In a Geiger-Muller counter which is the signal produced? If we imagine the tube as a capacitor, we intially have a certain voltage between the plates (the cilinder and the filament). But when the electrons arrive at the anode and they are collected by it, the voltage falls? And slowly the positive ions will reach the cathode, so they invert the voltage? Which is the correct behaviour of the signal produced?


When the gas in the tube becomes conducting the usual method of detection is to have a resistor connected between the cathode and the negative end of the power supply.
This converts the current pulse into a voltage pulse which can then be detected and counted.
This method of detection of a current pulse at the cathode is preferred to the detection of a pulse in the anode part of the circuit.
One reason is that the stray capacitance of the detecting circuitry will have less effect on the characteristics of the tube when connected to the much larger cathode as compared with the relatively small anode.

On hitting the cathode the slower moving positive ions are neutralised but at the same time can emit secondary electrons. On gaining electrons the ions can become excited atoms (some of the electrons are above the ground state configuration) and emit photons which could start the avalanche process again as can the secondary electrons.
So as well as argon gas in the GM-tube there is a small amount of quenching agent present (bromine or chlorine) and the atoms of the quenching are more easily ionised than argon because of their lower ionising potential.
The argon ions can take electrons from quenching agent atoms and when they hit the cathode there is less chance of the emission of secondary electrons.
There is also the possibility of the emission of photons by an argon ion after it becomes an excited argon atom. These photons could also trigger further avalanches (pulses). The photons emitted by the excited quenching agent atoms are of smaller energy and thus less likely to produce further avalanches.


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.