Topic 3 Particle Model of Light
|
Key Words 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.
·
Reflection
·
Refraction
·
Diffraction
· Polarisation
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.

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.

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
was observed when the electroscope was positively charged.
The results were:
|
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 a particle behaviour to
light.
| 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 – 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
l
| 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
charge on an electron is 1.6 × 10-19 C, so 1 eV = 1.6 × 10-19 J.
| 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.
| A photon has an energy of 10.3 eV. What is its wavelength? Where on the electromagnetic spectrum would this be? |
| Now try Topic Quiz 3 | Home | Physics AS | Physics Module 1 |