In a mass spectrometer used for measuring the masses of ions, the...

The radius of the orbit in which ions moving is determined by the relation as given below $\frac{m v^2}{R}=q v B$ where $m$ is the mass, $v$ is velocity, $q$ is charge of ion and $B$ is the flux density of the magnetic field, so that $q v B$ is the magnetic force acting on the ion, and $\frac{m v^2}{R}$ is the centripetal force on the ion moving in a curved path of radius $R$. The angular frequency of rotation of the ions about the vertical field $B$ is given by $\omega=\frac{v}{R}=\frac{q B}{m}=2 \pi v$ where $v$ is frequency. Energy of ion is given by or $$\begin{aligned} E & =\frac{1}{2} m v^2 \\ & =\frac{1}{2} m(R \omega)^2 \\ & =\frac{1}{2} m R^2 B^2 \frac{q^2}{m^2} \\ E & =\frac{1}{2} \frac{R^2 B^2 q^2}{m} \ldots(i) \end{aligned}$$ If ions are accelerated by electric potential $V$, then energy attained by ions $E=q V\ldots(ii)$ From Eqs. (i) and (ii), we get $$\begin{aligned} q V & =\frac{1}{2} \frac{R^2 B^2 q^2}{m} \\ or\frac{q}{m} & =\frac{2 V}{R^2 B^2} \end{aligned}$$ If $V and B$ are kept constant, then $\frac{q}{m} \propto \frac{1}{R^2}$