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ExamsJEE AdvancedPhysics

In a cyclotron, alpha particles (doubly ionized helium atoms, charge = 2e) are accelerated. The radius of the largest circular orbit is 0.5 m and the magnetic flux density is 1 T. What is the greatest kinetic energy (in MeV) to which these particles can be accelerated? If this energy is x MeV, find the value of 10x. (Mass of helium nucleus = 6.4 * 10⁻²⁷ kg, e = 1.6 * 10⁻¹⁹ C)

  1. 100
  2. 125
  3. 150
  4. 200

Correct answer: 125

Solution

In a cyclotron the radius of circular motion r = mv/(qB), so v = qBr/m and KE = (1/2)mv² = q²*B²*r²/(2m). With q = 2*(1.6e-19) = 3.2e-19 C, B = 1 T, r = 0.5 m, m = 6.4e-27 kg: KE = (3.2e-19)²*(1)²*(0.5)² / (2*6.4e-27) = 1.024e-37 * 0.25 / 1.28e-26 = 2.56e-38 / 1.28e-26 = 2.0e-12 J. Converting to MeV: 2.0e-12 / 1.6e-13 = 12.5 MeV. So x = 12.5, and 10x = 125.

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