Question

Discuss the principle and action of a Bainbridge mass spectrometer to determine the isotopic masses

Answer 1

Bainbridge mass spectrometer is an instrument used for the accurate determination of atomic masses. A schematic diagram of this spectrometer is shown in Fig.
Atoms with one or more electrons removed, have a net positive charge and they become positive ions. A beam of positive ions produced in a discharge tube is collinaed into a fine beam by two narrow slits $S_1$ and $S_2$. This fine beam enters into a velocity selector. The velocity selector allows the ions of a p articular velocity to come out of it, by the combined action of an electric and a magnetic field. The velocity selector consists of two plane parallel plates $P_1$ and $P_2$, which produces a uniform electric field $E$ and an electromagnet, to produce uniform magnetic field $B$ (represented by the dotted circle). These two fields are at right angles to each other and to the direction of the beam.
The electric field and magnetic field are so adjusted that the deflection produced by one field is nullified by the other, so that the ions do not suffer any deflection within the velocity selector. Let $E$ and $B$ be the electric field intensity and magnetic induction respectively and $q$ be the charge of the positive ion. The force exerted by the electric field is equal to $qE$ and the force exerted by the magnetic field is equal to $Bqv$ where $v$ is the velocity of the positive ion.
$qE=Bqv$
$v=E/B$
Only those ions having this velocity $v$, pass out of the velocity selector and then through the slit $S_3$, to enter the evacuated chamber $D$. These positive ions having the same velocity are subjected to another strong uniform magnetic field of induction $B^{\prime }$ at right angles to the plane of the paper acting inwards. These ions are deflected along circular path of radius $R$ and strike the photographic plate. The fore due to magnetic field $B^{\prime }qv$ provides the centripetal force.
$B^{\prime }qv=\frac{m{v}^2}{R}$
$m=\frac{B^{\prime }qR}{v}$
Substituting $v=E/B$
$m=\frac{B{B}^{\prime }qR}{E}$
Ions with different masses trace semi-circular paths of different radii and produce dark lines on the plate. The distance between the opening of the chamber and the position of the dark line gives the diameter $2R$ from which radius $R$ can be calculated.
Since, $B,B^{\prime },E$ and $R$ are known, the mass of the positive ions and hence isotopic masses can be calculated.