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Cyclotron frequency and maximum energy

Short answer

Oscillator frequency f = qB/(2πm) ≈ 11.5 MHz. Maximum kinetic energy K_max = q²B²R²/(2m) ≈ 4.9×10⁻¹² J ≈ 30.5 MeV.

A particular cyclotron is designed with Dees of radius R = 75cm and with magnets that can provide a field of 1.5T. (a) To what frequency should the oscillator be set if deuterons are to be accelerated? (b) What is the maximum energy of deuterons that can be obtained? (Mass of deuterons is 3.32×10⁻²⁷ kg)

In a cyclotron, qvB = mv²/r gives r = mv/(qB). The orbital frequency is independent of speed and radius:

  • f = qB/(2πm)
  • Maximum kinetic energy at r = R: K_max = q²B²R²/(2m)

Given: R = 0.75 m, B = 1.5 T, deuteron m = 3.32×10⁻²⁷ kg, q = e = 1.602×10⁻¹⁹ C.

(a) Oscillator frequency

f = (1.602×10⁻¹⁹ × 1.5) / (2π × 3.32×10⁻²⁷)
  = 2.403×10⁻¹⁹ / 2.086×10⁻²⁶
  = 1.15×10⁷ Hz

(b) Maximum energy

v_max = qBR/m = (1.602×10⁻¹⁹ × 1.5 × 0.75) / 3.32×10⁻²⁷ = 5.43×10⁷ m/s
K_max = ½ m v² = ½ × 3.32×10⁻²⁷ × (5.43×10⁷)²
      = 4.89×10⁻¹² J
      = 4.89×10⁻¹² / 1.602×10⁻¹³ MeV
      = 30.5 MeV

Answer: (a) f ≈ 11.5 MHz; (b) K_max ≈ 4.9×10⁻¹² J ≈ 30.5 MeV

✓ Answer checked by Claude Opus 5.5 (high effort)

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