PI^(2)mr^(2)f^(2)` `thereforeK_(max)=2pi^(2)m(r_(max))^(2)f^(2)` `thereforeK_(max)=2pi^(2)mR^(2)f^(2)` `(becauser_(max)=R="RADIUS of DEE")` `thereforeK_(max)=(2)(3.14)^(2)(1.67xx10^(-27))(0.6)^(2)(10xx10^(6))^(2)` `thereforeK_(max)=1.186xx10^(-12)J` `thereforeK_(max)=(1.86xx10^(-12))/(1.6xx10^(-19))EV=7.412xx10^(6)eV` `thereforeK_(max)=(7.412xx10^(6))/(10^(6))MeV=7.412MeV` |
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