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

(N/A) The frequency of the radio frequency $(rf)$ field is given by $f = \frac{qB}{2\pi m}$. The time period of the particle in the cyclotron is $T = \frac{2\pi m}{qB}$.
If the frequency of the $(rf)$ field is doubled,the oscillator frequency becomes $2f$,which means the time period of the $(rf)$ field becomes $T' = \frac{T}{2}$.
Since the time taken by the particle to complete a semi-circle inside a dee is $t = \frac{T}{2}$,the particle will now reach the gap between the dees before the electric field has completed its cycle.
Consequently,the particle will not be accelerated consistently at the gap,as the electric field polarity will not match the arrival of the particle. This disrupts the resonance condition required for the cyclotron to function,and the particle will not gain energy effectively.