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A proton is accelerating in a cyclotron having external magnetic field of 0.7 T. If the radii of its dees is 1.8 m, then its maximum kinetic energy (in MeV) is (approximately) (mass of a proton is $1.67 \times 10^{- 27} kg$ )

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Detailed Solution

$K_{max} = \frac{q^{2} B^{2} R^{2}}{2 m} = \frac{{(1 .6 \times 10^{- 19})}^{2} \times (0 .7)^{2} \times (1 .8)^{2}}{2 \times 1 .67 \times 10^{- 27}} J = 76 MeV$

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