Quantum Physics Tutorial 1 - The Evidence for Light as a Photon
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 do the experiment like this:
1. We charge the electroscope with a negative charge.
2. We expose the reactive metal to light of a long wavelength, e.g. red.
3. We observe that there is no effect, however bright the light.
4. We then expose the metal to short wavelength light, e.g. UV.
5. This time we see that the gold leaf drops down, showing that the electroscope is losing charge.
6. It does not matter how bright or dim the UV light is.
7. No effect is observed when the electroscope is positively charged.
If different reactive metals are used, the following results are observed:
|
Metal |
X-rays |
Ultra-Violet |
Blue Light |
Red Light |
Magnesium |
P |
O |
O |
O |
|
Zinc |
P |
P |
O |
O |
|
Sodium |
P |
P |
P |
O |
|
Caesium |
P |
P |
P |
P |
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 particle behaviour in light.
These findings led to the notion of light being tiny little packets of wave energy called photons.
If light were a wave we would see electrons being knocked off by very bright red light. Dim violet light would not remove a photoelectron.
| Question 1 | Why do these results suggest that light is not a wave? |
|
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 (a Latin word meaning “how much”.)
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
This gives us:
Wavelengths of light is measured in nanometres (nm) where 1 × 10-9 m. For example, 635 nm = 635 × 10-9.
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.
| Question 2 | What is the photon energy of light wavelength 350 nm? |
|
The charge on an electron is 1.6 × 10-19 C, so 1 eV = 1.6 × 10-19 J.
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.
We do this by multiplying by 1.6 × 10-19 J.
To convert joules to electron volts we divide by 1.6 × 10-19 J.
| Question 3 | Convert the answer to Question 2 to electron volts |
|
| Question 4 | A photon has energy of 10.3 eV. What is its wavelength? Where on the electromagnetic spectrum would this be? |
|
