Wave-Particle Duality, Photoelectric Effect, Photon, Quanta, Threshold Frequency, eV
Light as a Particle?
Historically there had been a lot of controversy about the wave nature of light, as proposed by the Dutch physicist Hans Huygens, against the corpuscular model as proposed by the headstrong Isaac Newton. The concept of wave-particle duality was the start of modern physics in the middle to late Nineteenth Century.
We know that light shows wave properties such as:
However it can also be shown to have particulate properties as well. Consider this model:
If we spray just a short burst, we get just a few spots on the screen:
The longer we spray, the more spots appear until the whole area is covered in paint:
When using a spray can, we don’t notice any diffraction effects as the particles pass through the stencil. Hardly surprising as the paint droplets are particles, not waves.
Now, if we expose a piece of photographic paper to a short burst of light we will see:
The intensity of the image on a photographic plate increases the longer the paper is exposed for. That intensity is determined by the number of silver grains deposited. We see that the pattern of silver grains deposited is random. It seems that the light that deposited the grains was actually made of particles.
The debate raged on until the discovery in the late nineteenth century with the discovery of the photoelectric effect.
The Photoelectric Effect.
The concept of wave-particle duality was the start of modern physics in the middle to late Nineteenth Century. We can show the photoelectric effect with apparatus like this:
- We charge the electroscope with a negative charge.
- We expose the reactive metal to light of a long wavelength, e.g. red.
- We observe that there is no effect, however bright the light.
- We then expose the metal to short wavelength light, e.g. UV.
- This time we see that the gold leaf drops down, showing that the electroscope is losing charge.
- It does not matter how bright or dim the UV light is.
- No effect was observed when the electroscope was positively charged.
The results were:
|Metal||X-rays||Ultra-Violet||Blue Light||Red Light|
This led to the conclusion that:
- Electrons were being knocked off. Reactive metals have outer shell electrons that can be removed easily.
- Red light would not show this effect however bright it was. So the amplitude of the light wave was not important. Red light only worked for caesium, which is a very reactive metal.
- There was a threshold frequency at which this phenomenon started to occur. Light waves with a frequency higher than this (shorter wavelength) always showed the effect, whatever the brightness; light waves with a lower frequency never showed it.
- The more reactive the metal, the lower was the threshold frequency.
- This indicated a particle behaviour to light.
Question 1 . Why do these results suggest that light is not a wave?
These findings led to the notion of light being tiny little packets of wave energy called photons.
Further work by Max Planck in 1900 produced the Photon Model of Electromagnetic Radiation. We can sum this up in the following points:
- Light and other electromagnetic radiation is emitted in bursts of energy. We say that it is quantised.
- The packets of energy, photons, travel in straight lines.
- When an atom emits a photon its energy changes by an amount equal to the photon energy.
- The energy changes are discrete amounts or quanta.
- The frequency of the light and the energy are related by a simple equation:
E = hf
[E – energy in J; h – Planck’s Constant; f – frequency of the radiation in Hz]
The constant h is Planck’s Constant with the value 6.6 ´ 10–34 Js (joule seconds, NOT joules per second).
We can combine the equation above with the wave equation:
E = hf and c = fl
Ž E = hc
Question 2 What is the photon energy of light wavelength 350 nm?
The joule is the SI unit for energy. However atomic physicists find the joule far too big and clumsy. (You would not measure the width of your desk in kilometres.) So they use a unit called the electron volt (eV). The electron volt is the amount of energy used when a charge of electronic charge passes through a potential difference of 1 volt.
The charge on an electron is 1.6 × 10-19 C, so 1 eV = 1.6 × 10-19 J.
Question 3 Convert your answer to Question 2 to electron volts.
Electron volts are almost always used in atomic and nuclear physics, but before using equations like E = hf, the energies MUST be converted to joules. This is a very common bear trap.
Question 4. A photon has an energy of 10.3 eV. What is its wavelength? Where on the electromagnetic spectrum would this be? ANSWER