In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential $V$ and then made to describe semicircular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant, the ratio $\left(\frac{\text { charge on the ion }}{\text { mass of the ion }}\right)$ will be proportional to

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 $$ \begin{aligned} E &=\frac{1}{2} m v^{2}=\frac{1}{2} m(R \omega)^{2} \\ &=\frac{1}{2} m R^{2} B^{2} \frac{q^{2}}{m^{2}} \\ E &=\frac{1}{2} \frac{R^{2} B^{2} q^{2}}{m} \end{aligned} $$ ... (i) If ions are accelerated by electric potential $V$, then energy attained by ions $$ E=q V ...(ii) $$ From Eqs. (i) and (ii), we get $$ q V=\frac{1}{2} \frac{R^{2} B^{2} q^{2}}{m} $$ or $$ \frac{q}{m}=\frac{2 V}{R^{2} B^{2}} $$ If $V$ and $B$ are kept constant, then $$ \frac{q}{m} \propto \frac{1}{R^{2}} $$