In a cyclotron experiment, if we assume that | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The maximum kinetic energy $KE_{\max}$ of a particle in a cyclotron is given by the formula $KE_{\max} = \frac{q^2 B^2 R^2}{2m}$, where $q$ is the

Vedclass · Accepted Answer

Deuteron

Vedclass · Answer

(C) The maximum kinetic energy $KE_{\max}$ of a particle in a cyclotron is given by the formula $KE_{\max} = \frac{q^2 B^2 R^2}{2m}$, where $q$ is the charge, $B$ is the magnetic field, $R$ is the radius of the dees, and $m$ is the mass of the particle.Since $B$ and $R$ are constant for a given cyclotron, $KE_{\max} \propto \frac{q^2}{m}$.For a proton $(p)$: $q = e, m = m_p \implies \frac{q^2}{m} = \frac{e^2}{m_p}$.For a deuteron $(d)$: $q = e, m = 2m_p \implies \frac{q^2}{m} = \frac{e^2}{2m_p}$.For an alpha particle $(\alpha)$: $q = 2e, m = 4m_p \implies \frac{q^2}{m} = \frac{4e^2}{4m_p} = \frac{e^2}{m_p}$.Comparing the values, the ratio $\frac{q^2}{m}$ is minimum for the deuteron. Therefore, the minimum kinetic energy gained is for the deuteron.

In a cyclotron experiment, if we assume that the maximum radius attained by any charged particle is equal to the radius of the dees, then the minimum kinetic energy gained by the particles will be for

Similar Questions

$A$ cyclotron is used to accelerate

The maximum velocity to which a proton can be accelerated in a cyclotron of $10\, MHz$ frequency and radius $50\, cm$ is

In the cyclotron, as the radius of the circular path of the charged particle increases, ($\omega$ = angular velocity, $V$ = linear velocity)

The cyclotron frequency of an electron gyrating in a magnetic field of $1\, T$ is approximately

Vedclass Test Series

Exam Paper Generator

Online Exam Module