Exploring Other Metallic Elements with a Lustrous Appearance

[Loren Booda]

Besides gold, copper and tin, what metallic elements can have an other than silvered appearance, and what is the explanation for this rarity?

 

[Gokul43201]

Zinc is bluish-grey. I believe, so are Tantalum and Osmium, though I've never actually seen them. And iron and molybdenum and not silvery at all - just grey.

The silvery, white color is just a result of the dependence of the transmission coefficient on frequency. Due to the large number (density) of free electrons in most metals, the dielectric constant, K is negative (up to a certain frequency called the plasma frequency). A negative K gives rise to an imaginary wave number, ( n ~ sqrt(K) ) and hence 'no' transmission (due to an exponentially decaying field within the metal). The plasma frequency for most metals is somewhere in the far UV range. As a result, metals reflect equally, all frequencies in the visible range. And if something produces a fairly uniform distribution of the colors, it looks...White. That's why the silvery white color is very prevalent among the transition metals.

Sometimes, you have a low Plasma Frequency, say, in the near UV, or even Blue-Violet range. In that case, the metal will transmit such frequencies and up (ie.: it will not reflect them). So the reflected spectrum will be weighted more at the low frequency (yellow-orange-red) side, making it looking reddish (Cu) or yellowish (Au).

In short, the color of the metal depends on the frequency dependence of the dielectric constant in the Visible range.

Tin, looks quite silvery white to me Lead is a greyish-black - but, lead (like tin) is not a transition metal.

 

[Gokul43201]

I believe this thread will fare better in Atomic/Solid State Physics. Perhaps, the uberlords will consider moving it ?

 

[Nenad]

good explination, its too advanced, try to use laymans terms.

 

[Gokul43201]

I'm not sure there is a very good layman's answer. Perhaps someone in Atomic/Solid state forum might do better. This is really not the best place for this question to reside.

My answer itself is only an educated guess...

 

[broomfieldjay]

"The color in the transition metals (d-block) is predominantly due to the splitting of the d shell orbitals into slightly different energy levels. As a result, certain wavelengths of energy can be absorbed by the d-block elements (with electrons jumping between these slightly different energy levels), resulting in the complement color being visible. As electrons in the outer d-shell can absorb visible wavelengths of light, transition metals appear brightly colored."


The model mentioned above involving plasma frequency works well for many optical aspects of metals especially Alkali metals like Sodium but fails for properties involving band transitions which is what causes the color of transition metals.

 

[Gokul43201]

"The color in the transition metals (d-block) is predominantly due to the splitting of the d shell orbitals into slightly different energy levels. As a result, certain wavelengths of energy can be absorbed by the d-block elements (with electrons jumping between these slightly different energy levels), resulting in the complement color being visible. As electrons in the outer d-shell can absorb visible wavelengths of light, transition metals appear brightly colored."


The model mentioned above involving plasma frequency works well for many optical aspects of metals especially Alkali metals like Sodium but fails for properties involving band transitions which is what causes the color of transition metals.

I believe your explanation is valid for the transition metal salts, and not the metals themselves. In a metal valence electrons occupy a band - there are no distinct d-orbital levels.

Transition metal salts are brightly colored - blue, green, red, orange, yellow, etc. - while the metals themselves are mostly about the same color as alkali or alkaline Earth metals, or even the p-block metals and metalloids. This was point originally raised by Loren.

 

[osskall]

I believe this thread will fare better in Atomic/Solid State Physics. Perhaps, the uberlords will consider moving it ?

Maybe one then also should change the Topic "Atoms, Molecules & Solids / From the periodic table to condensed matter"? The metallic luster question was a good example of what I hoped to find under this subject title...

 

[ZapperZ]

Er... solid state physics IS a part of condensed matter physics. This thread is in the section where it is supposed to be.

Zz.

 

[NoTime]

I suspect that it is a lattice effect (the way the atoms are arranged).
A lattice, with the proper arangement, can selectively pass a particular frequency (in this case color).

For example a thin sheet of gold will allow light to pass through (Blue IIRC).
So in this case the blue gets transmitted to the interior of the gold sample, where it eventually gets adsorbed, while the red and green light frequency bands get reflected. Red and green light combine to view as yellow.
Light color mixing is not the same as paint color mixing.

