Singly charged ions `He^(+)` are accelerated in a cyclotron so that their maximum orbital radiys is `r = 60 cm`. The frequency of a cyclotron's oscillator is equal to `v = 10.0 MHz`, the effective accelearating volatage across the deos is `V = 50 kV`. Neglecting the gap between the does, find:
(a) the total time of acceleration of the ion,
(b) the appoximate distance covered by the ion in the process of its acceleration.

Explanation

(a) The total time of acceleration is,
`t = (1)/(2v).n`,
where `n` is the number of passage of the Does.
But, `T = "neV" = (B^(2) e^(2) r^(2))/(2m)`
or, `n = (B^(2) e r^(2))/(2 mV)`
So, `t = (pi)/(eB//m) xx (B^(2) er^(2))/(2 mV) = (pi B r^(2))/(2V) = (pi^(2) mv r^(2))/(eV) = 30 mu s`
(b) The distance covered is, `s = sum v_(n). (1)/(2v)`
But, `v_(n) = sqrt((2eV)/(m)) sqrt(n)`
So, `s = sqrt((eV)/(2mv^(2))) sum sqrt(n) = sqrt((eV)/(2 mv^(2))) f sqrt(n) dn = sqrt((eV)/(2mv^(2))) (2)/(3) n^(3//2)`
But, `n = (B^(2) e^(2) r^(2))/(2 eV m) = (2 pi^(2) m v^(2) r^(2))/(eV)`
Thus, `s = (4pi^(3)n v^(2) mr^(2))/(3eV) = 1.24 km`