If the maximum value of accelerating potential provided by a radio frequency oscillator be `25kV`, find the number of revolutions made by a proton in a cyclotron to achieve one sixth of the speed of light. Mass of proton `=1*67xx10^(-27)kg`.

Explanation

In a cyclotron, when a proton crosses a region of potential difference V, the energy gained is `eV`. In cyclotron, in one revolution, the proton crosses the gap twice. So the energy gained by proton in each revolution `=2eV`.
Let the proton makes n revolutions before emerging from dees. The gain in its kinetic energy is
`E_k=1/2mv^2=nxx2eV`
or `n=(mv^2)/(4eV)=(m(c//6)^2)/(4eV)`
or `n=(mc^2)/(4xx36xxeV)`
`=(1*67xx10^(-27)xx(3xx10^8)^2)/(4xx36xx(1*6xx10^(-19))xx(25xx10^3))`
`~~261 revolutions`