by Din » Fri Dec 16, 2022 11:48 am
The solubility of potassium iodide in ethyl alcohol decreases as the concentration of ethyl alcohol increases*. Since the way the iodide reduces printout is by replacement of the bromide ion in emulsions with the iodide ion, the large number of bromide ions still in the emulsion (depending, of course, on the number of grains that have not been reduced), means you'd need quite a large amount of ethyl alcohol to dissolve sufficient potassium iodide. This may be counter-productive to the aim of reducing cost.
It's possible to calculate it all out, but I suggest you try it out first. I'd cut an exposed and developed plate in half, and use your method on one half. I'd put both halves under a bank of fluorescent lights (fluorescent because there's more blue in the spectrum), and examine both halves after, say, three/four hours. An even more objective test would be to illuminate both halves to laser light, and take a reading of the intensity of the light after passing through the plates. If successful, the intensity of one half, say I(1), of the printed out half would be less than the intensity of the other half, say I(2). Then, the amount of printout, P, is given by
P = {(I(2) - I(1)}/I(1).
*
https://ui.adsabs.harvard.edu/abs/2009R ... V/abstract
The solubility of potassium iodide in ethyl alcohol decreases as the concentration of ethyl alcohol increases*. Since the way the iodide reduces printout is by replacement of the bromide ion in emulsions with the iodide ion, the large number of bromide ions still in the emulsion (depending, of course, on the number of grains that have not been reduced), means you'd need quite a large amount of ethyl alcohol to dissolve sufficient potassium iodide. This may be counter-productive to the aim of reducing cost.
It's possible to calculate it all out, but I suggest you try it out first. I'd cut an exposed and developed plate in half, and use your method on one half. I'd put both halves under a bank of fluorescent lights (fluorescent because there's more blue in the spectrum), and examine both halves after, say, three/four hours. An even more objective test would be to illuminate both halves to laser light, and take a reading of the intensity of the light after passing through the plates. If successful, the intensity of one half, say I(1), of the printed out half would be less than the intensity of the other half, say I(2). Then, the amount of printout, P, is given by
P = {(I(2) - I(1)}/I(1).
*https://ui.adsabs.harvard.edu/abs/2009RJPCA..83.1896V/abstract