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Cyclotron proton frequency and energy

Short answer

Oscillator frequency f = qB/(2πm) ≈ 1.83×10⁷ Hz (18.3 MHz). Kinetic energy K = ½m(qBr/m)² ≈ 2.76×10⁻¹² J = 1.72×10⁷ eV ≈ 17.2 MeV.

In a certain cyclotron a proton moves in a circle of radius 0.5m. The magnitude of the magnetic field is 1.20T. What is the oscillator frequency? What is the kinetic energy of the proton in eV?

In a cyclotron, qvB = mv²/r. The proton's circulation frequency is f = qB/(2πm), independent of speed. The speed at radius r is v = qBr/m.

Given: r = 0.5 m, B = 1.20 T, m_p = 1.673×10⁻²⁷ kg, q = 1.602×10⁻¹⁹ C.

Oscillator frequency

f = (1.602×10⁻¹⁹ × 1.20) / (2π × 1.673×10⁻²⁷)
  = 1.9224×10⁻¹⁹ / 1.0511×10⁻²⁶
  = 1.83×10⁷ Hz

Kinetic energy

v = qBr/m = (1.602×10⁻¹⁹ × 1.20 × 0.5) / 1.673×10⁻²⁷ = 5.74×10⁷ m/s
K = ½ m v² = ½ × 1.673×10⁻²⁷ × (5.74×10⁷)² = 2.76×10⁻¹² J
K = 2.76×10⁻¹² / 1.602×10⁻¹⁹ = 1.72×10⁷ eV

Answer: f ≈ 18.3 MHz; K ≈ 1.72×10⁷ eV = 17.2 MeV (non-relativistic, since v ≈ 0.19c).

✓ Answer checked by Claude Opus 5.5 (high effort)

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