i.e., $\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{qvB}.........(1)$

where $m=$ mass of the ion, $v=$ velocity, $q=$ charge of ion, $\mathrm{B}=$ flux density of the magnetic field.

we have, $\mathrm{v}=\mathrm{R} \omega$

or $\omega=\frac{\mathrm{v}}{\mathrm{R}}=\frac{\mathrm{qB}}{\mathrm{m}} \quad(\mathrm{From}(1))$

Energy of ion is given by,

$\mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{m}(\mathrm{R} \omega)^{2}=\frac{1}{2} \mathrm{mR}^{2} \frac{\mathrm{q}^{2} \mathrm{B}^{2}}{\mathrm{m}^{2}}$

or $\mathrm{E}=\frac{1}{2} \frac{\mathrm{R}^{2} \mathrm{B}^{2} \mathrm{q}^{2}}{\mathrm{m}}.........(2)$

If ions are accelerated by electric potential $V$, the energy attained by ions,

$E=q V.........(3)$

From eqns $( 2)$ and $( 3)$

$\mathrm{qV}=\frac{1}{2} \frac{\mathrm{R}^{2} \mathrm{B}^{2} \mathrm{q}^{2}}{\mathrm{m}}$ or $\left(\frac{\mathrm{q}}{\mathrm{m}}\right)=\frac{2 \mathrm{V}}{\mathrm{R}^{2} \mathrm{B}^{2}}$

i.e., $\left(\frac{\mathrm{q}}{\mathrm{m}}\right) \propto \frac{1}{\mathrm{R}^{2}}(\text { If } \mathrm{V} \text { and } \mathrm{B} \text { are const. })$