$A$ cyclotron is used to accelerate protons $\left({ }_{1}^{1} H\right)$,deuterons $\left({ }_{1}^{2} H\right)$,and $\alpha$-particles $\left({ }_{2}^{4} He\right)$. While exiting under similar conditions,the minimum kinetic energy $(KE)$ is gained by:

Cyclotron is used to accelerate

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

$A$ proton is accelerating in a cyclotron where the applied magnetic field is $2 \,T$. If the potential gap is effectively $100 \,kV$, then how many revolutions does the proton have to make between the "dees" to acquire a kinetic energy of $20 \,MeV$?

Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency $(rf)$ field were doubled.

$A$ cyclotron's oscillator frequency is $10 \; MHz$. What should be the operating magnetic field for accelerating protons? If the radius of its 'dees' is $60 \; cm$,what is the kinetic energy (in $MeV$) of the proton beam produced by the accelerator? $(e = 1.60 \times 10^{-19} \; C, m_p = 1.67 \times 10^{-27} \; kg, 1 \; MeV = 1.6 \times 10^{-13} \; J)$