In the Dubna heavy ion cyclotron neon ions are accelerated to an energy of 100 MeV. The diameter of the dees is 310 cm, the magnetic field induction in the gap is 1.1 T the accelerating potential is 300 kV. Find the degree of ionization of a neon atom, the total number of revolutions of an ion in the process of its acceleration and the frequency of the change in polarity of the accelerating field.

Answer

`4; 41 ; 3.0 MHz`

Explanation

First we find the charge of the ion using the condition `mv^2//R=quB`. We have `q="mu/BR"`. Taking account of the fact that heavy ions (in the energy range of hundreds of mogaelecton-volts move at nonrelativistic speeds, we have `m=A xx 1.66 xx 10^(-2)`, where
A 20.18 amu is the atomic mass of neon. The momentum of the ion is found from its kinetic energy `"mu"=sqrt(2mK)`. Hence the charge of the ion.
`q=sqrt(2mK)/(BR)=sqrt(2 xx 20.18 xx 1.66 xx 10^(-27) xx 100 xx 1.6 xx 10^(-13))/(1.55 xx 1.1)`
`=6.6 xx 10^(-19)C`
Dividing by the electron charge we find teh neon ions to be ionized quadruply.
The total number of revolution as ion makes is equal to its kinetic energy divided by the energy acquired in the process of pasing twice through the accelerating gap:
`N=(K)/(2q Psi)=(100 xx 10^(8))/(2 xx 4 xx 300 xx 10^(3))=42`
The frequency of the change in polarity of the accelerating fieldis equal to the circular frequency of the ion.
`=u/(2pi R)=(q B)/(2pim)`