Question on Chemistry of Color Film

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lensman_nh

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This is a question for the chemists here.

I've been doing some reading on the roles of bleach and fixer in color processing, and as these things tend to lead to, I ended up down a rabbit hole.

What I have been able to deduce is that a photon hitting a silver halide crystal can liberate an electron that migrates to an electron trap where it reduces the silver ion to metallic silver. Development further reduces the halides to metallic silver.

The bleaching process removes the silver from the image, leaving only the dye clouds from each color emulsion layer. Bypassing the bleach leaves the silver in the film giving a dull, gritty, look beloved by some directors.

Fixing removes any remaining silver halides. B&W processing has no bleach so you are left with metallic silver.

I ended up with 2 questions I couldn't quite get my head around.

Firstly, why does the developer only reduce those areas where the free electron has reduced the halide to silver already? In my very limited experience reduction has been an all or nothing experience, so I would naively expect fogging not latent image development.

Secondly, why does the dye sensitize the silver halide grain? In my very basic understanding the free electron gets knocked off when the photon interacts with the halide atom, and I'm at a loss to understand why the dye makes that more likely.

Thanks

J.
 

Rudeofus

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Regarding dye sensitizers: typical silver halide crystals require energetic photons to liberate electrons, which means they are sensitive to blue (Silver Iodide) or violet light (Silver Iodide, Silver Bromide, Silver Chloride). A sensitizing dye adsorbs to the silver halide crystal and interacts with a specific (typically longer) light wavelength range in such a way, that these longer wavelengths can still create a latent image. BTW this is used in black&white and color film!
 

koraks

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Firstly, why does the developer only reduce those areas where the free electron has reduced the halide to silver already? In my very limited experience reduction has been an all or nothing experience, so I would naively expect fogging not latent image development.
That's why a development agent is used that selectively develops only exposed silver halide particles. Strong reducers would indeed reduce all particles.

Secondly, why does the dye sensitize the silver halide grain? In my very basic understanding the free electron gets knocked off when the photon interacts with the halide atom, and I'm at a loss to understand why the dye makes that more likely.
The dye doesn't make it more likely, but rather less likely for photons that are outside the desired spectrum. By itself, silver halides are sensitive only to UV light, and sensitizing dyes can stretch this sensitivity towards longer wavelengths. (As to the exact mechanism, I'm not sure and I have not yet read a for me understandable explanation of this, although I vaguely remember coming across some papers about it). Note that sensitivity for the shorter wavelengths remains in place. For that reason, the different sensitivity peaks of each color layer are realized through the combination of sensitizing dyes and filter layers between the emulsion layers. This is why the top layer is sensitive to blue light (not sensitized for green and red), the green-sensitive layer is in the middle underneath a yellow filter (this blocks any blue light passing through the top layer which would otherwise also expose the 'green/magenta' layer!) and the red-sensitive layer is at the bottom, underneath yet another filter layer which blocks green and blue.

This is the simplified version; in a real film product, there are typically (as I understand) at least 9 emulsions, 3 for each color layer, each with different speed and grain size, and all of these emulsions are precisely tuned to each other to get good linearity within each color and good tracking between the colors.

Photo_Engineer can doubtlessly explain this in more detail (and correct any errors I may have made above).
 
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lensman_nh

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That's why a development agent is used that selectively develops only exposed silver halide particles. Strong reducers would indeed reduce all particles.

But there's the paradox. The enthalpy of reaction required (or liberated) to reduce the halide to a silver atom is the same. So a weak reducing agent would, in the absence of any other factor, cause a slow rise in the level of fog somewhat uniformly. Base fog does indeed also rise across the board, and the effect is seen most when push processing.

But there is something about the presence of the initial latent silver image that predisposes the reaction to preferentially occur in those areas. It is almost like it's a catalyzing effect, where the presence of silver initially causes the reaction to proceed preferentailly in that area. Is it indeed a catalytic effect, or some quantum level surface effect more akin to electron tunneling that somehow modifies the enthalpy of reaction.

J.
 

koraks

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I wouldn't know, that's really beyond my present knowledge. It does sound plausible (and in the end a catalytic effect could probably explained in terms of a quantum effect I guess...)
 

Alan Johnson

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On exposure an electron is freed from a bromide ion leaving a positive hole and an interstitial silver ion. According to the theory of Mott-Gurney, the generation of the electron may be followed by its being trapped at a remote site where its energy is lower than that in the excited state..
A mobile silver ion then combines with it to form a silver atom, which makes a local electron trap and makes more likely the capture of other loose electrons (nucleation process).More electrons may be captured leading to formation of a sensitivity center,usually at a crystal surface defect. Such electron traps may be increased in effectiveness by chemical sensitisation, generally sulfide, gold.
In my miss-spent youth I went to a lecture by Nevill Mott, nice old guy, even then he was a bit of a celebrity.
 

Rudeofus

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But there's the paradox. The enthalpy of reaction required (or liberated) to reduce the halide to a silver atom is the same.
Nope, it's not. A single metallic silver atom is neither stable nor developable, but as soon as you put more than two silver atoms together (and they have a habit of coming together if they are created at roughly the same time and location), you get very different enthalpies for development of further silver. As you increase the reduction potential of the developer, it will be strong enough to develop smaller and smaller latent image centers until, at a certain reduction potential it will develop any silver ion it find - this would no longer count as photographic development. Note, that these energy limits are not sharp limits: thermal energy blurs the line, and some latent image centers may be located at different sites of a silver halide crystal. Therefore even a weak developer will slowly develop smaller latent image centers and even create fog, if you give it enough time.
 

Rudeofus

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In his book "Photographic Sensitivity Theory and Mechanisms", Tadaaki Tani published a graph of required reduction potential dependent of latent image cluster size:

TadaakiTani_DevelopmentOfLatentImageCenters.png


The label 1/d means inverse diameter of latent image center in µm, and a number estimate for silver atoms is printed on the top rim of the frame. Ag7 denotes a cluster comprising 7 silver atoms.
 
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lensman_nh

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Nope, it's not. A single metallic silver atom is neither stable nor developable, but as soon as you put more than two silver atoms together (and they have a habit of coming together if they are created at roughly the same time and location), you get very different enthalpies for development of further silver

Thank you! That fixes my faulty staring assumptions, and now the development process makes sense. I've found a couple of used copies of Photographic Sensitivity Theory and Mechanisms and I'm very tempted to get one.

I figure the moment that you stop educating yourself and learning, you die.

J.
 

lantau

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I never specifically looked it up for the usual b/w developers, but I did for the color developers and I assume it's the same. For our chosen development agents the reduction of silver ions to metal is catalysed by metallic silver. Without it nothing will happen in the unexposed parts.

Only where there are silver clusters (the latent image) will the reaction occur. More silver is produced, further sustaining the reaction. So it starts at the nucleus, spreading out from there. More nuclei from higher exposure will translate into more activity there.
 
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