3
$\begingroup$

After a molecule is ionised, the ion is accelerated by an electric plate up to a certain kinetic energy by the attraction of the ion to the electric plate. However, after the ion has passed the electric plate, won't the plate still exert an attractive force on the ion? If so, by the attraction between the ion and the electric plate, the kinetic energy of the ion were attenuated, perhaps even to the point where there would not be a net gain in kinetic energy of the ion.

$\endgroup$
4
$\begingroup$

The acceleration of the freshly generated ions is stepwise (similar to a linear particle accelerator), i.e. the beam of ions is guided an array of electrodes with a small opening in their centre, as schematically depicted below:

enter image description here

(image source)

Hence when ions approach the first accelerator electrode from the left hand side, their interaction with the subsequent accelerating electrode already phases in. Ions passing the slit of the first accelerating electrode will experience an even stronger accelerating force by the subsequent accelerating electrode. Eventually, their momentum will increase further; and on the same time, the ions preferentially will keep their direction of propagation towards the mass dispersive element (here: the magnet), regardless the attractive force between them and the last accelerating electrode.

$\endgroup$
2
$\begingroup$

Your question since an ion is attracted to the second plate why doesn't this plate also attract after the ion has passed through it and bring it to a halt.

The ions are initially accelerated by a potential difference, which means there are different voltages between two metal plates separated by some distance. As a result the potential energy present due to the voltage between two plates is converted into kinetic energy of the ion. In energy terms if at the exit plate, leading into the spectrometer, the energy is $E_x=mv_x^2/2$ and initially the ion energy is $E_0=mv_0^2/2$ then $E_x = E_0 +W$ where $W$ is the work term, i.e. the kinetic energy added to the ion. This is $W=\pm qEd$ where q is the ion's charge, E field strength in volts/cm and d the separation of the two plates, i.e the distance over which the acceleration takes place. If W is positive the ion is accelerated, if not decelerated.

Thus you can see that if the potential were symmetrical, for example with three plates of potential +V -V +V through which the ion travels in sequence it would be brought to the same speed as initially at the final plate. But suppose that after the second plate the field ( volts/cm ) is arranged to be very small or even zero then the ion will travel at constant speed (ignoring the effect of gravity) into the rest of the spectrometer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.