1. Optical Properties

2. Write 6 bullet points to explain this diagram
Electron Transitions
Write 6 bullet points to explain this diagram

3. Electron Transitions
An electron in an isolated atom can only occupy discrete energy levels.
Electron transitions can take place if light energy is absorbed from a photon.
The energy difference between levels, ΔE must exactly equal the energy of the photon.
ΔE = hf
When the electron falls back , energy is released.
Some materials emit energy in the frequency range of visible light = luminescence

4. Optical communication
λ = 0.4 μm High energy
Optical communication
λ=1.55 μm
Low Energy

6. Transparency and opacity
Aim
To use band theory to describe why insulators absorb photons with a given range of energies but fail to absorb photons with energies below this level
Explain why this makes insulators transparent to visible light
Use E=hf to determine energy condition for transparency of insulator

7. Why are Insulators Transparent?
Insulators have an empty conduction band and forbidden gap > 2 eV.
If incoming photon energy > 2eV electrons can jump up to the conduction band.
Photons of energy < 2eV will pass through without being absorbed.
For a material to be transparent, all visible light must pass through.
Using E = hf, we can determine what band gap is necessary for visible light.

8. Transparent to the visible spectrum?
Longest λ (lowest energy) is near infrared = 1.55 μm
Minimum photon energy, E = hc / λ
= 0.8 eV (1 eV = 1.60 x J)
An insulator with a band gap of 0.8 eV will absorb near infra-red light of λ = 1.55 μm and all wavelengths shorter (higher energy) than this.
It will not absorb photons with energies smaller than 0.8 eV.
To transmit visible λ (0.4 μm – 0.7 μm), the band gap must be larger.
Shortest λ (highest energy) is violet = 0.4 μm
Maximum photon energy, E = hc / λ
= 3.1 eV
If the band gap is between 0.80 eV and 3.1 eV part of the visible light spectrum will be absorbed, making the material appear coloured.

9. What colour is cadmium sulphide?
Eg = 2.40 eV
What frequencies of light will it transmit and how does this explain its colour?
Energy of photon, E = hf
f = 5.8 x 1014 Hz
Cadmium sulphide will transmit all f < 5.8 x 1014 Hz and absorb f > 5.8 x 1014 Hz.
Associated λ = 0.52 μm
Cadmium sulphide will absorb all λ shorter than this, cutting out violet, blue and most green light. The result will be a yellow-orange colour.

10. Aim
Use band theory to explain why metals are opaque to infra-red and visible light

11. Why are Metals Opaque and Shiny?
In a metal, only the lower levels in the conduction band are filled.
The finely spaced levels mean that electrons can accept photons with low energies, i.e. low frequencies.
As a result, metals are opaque to radio waves, microwaves, infrared and visible light.
At the high end of the spectrum they are transparent. Most metals become transparent in the middle of UV range at a specific frequency.
Most visible light is absorbed in the outer surface of the metal. Light is absorbed and re-emitted as photons of the same wavelength, giving them a silvery lustre.
Many metals show selective absorption of certain wavelengths therefore, selective reflection of the same wavelengths. Thus copper has a red-orange colour and gold appears yellow.

12. Sentence Match Up
Metals are generally opaque and shiny. We know that in a metal there are electrons permanently in the conduction band, but generally these occupy the lower levels of the conduction band. As the energy levels within the band are very close together, photons of very low energies (long wavelengths) can be absorbed. The electrons can also be promoted up to much higher unoccupied levels by single photons up to the UV frequencies. This makes metals opaque to all frequencies up to ultra-violet, but they cannot absorb higher energy photons so they are transparent to X-rays and gamma rays.
Once a photon has been absorbed, the energy can either be dissipated as internal energy of the atoms, or it can be re-emitted in a series of steps, or as one big step. In metals, most of the light absorbed is re-emitted across all wavelengths, meaning metals have a reflective, silvery appearance. Some metals also show selective absorption of certain wavelengths, and reflect this wavelength more strongly too, leading to a shiny coloured appearance (e.g. copper looks orangey).