Question

If the radius of the dees of cyclotron is \(r\), then the kinetic energy of a proton of mass \(m\) accelerated by the cyclotron at an oscillating frequency \(\nu\) is

• A. \(4\pi^2 m^2 \nu^2 r^2\)
• B. \(4\pi^2 m \nu^2 r^2\)
• C. \(2\pi^2 m \nu^2 r^2\)
• D. \(\pi^2 m \nu^2 r^2\)
• E. \(\pi^2 m^2 \nu^2 r^2\)

Hint:

The maximum kinetic energy of a particle in a cyclotron depends only on the magnetic field, the charge, the radius of the dees, and the mass. Using frequency \(\nu\) is a common way to express this without needing the magnetic field \(B\) explicitly.

Answer 1

The Correct Option is C

Concept: In a cyclotron, charged particles are accelerated to high speeds using an oscillating electric field and a static magnetic field.
• Velocity (\(v\)): For a particle of mass \(m\) and charge \(q\) moving in a radius \(r\) with frequency \(\nu\), the linear velocity is \( v = \omega r = 2\pi \nu r \).
• Kinetic Energy (\(KE\)): The energy is given by the standard formula \( KE = \frac{1}{2} m v^2 \).

Step 1: Express velocity in terms of frequency.
The oscillating frequency \(\nu\) of the electric field must match the cyclotron frequency of the particle. The relationship between linear velocity \(v\), radius \(r\), and frequency \(\nu\) is: \[ v = 2\pi \nu r \]

Step 2: Calculate the Kinetic Energy.
Substitute the expression for \(v\) into the kinetic energy formula: \[ KE = \frac{1}{2} m (2\pi \nu r)^2 \] \[ KE = \frac{1}{2} m (4\pi^2 \nu^2 r^2) \] \[ KE = 2\pi^2 m \nu^2 r^2 \]