Smoke alarms are a necessary part of many modern day buildings, especially in laboratories and of course, the family home.

According to the World Nuclear Association page Smoke Detectors and Americium (2014), a major part of the operation of a smoke detector is when the Americium-241 emits alpha particles collide with atmospheric oxygen and nitrogen, ionising them - which form part of an electric current between two electrodes.

When there is smoke (from the article) - bolding mine:

When smoke enters the space between the electrodes, the smoke particles attach to the charged ions, neutralizing them. This causes the number of ions present – and therefore the electric current – to fall, which sets off an alarm.

What is the chemical process that causes smoke to neutralise the charged oxygen and nitrogen ions in smoke alarms?


1 Answer 1


From this paper, found the mechanism for neutralisation,

Smoke particle charge interaction

When smoke particles enter in the ionization chamber, they are subjected to the electric field generated by electrodes but also to the one generated by the charges. Close to the charges, the electric field is highly divergent resulting in dielectrophoretic force acting on smoke particles. The force tends to bind the smoke particles to the charges. When smoke particles are considered as isotropic neutral dielectric bodies linearly and homogeneously polarisable, dielectrophoretic force F can be expressed as

$ F = \frac{1}{2} \alpha v \overrightarrow{grad} |E^{2} |$

where $\alpha$ depends on smoke particle polarizability, $v$ is the smoke particle volume and $E$ is the electric field produced by the charges [11]. Coefficient $\alpha$ is typically of the order of $\epsilon_{0}/3$ [12] where $\epsilon_{0}$ is the dielectric constant of air. Volume $v$ is approximately $4 \mu m^{3} $ for a 1-$\mu m $ diameter smoke particle. Amplitude of electric field $E$ produced by charges is given by coulomb’s law:

$ E = \frac{1}{4\pi \epsilon_{0}} \frac{q}{r^{2}}$

in which $r$ is the distance between the charge and the smoke particle. Due to the divergence of the field, the dielectrophoretic force decrease as $\frac{1}{r^{5}}$. If a smoke particle is close to a charge then they can bind. If the smoke particle is relatively far from the charge, the time of interaction may be relatively long and it is necessary that the charge drifts sufficiently slowly in the chamber to optimize the interaction. Indeed if the charge passes too rapidly there will be not enough time for the smoke particle to bind and thus nothing would be detected. In fact the capture radius reduces with the electric field and is of the order of 10 $\mu m $ for $E= 1 V/mm$.

When the smoke particles attach to the free charges due to electrophoresis force, the charges become heavy bodies then the charge mobility $\mu$ decreases resulting in a drop of the electric current measured between the electrodes of the ionization chamber. Even if ionic detectors are very efficient, according to new standards they are no longer authorised in almost all of the developed countries because of the radioactive sources they include for generating the ions. Such radioactive sources are indeed difficult to collect and to recycle.


Your Answer

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