# Why do metals have high optical reflectivity?

What's the reason behind metals' high optical reflectivity?

You are asking why metals have highly optical reflectivity.

Now this has too a little bit to do with why metals are often silver. The same goes why we use aluminum in our mirrors.

When a photon interacts with an atom in the metal, three things can happen:

1. elastic scattering, the photon keeps its energy level and phase and changes angle

2. inelastic scattering, the photon gives part of its energy to the atoms and molecules of the metal, and changes angle

3. absorption, the photon gives all its energy to the atom in the metal, the absorbing electron moves to a higher energy level as per QM

Now in the case of metal, the ratio of the three is different:

1. elastic scattering, this is what has the highest ratio, most of the photons get elastically scattered. This is why mirrors create a mirror image, keeping the energy level and phase of photons. This creates specular reflection.

2. inelastic scattering, now in the case of metals, this has a lower ratio, lesser photons get inelasstically scattered, this heats up the metal, gives the energy of the photons to the vibrational kinetic energy of the molecules, thermal energy

3. absorption, this gives metals usually their silver color, because as per QM, in metals d electrons absorb visible light, and jump to d orbitals. Now in metals, like silver, the bang gap between s and d orbitals is too big, and all visible wavelength photons' energy level is too little to be absorbed. So most of the visible wavelength photons cannot be absorbed, they get reflected. Since all visible wavelength photons get reflected, silver has no color of its own, it looks shiny, silver color.

So metals are highly reflective, because:

1. most of the photons get elastically scattered, that is reflection

2. lesser number of photons get inelastically scattered, these heat up the metal

3. very little number of photons get absorbed in the visible range, most of these get reflected and that gives metals a shiny color

• Why most of the photons get elastically scattered in metals? Why the number of photons get inelastically scattered and absorbed are very low? – Ahmed Estiak Aug 27 '18 at 5:49
• Well, you are talking about the visible wavelengths, and why metal is shiny. So the visible wavelengths need a certain band gap (energy level difference) between the s and d electron orbitals. Now in metals like silver, this bang gap is very big. So, only photons with very high energy can be absorbed. So none of the visible wavelength photons have enough energy to be absorbed. Since they cannot be absorbed, they will be reflected (elastically scattered). – Árpád Szendrei Aug 27 '18 at 5:58
• Now we are talking about visible wavelengths. With inelastic scattering, usually this is deeper into the metal, and visible wavelengths cannot do that, so very little visible wavelengths will be inelastically scattered. They will be elastically scattered from the surface. – Árpád Szendrei Aug 27 '18 at 6:00
• Now non-visible wavelengths will be inelastically scattered at a higher ratio, and those will heat up the metal. – Árpád Szendrei Aug 27 '18 at 6:01

I assume you’re referring to the visible range of the spectrum, and so the answer to this question comes down to essentially three things:

1. There are a lot of free electrons in a metal.
2. These electrons scatter off of themselves, defects, and lattice vibrations a lot (but not too much).
3. There is insignificant absorption via interband transitions in the visible range.

Facts (1) and (2) lead to a large, Drude-like conductivity of the metal, which in the visible range is primarily imaginary (meaning the oscillating electrical current in the metal excited by the light is essentially $\pi/2$ out of phase with the light). Thus, the re-emitted light is completely out of phase, and so the light’s electric field basically goes to zero at the metal surface. This condition only applies if there is very little auxiliary absorption (i.e. if fact (3) is true, which it is for silver, say, but not for gold in the blue/green part of the spectrum, which is why gold has its color).

Since the light field goes (close) to zero at the surface, it has very little penetration into metal at all, meaning the vast majority of the light power is reflected. This can be understood as the free electrons in the metal moving at the surface to effectively screen out the light field from getting into the bulk, and, in doing so, re-radiating the power outwards again.

Given that light is an oscillating, travelling EM field, upon reaching the surface of a metal, it will cause the electrons to vibrate. Due to the nature of metallic bonding, there are plenty of delocalized, mobile electrons to respond to the incident light. And since a moving electric charge creates a moving EM field, the result would be an emitted EM wave of the same wavelength(s) as the incident wave. That would be the reflected wave. In other words, the incident photon is absorbed and a new photon is emitted. This is a very simplistic answer to the question, but hey, I taught high school physics.

• Good answer. Clear and simple! – Gilbert May 14 at 4:40

At low frequency for substances of high conductivity the reflection coefficient is close to unity and so essentially all the energy is reflected.This is why metals i.e.good conductors of electricity are opaque to light.The little energy that flows into the metal is rapidly dissipated by the heat loss associated with the induced current.

• Rajendra Pd, I don't think this answered the question. There are interactions between incoming light and the free electrons in the conductor. Can you answer within the context of those interactions? – David White Aug 26 '18 at 17:54
• Optical frequencies, of the order of a petahertz, are not considered low. – my2cts Aug 26 '18 at 18:22
• The free electrons in metals oscillate due to interaction with the incident light wave and give rise to a radiation field.The forward wave in this field interferes with the original wave destructively and gives rise to a small transmission.The reflected wave combines with the incident wave and forms a standing wave with a node on the surface. – Rajendra Pd Aug 26 '18 at 18:49
• @Rajendra You should add that extra info to your answer, don't bury it down here in the comments. – PM 2Ring May 13 at 4:17