It has been shown that it is necessary to formulate two theories of development, since we do not know whether the unchanged or the changed silver compounds on the plate are attacked the more rapidly by the developer. The general law which has been found to apply in all such cases is that the less active substance is unaffected, whilst the more active one is attacked at an increased rate, which increase is directly proportioned to the difference in chemical activity between the two substances. On the supposition that the photo salt is attacked less rapidly than the unchanged silver halide, the simplest explanation of development lies along the lines laid down by Desalme.1
Desalme's Theory of Development is electrolytic. It is found that whenever two substances that differ either chemically or physically, are placed in electrical contact, a difference of potential is set up between them. (In the case of an exposed plate this potential difference between:
1 We will assume the reader to be familiar with the concept of ions and will adopt the ionic mode of expressing chemical change. We must therefore mention what ions are present in the developing solution.
The film supplies bromidions and argentions. (The solubility of the halides of silver, though too small to be detected by the balance, is proved by the increase in the conductivity of water in which they are shaken up, and the number of ions present in unit volume, estimated by measuring this increase.) Be it remembered that equilibrium will be established in the case of a solid in contact with water when the number of free ions in solution reaches a definite value; and if any be removed, a fresh number will immediately enter into solution to restore equilibrium.
The alkaline hydrates will yield sodions (or potassions) and hydroxidions. Alkaline carbonates and even phosphates also yield these ions, owing to the hydrolysis occurring when they enter into solution.
Na2C03 + H20 = NaOH + NaHC03 Na3PO4 + H,0 = NaOH + Na„HPO4 the photo-salt1 and the unchanged AgX 2 will be increased by the action of the electrolyte.)
If this E. M. F. be high enough electrolysis will commence. But, immediately, owing to the accumulation of anions and cations in opposite regions, an equal back E. M. F. will spring into existence, and cause polarisation. Electrolysis ceases. To re-establish it, a depolariser is needed. It will be enough to remove the bromidion, by causing it to enter into chemical combination.
It has generally been assumed that the bromidion combines with the developer. This is true in the case of ferrous oxalate :
3(COO)2Fe + 3 AgX = FeX3 + (COO)3Fe2+ 3Ag but untrue for the usual organic developers. It was thought that the halogen attacks the organic developer, forming a halogen substitution compound, with liberation of HX, which was neutralised by the alkali, whereby the action of HX on the silver of the image was prevented. But if this were so, then we should find that bodies which react most readily with the halogens (such as the monohydrophenols) would make very good developers; but such is not the case, these compounds do not develop. Furthermore, if the bromidion combined with the developer, we should expect unsaturated compounds, such as cinnamic acid C6H5 • CH : CH • COOH which adds on bromine directly to form C6H5 • CHBr • CHBr . COOH, without evolution of HBr, would develop without alkali. Such compounds do not develop.
Since therefore, no HX is liberated, the alkali must play another part. On running our eye over a list of organic developers we find aromatic compounds containing at least two hydroxyl groups, or both hydroxyl and amido groups, joined directly to the benzene nucleus, and are struck by the fact that such compounds are sensitive to hydroxidion.
1 By which we mean the substance, whatever it may be, forming the latent image.
2 X is used indifferently to indicate chlorine, bromine, or iodine, and x for the ion.
We now see daylight. Hydroxidion combines with the developer and loses its charge. The sodion to which it was joined is now free to react with the bromidion:
+ + - + - +
Na + AgBr = NaBr + Ag and the argention liberated will turn into metallic silver :
Ag -> Ag.
If we assume, that the photo salts are attacked more readily than the unchanged AgX, then the chemistry of the process appears simpler, though the final products are the same. The first action will be the disappearance of the photo salt. Heat will be liberated.1 This heat will be taken up by the neighbouring particles of AgX, and the action of the developer on those particles will be hastened in proportion to the amount of heat received (which will depend on the local amount of photo salt) in obedience to the law that the rate of a chemical change is doubled (roughly) for every rise of 10°C in temperature. Thermo-chemical calculations show that the action of the developing solution on AgX is exothermic.2 Once started, it will go on of itself, at an increasing rate, the photo salt merely imparting an initial acceleration.
Meanwhile, however, heat is travelling along the film by conduction, and by local convection currents, and hence as development proceeds the action slowly spreads to other parts of the plate, and will ultimately involve the whole film. In this light the solution of the problem regarding the action of the photo salt is as simple as the starting of a furnace by the aid of a safety match; though the chemical changes are more involved.