# Why are most metals gray/silver?

Why do most metals (iron, tin, aluminum, lead, zinc, tungsten, nickel, etc.) appear silver or gray in color? (What atomic characteristics determine the color?)

What makes copper and gold have different colors?

• Comment to the question (v2): Is there an aspect of the question you in particular are interested in, that is not answered by simple Google searches, such as, e.g., Wikipedia or this web page? – Qmechanic Jul 26 '13 at 18:14
• This may be helpful: fourmilab.ch/documents/golden_glow – Ben Crowell Jul 26 '13 at 21:08
• Why is silver 'close' to white, as suggested in most answers? – mehfoos Jul 26 '13 at 21:09
• @Chris Here is another answer I had found on the web funtrivia.com/askft/Question49344.html . Checking the periodic table is another thing I did, as I do often with such questions, with the same conclusion as you (though I am mostly incompetent on such matters). I was just wondering why nobody, including the OP, had made any such comment. Then, now, the OP has 23 votes, while the 3 answers together have only 18. Obviously, the question is intriguing, especially after checking the periodic table. Maybe I do not understand the purpose of votes, but I would expect more participation. – babou Jul 27 '13 at 20:18
• I asked this question in "electronics" and "physiscs" both as I thought maybe device people in electronics may contribute from a different point of view (than physics people). Does anyone know how to merge them? – alifornia Jul 28 '13 at 21:32

Why do most metals appear silver in color, with gold being an exception?

It is hardly surprising that the answer to this question relies heavily on quantum theory, but most people will be surprised to hear that the full answer brings relativistic considerations into the picture. So we are talking quantum relativistic effects.

The quantum bit of the story tells us that the colour of metals such as silver and gold is a direct consequence of the absorption of photons by d electrons. This photon absorption results in d electrons jumping to s orbitals. Typically, and certainly for silver, the 4d→5s transition has a large energy separation requiring ultraviolet photons to enable the transition. Therefore, photons with frequencies in the visible band have insufficient energy to be absorbed. With all visible frequencies reflected, silver has no colour of its own: it's reflective, an appearance we refer to as 'silvery'.

Now the relativistic bit. It is important to realize that electrons in the s orbitals have a much higher likelihood of being in the neighborhood of the nucleus. Classically speaking, being close to the nucleus means higher velocities (cf speed of inner planets in solar system with that of the outer planets).

For gold (with atomic number 79 and hence a highly charged nucleus) this classical picture translates into relativistic speeds for electrons in s orbitals. As a result, a relativistic contraction applies to the s orbitals of gold, which causes their energy levels to shift closer to those of the d orbitals (which are localized away from the nucleus and classically speaking have lower speeds and therefore less affected by relativity). This shifts the light absorption (for gold primarily due to the 5d→6s transition) from the ultraviolet down to the lower frequency blue range. So gold tends to absorb blue light while it reflects the rest of the visible spectrum. This causes the yellowish hue we call 'golden'.

Reflectivity as function of wavelength. Purple/blue light corresponds to 400 - 500 nm, the red end of the visible spectrum to about 700 nm.

• Thanks for the clear explanation. So, does this mean that the color of copper is unrelated to the color of gold, despite their being is the same column of the periodic table ? – babou Jul 27 '13 at 20:42
• @babou - indeed, the color of copper can be understood non-relativistically. Apparently, in copper the filled 3d is less shielded by the s and p subshells. desy.de/user/projects/Physics/Relativity/SR/gold_color.html – Johannes Jul 28 '13 at 4:49
• Two questions: 1-Does "relativistic effects" mean including Spin-Orbit interaction? 2-I am confused about the role of lattice. Does lattice play a role here? – richard Aug 1 '13 at 10:34
• Copper and gold are yellowish because d-band transitions absorb blue. That is why the color is what it is. The d-bands are where they are due to relativistic effects. Whether or not the argument is transitive ... "gold and copper are colored due to relativistic effects" ... (I know you did not say this, but hold on for a minute) is semantics, and arguable. For example, one could equally say that the color is what it is because Planck's constant has its particular value. So while I greatly appreciate your correct answer, in my opinion the best answer is "because of d-band transitions" – garyp Jun 3 '14 at 14:49
• How does relativity come into play, since the electrons aren't actually moving classically? – Michael Dec 17 '14 at 19:32

D electrons in metals allow optical transitions in the visible regime. Visible light can be absorbed by elements having unbound valence electrons in the d shell. So

