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 $\left(\frac{\mathrm{charge} \mathrm{on} \mathrm{the} \mathrm{ion}}{\mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{ion}}\right)$ will be proportional to

• $\frac{1}{\mathrm{R}}$

• $\frac{1}{{\mathrm{R}}^2}$

• R²

• R

B.

$\frac{1}{{\mathrm{R}}^2}$

The radius of the orbit in which ions moving is determined by the relation as given below

$\frac{{\mathrm{mv}}^2}{\mathrm{R}} = \mathrm{qvB}$

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 qvB is the magnetic force acting on the ion, and $\frac{{\mathrm{mv}}^2}{\mathrm{R}}$ 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

$$\begin{gathered} \mathrm{\omega } = \frac{\mathrm{v}}{\mathrm{R}} \\ = \frac{\mathrm{qB}}{\mathrm{m}} \\ \mathrm{\omega }= 2\mathrm{\pi v} \end{gathered}$$

where ν is the frequency.

Energy of ion is given by

$$\begin{gathered} \mathrm{E} = \frac{1}{2}{\mathrm{m\nu }}^2 \\ =\frac{1}{2}\mathrm{m} \left({\mathrm{R\omega }}^2\right) \\ \Rightarrow \mathrm{E} = \frac{1}{2}\mathrm{m} {\mathrm{R}}^2 {\mathrm{B}}^2 \frac{{\mathrm{q}}^2}{{\mathrm{m}}^2} \\ \Rightarrow \mathrm{E} = \frac{1}{2}\frac{{\mathrm{R}}^{2 }{\mathrm{B}}^2 {\mathrm{q}}^2}{\mathrm{m}} ....\left(\mathrm{i}\right) \end{gathered}$$

If ions are accelerated by electric potential V, then energy attained by ions

E = qV                                ....  (ii)

from equation (i) and (ii) we get

$$\begin{gathered} \mathrm{q} \mathrm{V} = \frac{1}{2}\frac{{\mathrm{R}}^2 {\mathrm{B}}^2 {\mathrm{q}}^2}{\mathrm{m}} \\ \Rightarrow \frac{\mathrm{q}}{\mathrm{m}} = \frac{2 \mathrm{V}}{{\mathrm{R}}^2 {\mathrm{B}}^2} \end{gathered}$$

If V and B are kept constant

$\frac{\mathrm{q}}{\mathrm{m}} \propto \frac{1}{{\mathrm{R}}^2}$