Question

A cyclotron is used to accelerate protons (\({}^1_1\text{H}\)​), Deuterons (\({}^2_1\text{H}\)​) and α-particles (\({}^4_2\text{He}\)). While exiting under similar conditions, the minimum K.E. is gained by

• A. α-particle
• B. Proton
• C. Deuteron
• D. Same for all

Answer 1

The Correct Option is C

In a cyclotron, the kinetic energy gained by a particle is given by the relation: \[ K.E. = \frac{q^2 B^2 r^2}{2m} \] Where:
\(q\) is the charge of the particle,
\(B\) is the magnetic field strength,
\(r\) is the radius of the circular path,
\(m\) is the mass of the particle. For a given magnetic field strength and radius, the kinetic energy gained by the particle depends on its mass. Since the proton, deuteron, and \(\alpha\)-particle have different masses, they will experience different accelerations. The proton has the least mass, followed by the deuteron and then the \(\alpha\)-particle. The minimum kinetic energy is gained by the particle with the smallest mass because it requires the least force to accelerate to a given velocity. Thus, the minimum K.E. is gained by the deuteron, as it has a mass between the proton and \(\alpha\)-particle.

The correct option is (C): \(K.E. =\frac{q^2}{m}\)