Chemistry: optical d->s$^2$ transition

• Iron [Ar] 3d$^6$ 4s$^2$
• Tin [Kr] 4d$^{10}$ 5s$^2$ 5p (full d shell)
• Aluminium [Ne] 3s$^2$ 3p$^1$ (is a special case: no d valence electrons, but Aluminium reflectivity. I have no other explanation than the calculation of Fresnel equations. However I can't grasp the reason for this distinction.)
• Lead [Xe] 4f$^{14}$ 5d$^{10}$ 6s$^2$ 6p$^2$ (full d shell)
• Zinc [Ar] 3d$^{10}$ 4s$^2$ (full d shell)
• Tungsten [Xe] 4f$^{14}$ 5d$^4$ 6s$^2$
• Nickel [Ar] 4s$^2$ 3d$^8$ or 4s$^2$ 3d$^9$
• Copper [Ar] 3$d^{10}$ 4$\mathbf{s^1}$ (one s and full d shell)
• Gold [Xe] 4f$^{14}$ 5d$^{10}$ 6$\mathbf{s^1}$ (one s and full d shell)

The shiny metals, except aluminium, have d electrons. A single s electron and a full d shell hint at an important d to s$^2$ orbital transition in the visible spectrum. A full s shell is energetically preferred. There seems to be no explanation for the colored appearance of gold and copper, other than a distinctive electron configuration - at least chemistry does not provide an answer.

Physics: sign change of $\epsilon(\lambda)$ near blue

If the absorbed light is reemitted (in fact reflected) for the whole visible spectrum, the metal appears shiny as a mirror. In fact, our bathroom mirrors are made of an aluminum backside coated glass.

Here physics has to explain more than just "is there a d valence electron". A second more physical reason doesn't describe its origin: Reflectivity, out of the Fresnel equations using $$n=\sqrt{\epsilon_r\cdot \mu_r}\qquad\text{with}\qquad \epsilon_r=1-\frac{n_e e^2}{\epsilon_0m\omega^2}\qquad\text{with a sign change at}\qquad \omega=\omega_p$$

out of the Drude free electron gas model for electrons (and density of electrons $n_e$), is high through the whole visible spectrum for these metals. This sign change at $\omega=\omega_p$, plasma frequency is the reason for a changing $\epsilon_r$, therefore a changing refractive index $n$, due to the Fresnel equations, a changing reflectivity. If this change happens to be in the visible spectrum, then there are colored reflections like gold. Blue absorption of gold happens, because special relativity has to be taken into account for this heavy element. See top answer. Copper and Gold don't have a high reflectivity for blue ($\approx 475\,$nm).

• "There seems no explanation for colored appearence of gold and copper" -- actually there is (see my answer). – Johannes Jul 27 '13 at 5:07
• @StefanBischof, That was a good rewrite. You clearly connected the blue frequency absorption of gold with the special relativity effect. Thanks. I'll take down my comment, as it is no longer relevant. – Thomas Lee Abshier ND Jul 6 '18 at 20:10
• @StefanBischof, in your sentence " A second more physical reason 'doesn't' describe its origin: ..." Paraphrased, I think you said, that these equations and model "don't" describe the origin of metallic reflectivity. Would you please elaborate/explain your point? Thanks. – Thomas Lee Abshier ND Jul 6 '18 at 21:35
• @ThomasLeeAbshierND Fresnel equations are based on $\epsilon(\lambda)$ and $\mu$. They are practical for optics design. The physical reason is hidden in e.g. special relativity and Drude model. – Stefan Bischof Jul 7 '18 at 9:54
• @StefanBischof, I understand. In summary: The underlying physics producing the effect of frequency-dependent gold and copper light reflection is not overtly/explicitly considered in the Fresnel equations (which are used in high-level design/practical engineering). If we wish to understand the mechanism underlying these effects, we must go deeper and derive the $\epsilon (\lambda)$ and $\mu$ from even more basic theory, namely the Drude model and SR. And in particular, the computation of the plasma frequency, above which light goes from being reflected to absorbed. Is this what you meant? – Thomas Lee Abshier ND Jul 8 '18 at 1:14

"The color of metals can be explained by band theory, which assumes that overlapping energy levels form bands.

In metallic substances, empty conduction bands can overlap with valence bands containing electrons. The electrons of a particular atoms are able to move to a higher-level state, with little or no additional energy. The outer electrons are said to be "free," and ready to move in the presence of an electric field.

The highest energy level occupied by electrons is called the Fermi energy, Fermi level, or Fermi surface.

Above the Fermi level, energy levels are empty (empty at absolute zero), and can accept excited electrons. The surface of a metal can absorb all wavelengths of incident light, and excited electrons jump to a higher unoccupied energy level. These electrons can just as easily fall to the original energy level (after a short time) and emit a photon of light of the same wavelength.

So, most of the incident light is immediately re-emitted at the surface, creating the metallic luster we see in gold, silver, copper, and other metals. This is why most metals are white or silver, and a smooth surface will be highly reflective, since it does not allow light to penetrate deeply.

If the efficiency of absorption and re-emission is approximately equal at all optical energies, then all the different colors in white light will be reflected equally well. This leads to the silver color of polished iron and silver surfaces.

For most metals, a single continuous band extends from valence energies to 'free' energies. The available electrons fill the band structure to the level of the Fermi surface.