 

[ZapperZ]

I suspect that it is a lattice effect (the way the atoms are arranged).
A lattice, with the proper arangement, can selectively pass a particular frequency (in this case color).

For example a thin sheet of gold will allow light to pass through (Blue IIRC).
So in this case the blue gets transmitted to the interior of the gold sample, where it eventually gets adsorbed, while the red and green light frequency bands get reflected. Red and green light combine to view as yellow.
Light color mixing is not the same as paint color mixing.

Note that the "thiness" of a thin film has nothing to do with the lattice structure. The reason why a thin film of metal can allow some light to pass through is because there is a finite penetration depth for a particular frequency of light. If the film is less than the penetration depth, then that light will pass through. So this has nothing to do directly with the lattice structure - if we are comparing film and bulk having that identical structure.

Zz.

 

[NoTime]

Note that the "thiness" of a thin film has nothing to do with the lattice structure. The reason why a thin film of metal can allow some light to pass through is because there is a finite penetration depth for a particular frequency of light. If the film is less than the penetration depth, then that light will pass through. So this has nothing to do directly with the lattice structure - if we are comparing film and bulk having that identical structure.

Zz.

I didn't say that the thinness of the film affected the lattice structure. Actually, I would be saying lattice structure was not affected.
I also didn't say that I know this to be the mechanism behind metal color.

OTOH I also don't know if you are saying this is wrong.
If you are saying it is wrong then what is the correct answer?

Also since lattices are known to be involved in selective transmission of EM then wouldn't "penetration depth for a particular frequency of light" be a lattice effect?
In either case I was just saying that the blue light just penetrates too deeply to get reflected, leaving yellow as the result.

 

[ZapperZ]

I didn't say that the thinness of the film affected the lattice structure. Actually, I would be saying lattice structure was not affected.
I also didn't say that I know this to be the mechanism behind metal color.

OTOH I also don't know if you are saying this is wrong.
If you are saying it is wrong then what is the correct answer?

Also since lattices are known to be involved in selective transmission of EM then wouldn't "penetration depth for a particular frequency of light" be a lattice effect?
In either case I was just saying that the blue light just penetrates too deeply to get reflected, leaving yellow as the result.

I responded because you said you "suspect the lattice", but yet your explanation was due to a thin film. The reason and explanation via the example you gave didn't match.

While the lattice certain play a role in optical conductivity (I've explained this ad nauseum on here - you can do a search), it isn't the reason for the penetration depth or skin depth for a metal, or why metals are highly reflective. The conduction electrons play the most significant role in that mechanism, not the lattice.

Zz.

 

[NoTime]

I responded because you said you "suspect the lattice", but yet your explanation was due to a thin film. The reason and explanation via the example you gave didn't match.

While the lattice certain play a role in optical conductivity (I've explained this ad nauseum on here - you can do a search), it isn't the reason for the penetration depth or skin depth for a metal, or why metals are highly reflective. The conduction electrons play the most significant role in that mechanism, not the lattice.

Zz.

Err, I think you read too quickly. I was using the thin film to demonstrate that the penetration depth was greater for blue.

Since all metals have conduction electrons, which are generally treated as a gas. Is it the lattice structure that eventually imposes the ordering on the gas that makes it exhibit different properties? Or is there some other ordering mechanism? Or a combination of different mechanisms?

 

[ZapperZ]

Err, I think you read too quickly. I was using the thin film to demonstrate that the penetration depth was greater for blue.

Since all metals have conduction electrons, which are generally treated as a gas. Is it the lattice structure that eventually imposes the ordering on the gas that makes it exhibit different properties? Or is there some other ordering mechanism? Or a combination of different mechanisms?

But what different properties are you talking about? The reflectivity for metals isn't a strong function of the lattice. Neither does the penetration depth. Almost all metals have roughly similar values for those within the visible range.

Zz.

