# What is the minimum number of metal atoms necessary to make a mirror?

I am still unsatisfied with what I have read so far about the physical basis of metallic mirrors reflectivity. In particular, I am skeptical about the idea that individual electrons serve as mirrors by bouncing back photons, and I am tempted to think that there is somewhere an interface (e.g. air-metal), or at least something due to a collective action of electrons, producing a flat continuous plane of reflection to front waves (any photon-based explanation seems flawed to me). I am not merely asking about how many micrometers of metal are sufficient, but about atomic layout, down to the question of how conduction electrons might participate differently than other shells to reflection. Please do not answer if you cannot go down this level of detail, thank you in advance. Also, I am looking for plain English descriptions of physical processes, not math recipes (but feel free to combine). The point is to understand the minimum configuration of metal atoms (say silver, for example) that suffice to produce a mirror and why so.

EDIT: Using the popular definition of a mirror: a flat symmetric reflection of visible light, close to 100 percent. In other words, getting any light projected to the said mirror to reflect following usual optical rules of a bathroom mirror, although the question here is to know how small it can be.

EDIT2: PM 2Ring wrote a comment that is perfectly in line with what I am actually looking for: Yes, conduction electrons are the key, so I think you need enough atoms for there to be a "sea" of them, but I have no idea how small a metal crystal can be and still support a sea of conduction electrons that will give reflection that behaves as we observe with macroscopic mirrors.

So if someone could provide an answer to « how small a metal crystal can be and still support a sea of conduction electrons that will give reflection that behaves as we observe with macroscopic mirrors », that would answer my question.

EDIT3: unfortunately, the good stuff was moved from comments to chat. If someone wants to build on them and create a complete answer, be my guest. This question is about the particles level description of reflectivity. Unlike what KF Gauss says, it is not that I don't want an answer to depend on wavelength. It is rather that this is not sufficient to say the width of the mirror should depend on it without detailing the density and configuration of particles in between. An analogy, so to speak, is a Faraday cage. A Faraday cage is not a continuous sheet of conductor. So what about visible light, can reflection happen with a density of particles reduced to only cover the «corners» of a square sized to incident wavelength?

• How would you define a mirror? – my2cts May 8 '19 at 20:47
• Using the popular definition of a mirror: a flat symmetric reflection of visible light, close to 100 percent. In other words, getting any light projected to the said mirror to reflect following usual optical rules of a bathroom mirror, although the question here is to know how small it can be. – Winston May 9 '19 at 10:18
• Yes, conduction electrons are the key, so I think you need enough atoms for there to be a "sea" of them, but I have no idea how small a metal crystal can be and still support a sea of conduction electrons that will give reflection that behaves as we observe with macroscopic mirrors. – PM 2Ring May 9 '19 at 10:48
• SE posts are version controlled, so please do not make your post look like a revision table, instead just seamlessly integrate the new material into the post. There is an edit history button at the bottom of the post for those interested in seeing what changed. – Kyle Kanos May 10 '19 at 11:27
• @Exocytosis as stated, there is a link at the bottom of the post that shows the edits made (yours is here). It's perfectly visible and there is zero need to indicate that changes were made in other manners. – Kyle Kanos May 10 '19 at 14:29

If you want a true mirror, then you definitely need to be in the regime where the mirror is much larger than the wavelength of light, otherwise diffraction and/or Mie scattering will make it impossible for you to see a reflection without a lot of distortion.

Because optical light has wavelengths on the order of 500nm, the smallest mirrors should be a few times larger, so 1-3 micrometers in lateral length. The thickness of the mirror should be at least twice as large as the optical skin depth (i.e. optical penetration depth), which is about 20nm.

All together we have a mirror with a rough size of 2$$\mu$$m x 2$$\mu$$m x 50nm. If this mirror is made of gold, then we would have a weight of $$4\times10^{-12}$$ grams, or roughly 12 billion atoms give or take an order of magnitude. For reference, an ordinary mirror weighs about $$10^{15}$$ times more.

If your question is simply asking: how many atoms does it take to become a metal? Then even having 100-1000 atoms show good electrical conductance, which translates to gold nanoparticles that are a few nanometers in size. Because this is deeply sub-wavelength in size, the exact shape does not matter much in what concerns this question. In other words, the electrons will form a sea no matter what the exact shape of the nanoparticle. So the "configuration" is not relevant.

Edit: the question seems to have contradictory requirements. On one hand OP wants a mirror with standard optical properties (i.e. obeys usual geometric optics to give a clear reflection), but on the other does not want the answer to depend on wavelength. You can't have it both ways, Maxwell's equations don't allow it.

Edit 2: There is some concern about how reflection can occur if mirrors are atomically rough (atoms not perfectly aligned to be flat). That is irrelevant as long as that everything is flat on lengths comparable to the wavelength of light. Considering that atomic spacings are about 1000 times smaller than optical wavelengths, the exact surface shape does not matter. Moreover it is the electrons, and not atoms, that cause the reflection of light.

As an aside, ordinary mirrors can be even more rough because we only expect to resolve millimeters at best and don't care about distortions below that scale.

• Comments are not for extended discussion; this conversation has been moved to chat. – Chris May 9 '19 at 12:20