If the efficiency decreases with increasing energy, as is the case for gold and copper, the reduced reflectivity at the blue end of the spectrum produces yellow and reddish colors.

Silver, gold and copper have similar electron configurations, but we perceive them as having quite distinct colors.

Gold fulfills all the requirements for an intense absorption of light with energy of 2.3 eV (from the 3d band to above the Fermi level). The color we see is yellow, as the corresponding wavelengths are re-emitted.

Copper has a strong absorption at a slightly lower energy, with orange being most strongly absorbed and re-emitted.

Silver. The absorption peak lies in the ultraviolet region, at about 4 eV. As a result, silver maintains high reflectivity evenly across the visible spectrum, and we see it as a pure white. The lower energies corresponding to the entire visible spectrum of color are equally absorbed and re-emitted making silver a good choice for mirror surfaces.

• This absorption and re-emission doesn't add up for polished surfaces that act as mirrors. The angle or the re-emission would not be a function of the incident angle, so temporary absorption and re-emission is not what is going on in a mirror. – Olin Lathrop Jul 28 '13 at 16:05
• @Olin: I've thinking about your statement for about a week now on how re-emission is not angle dependent. It makes sense but I am struggling to physically understand why this is so. Can you please expand on this? – Carlos Aug 18 '13 at 4:18
• @OlinLathrop the reflection is caused by the sea of fluid electrons that are the defining criteria of a metal. The light tries to pass through this negative-charged gas, and the E field acts on the electrons to cancel it out and conjur a new wave going the other way. Angle of reflection doesn't work right with single photons (it depends on phase instead). See Feynman's little book on QED. – JDługosz Aug 18 '15 at 17:09

This question has another interesting aspect which has more to do with neuroscience than physics: why do we perceive metals with a neutral colour (such as silver) as grey, even why they are shiny and therefore simply reflect the colours of their surroundings?

One answer is that such metals always have some roughness and therefore scatter light from a range of angles, and these rays typically have a range of wavelengths. The mixing of these wavelengths tends to desaturate the perceived colour, and moves it towards a neutral tone. However, some simple experiments suggest there's more to it than this. Even when the surface is reflecting one dominant colour our perception of the surface colour is grey.

The reason for this is connected with the way the brain processes colour information. Colour constancy ensures that our perception adjusts for colour bias in the ambient light conditions: we tend to perceive an object's intrinsic colour rather than the colour of the light reflected from it. The apparent greyness of metallic surfaces (both shiny and matt) seems to be an interesting variant of this phenomenon.

• – uhoh Dec 3 '18 at 23:39

Let's start with what "that thingy is X in colour" fundamentally means:

You notice silver is not one of the colours here. However, silver is a LOT like white, as we will see in a second.

There is another factor involved called specular vs diffuse reflection.

White reflects all wavelengths diffusely (the reflected rays go every which way). Silver (e.g., a mirror) reflects all wavelengths specularly (the reflected rays bounce off nicely).

Now, metals do not necessarily always look like mirrors - they are often bumpier than that, so their reflection is a little bit diffuse as opposed to totally specular.

Anyway, the point is that "silver colour" means "reflects all wavelengths specularly (more or less)".

Why do these metals reflect most visible light? Because they have lots of free electrons (that also happens to be why they're good conductors). When light (electromagnetic radiation) hits the surface of a metal, it gets absorbed by electrons orbiting the metal atoms, and re-emitted as the electrons fall back to a more stable configuration. The size of the band gaps determines which frequencies get absorbed and emitted.

A coloured metal like gold has most of these properties, but it absorbs just a little bit of radiation in the green-blue-violet area. So whatever it reflects out has a bit of green-blue light removed and the result looks (by subtraction) yellowish red.

A metal like lead also has most of these properties, but it absorbs a little more of the entire spectrum, so it looks grey.

PS This answer is provided by "Ian Pollock, Sci/Phil dilettante" at quora.com.

• This answer explains reflection. However there is no evidence, what physical phenomenon causes the color. -1 – Stefan Bischof Sep 21 '13 at 17:45
• If this answer comes from someone else, you should mark it as a community wiki. – Kyle Kanos Sep 25 '14 at 19:15

Metallic band structure allows absorption and re-emission of light as depicted on this site.

Metals are colored because the absorption and re-emission of light are dependent on wavelength. Gold and copper have low reflectivity at short wavelengths, and yellow and red are preferentially reflected. Silver has good reflectivity that does not vary with wavelength, and therefore appears very close to white.

• I assume the OP is asking for an atomic-level explanation of why the reflectivities behave this way, rather than a generic explanation of color. – Ben Crowell Jul 26 '13 at 22:01
• In the future, please put quoted material in quotation marks. (Stefan Bischof has done this in an edit, which is awaiting approval.) – Ben Crowell Jul 26 '13 at 23:12
• Will do. My mistake. – Michael Luciuk Jul 26 '13 at 23:19

## protected by Qmechanic♦Sep 25 '14 at 17:05

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