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Numericals · Q21

Q.In a cyclotron, protons are to be accelerated. The radius of its D is 60 cm and its oscillator frequency is 10 MHz. What will be the kinetic energy of the proton thus accelerated? (Proton mass =1.67×10−27=1.67\times10^{-27} kg, e=1.6×10−19e=1.6\times10^{-19} C, 1 eV=1.6×10−191\ \text{eV}=1.6\times10^{-19} J)

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At the exit radius R, mv=qBRmv=qBR (cyclotron formula), and the resonance condition gives B=2πmf/qB=2\pi mf/q (from f=qB/(2πm)f=qB/(2\pi m)). Substituting this B into the momentum relation: v=qBRm=q(2πmf/q)Rm=2πfRv=\dfrac{qBR}{m}=\dfrac{q(2\pi mf/q)R}{m}=2\pi fR. The kinetic energy is then KE=12mv2=12m(2πfR)2=2π2mf2R2KE=\tfrac12mv^2=\tfrac12m(2\pi fR)^2=2\pi^2mf^2R^2. Substituting m=1.67×10−27m=1.67\times10^{-27} kg, f=10×106f=10\times10^6 Hz, R=0.60R=0.60 m: $KE=2\pi^2(1.67\times10^{-27})(10^7) …

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