A cyclotron is working at a frequency of 10 MHz. If the radius of its dees is 60 cm. The maximum kinetic energy of accelerated proton will be :
(Take : e = 1.6 × 10⁻¹⁹ C, mₚ = 1.67 × 10⁻²⁷ kg)

Solution

Physics sir · 4 steps

You are not given B, and you do not need it: the cyclotron frequency gives the speed at the dee's rim.

1. Frequency fixes speed. The oscillator frequency equals the proton's revolution frequency, f = qB/(2π m). The proton leaves at radius R = mv/qB, so vₘₐₓ = qBR/m = 2π f R.

   Kₘₐₓ = 1/2 m vₘₐₓ² = 2π² m f² R²

2. Put in numbers. With f = 10⁷ Hz and R = 0.6 m, vₘₐₓ = 2π × 10⁷ × 0.6 = 3.77 × 10⁷ m s⁻¹. Then Kₘₐₓ = 1/2 × 1.67 × 10⁻²⁷ × (3.77 × 10⁷)² = 1.19 × 10⁻¹² J.
3. Convert to eV. Kₘₐₓ = (1.19 × 10⁻¹²)/(1.6 × 10⁻¹⁹) eV = 7.4 × 10⁶ eV.
4. So. The maximum kinetic energy of the proton is about 7.4 MeV, which is option (a).

The idea to keep. In a cyclotron, vₘₐₓ = 2π f R, so Kₘₐₓ = 2π² m f² R². If B is given instead, use Kₘₐₓ = q²B²R²/2m.

Answer from the NTA final answer key. Solution by PrepWiser.