Re: When a silver chloride sample is put in light, what happens to the sample?

Hello Ryah – this is a complex question of solid-state physics - and the subject of many high-powered papers and conferences over the years. Obviously you don’t want a long treatise on the subject, but a relatively easy-to-understand idea of what happens. Photography depends on this process, and uses not only silver chloride (AgCl) but the other halides – bromide (AgBr) and iodide (AgI) as well, often as mixtures.

Silver halides absorb light – from ultraviolet into the visible spectrum up to 500 or more nanometers. Pure silver chloride really only absorbs up to a little over 400 nm, which is barely in the visible part of the electromagnetic spectrum, so in practice dyes are added to the grains to absorb more visible light and transfer the energy absorbed direct to the grain. (This is called spectral sensitization).

When light is absorbed, the energy excites electrons within the atoms of the crystal from one “allowed” level to another – and physicists talk of exciting electrons from the valence band to the conduction band. The valence band is an energy state of the electron in which it is participating in the lattice bonding of the crystal structure – while the conduction band – as the name implies - allows the electron to be released from its original state and travel through the crystal lattice – to conduct electricity, e.g. The lattice is the array of silver ions and chloride ions. That is silver (Ag) atoms with a positive charge (having given up an electron), and an equal number of chloride ions which are chlorine atoms with a negative charge (having obtained the electron that the silver gave up). When the crystal absorbs the light energy, the electron on the chloride ion is raised in energy to conduction band level. Now, what might then happen is that this electron could relax back to its original state, giving off its energy in another form – heat for instance, or – instead - with its new found mobility – find another route to release its new-found energy. This could be to wander off elsewhere in the crystal and react with another silver ion to form a silver atom. Both things happen, and of course the trick in photography is to design the system (deliberate impurities etc.) such that the latter mechanism predominates. If the electron is lost in this way – it leaves behind a chloride radical which will react with anything it can – e.g. the impurity. So atoms of metallic silver are formed. If you expose a silver chloride crystal to light it starts off a light grey colour, and gradually darkens – because metallic silver is black in sub-divided form.

In photography, each crystal needs to form only a few atoms of silver on its surface to make it “developable”. The silver atoms catalyse the reduction of the whole grain by the developer (a reducing agent) so that it is all converted to metallic silver before neighbouring grains which had no exposure and therefore no silver atoms even start to get reduced. These silver specks or aggregates of silver atoms are called the latent image.

Here’s an easy experiment to do – get a roll of unexposed film – anything will do, and pull some out of the cassette into the light. Quickly put a coin on top of the film (on the matt side) and just leave it. After a while remove the coin and look at the film and see if there is any image there. How long does it take to form? Would this be a practical way to take pictures? How long do you normally expose the film to light when taking a photograph?

You could even do this with a new roll of film for your camera – just pull the leader out – a couple of inches is enough – probably the bit that’s not going to be used for pictures anyway, and you won’t waste anything. Most 35mm cameras cause this to happen when you load them before shutting the back and so that piece of film gets totally “fogged”. It all gets developed to black. 35mm is best - you can’t do this with the new APS films which self-load.

If you want to ask any more – just e-mail me (adamhh@dial.pipex.com)