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A proton is accelerating in a cyclotron where the applied magnetic field is 2T. If the potential gap is effectively 100 kV then how much revolutions the proton has to make between the “dees” to acquire a kinetic energy of 20 MeV?

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answer is A.

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Detailed Solution

KE 20 10⁺⁶ev
KE gain in each revolution = 2eV = 2e 10⁵v
No of revolution $= \frac{20 \times 10^{6}}{2 \times 10^{5}} = 100$

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