 

[NoTime]

Looks like I was reading too fast thist time

 

[Creator]

... Due to the large number (density) of free electrons in most metals, the dielectric constant, K is negative (up to a certain frequency called the plasma frequency). A negative K gives rise to an imaginary wave number, ( n ~ sqrt(K) ) and hence 'no' transmission (due to an exponentially decaying field within the metal). The plasma frequency for most metals is somewhere in the far UV range. ...

In short, the color of the metal depends on the frequency dependence of the dielectric constant in the Visible range.

.

Good answer, Gokul. And of course, it is typically stated that there is no tranmission below the plasma frequency. However, why does a larger free electron density turn K negative ? Would you say it the permittivity that is affected; what does neg.K imply and how does that relate to the exp. field decay ?

Creator

 

[broomfieldjay]

Luster is the glow of reflected light and is not necessarily related to color although the discussion here is interesting.

In layman’s terms metals reflect almost all incident light regardless of wavelength in the visible range and are therefore essentially colorless. The relatively loose electrons of the metal elements do absorb or capture radiant energy but quickly returns to their original energy level emitting the same energy they absorbed with little dissipation. So, when light strikes a metal surface all visible wavelengths are reflected with little or no alteration in color. (This is the gist of what was said before.)

 

[broomfieldjay]

Going further, the efficiency of this process depends on the selection rules that apply to the atomic orbitals from which the energy band had formed. If the efficiency of absorption and reemission is approximately equal at all optical energies, then the different colors in white light will be reflected equally well, thus leading to the “silvery” color of polished iron or silver surfaces. However, 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 results in yellow and reddish colors respectively. (Again, this does not differ much from what was said before.)

For example, when light falls onto a piece of iron, the electrons below the Fermi surface, the highest occupied level of electrons in a metal, can also become excited into higher energy levels in the band by absorbing the energy from the light producing electron-hole pairs. The light is so intensely absorbed that it can penetrate to a depth of only a few hundred atoms, typically less than a single wavelength. Since the metal is a conductor of electricity, this absorbed light, which is after all, an electromagnetic wave, will induce alternating electric currents on the metal surface. These currents immediately re-emit the light out of the metal, thus providing strong reflection of a metal surface.

 

[broomfieldjay]

This layman’s reasoning was carefully worded especially including the word ‘efficiency.’
The color of copper and gold are related to the lattice. When metal atoms form a solid crystal, the individual orbitals are thought to become bands because of the Pauli exclusion principle. The valence electrons occupy the conduction band which is often formed from the s-orbitals. By definition, a metal has a partially filled conduction band that can accommodate more electrons. The d-band is formed from the d-orbital.

For copper the valence electrons in the conduction band may absorb a photon of light with energy greater than 4 eV and jump to the next unoccupied band. However, the d-orbital band is just 2 eV below the conduction band which is not full. So this is a more ‘efficient’ process and light with energy above 2 eV (<4eV) is absorbed.

Light can be described by its energy range. Red is about 1.59 to 1.99 eV, orange is about 1.99 eV to 2.08 eV while yellow is 2.08 to 2.15, green 2.15 to 2.52. Blue, violet and UV are higher yet. eV = electron volt

So what happens is that much of the orange light is absorbed, and all yellow, green, blue, violet are absorbed leaving a little orange and red to be reflected back by the process already described.

 

[broomfieldjay]

I apologize for being so long and using several replies, but I’ve had some trouble with my computer.

Now, a formula for reflectivity in the visible region. The reflectivity for metals is almost one:
R = 1 – a small number. (Generally .95)

An approximate formula is:
R = 1 – 2/(plasma frequency * relaxation time)
The plasma frequency is related to constants and properties of the metal.

Plasma frequency = n* e^2/(permittivity of free space * electron mass)
n = concentration of electrons = Avogadro’s number * Z*metal’s density/atomic mass
Z = valence
Relaxation time depends on the metal, its around 10^-15 sec.

Part of the reason this is so difficult is that it’s really a quantum mechanical process and solid state physics, classical electromagnetism, and layman’s wording still get the gist of what’s happening. There is even a formula for the dielectric function that can show mathematically the transition of d-band electrons to the conduction band. If you are familiar with quantum mechanics, you know that probabilities are involved and the word ‘efficiency’ mentioned above was indeed carefully chosen.