Objectives:

• To understand the particulate nature (photon model) of electromagnetic radiation
• To identify a photon as a quantum of energy of electromagnetic radiation
• To able to define the energy of a photon;  $latex E = hf$  and  $E = \frac{hc}{\lambda}$
• To understand and use the electron volt (eV) as a unit of energy

You will be aware that light travels in the form of electromagnetic waves, you may also have read or heard somewhere that light is made up of photons, which are considered particles.

Electromagnetic radiation travels in the form of transverse waves as seen by the fact that it can be reflected, refracted, diffracted or even polarised.

The idea that light could also be considered a particle should be acceptable; after-all particles reflect off surfaces, they can change direction when travelling through different mediums, they can even spread out when travelling through gaps, additionally if you refer back to the polarisation page and the analogy of a string oscillating in a certain plane, the vibrations could cease to exist if there is a slit in the wrong orientation to it that the wave has to pass through – in other word particles could potentially be polarised.

So does light travel in the form of a wave or a particle, or both?

Light in fact does have characteristics of both waves and particles and therefore it is stated that it travels in both forms at the same time – wave-particle duality.

Electromagnetic radiation is a packet of energy and if it travels in the form of a photon, the photon must be pure energy (because light doesn’t have any mass). A photon is therefore considered to be a ‘packet’ of energy, (but a very small amount of energy) known as a quantum of energy.

Energy of a Photon

Throughout your GCSE’s and A level so far you have come across a whole range of equations for energy;

• $E_{g.p.e} = mgh$
• $E_{k} = \frac{1}{2}mv^{2}$
• $E_{e} = \frac{1}{2}kx^{2}$
• $\Delta E_{thermal} = mc \Delta \Theta$
• $E_{latent} = mL$

You may notice that each of these equations involve physical quantities (not forms of radiation) – even $E = mc^{2}$ involves the mass of a particle!

Light is defined by its frequency, since this is always constant no matter what medium it is in, so it makes sense that the energy of a photon is dependent on its frequency.
Consider the electromagnetic spectrum;

As shown in the EM spectrum, the frequency of radiation increases as the wavelength decreases (the closer to the gamma region the higher the frequency). Which of type of EM radiation has the most energy and which has the lowest? The answers are gamma rays and radio waves respectively as you will have known from your GCSE studies. So from this we can see that as frequency rises, the energy rises too.

The equation for the energy of a photon is;

$E = hf$

where
$E$  is the energy of the photon
$h$ is a constant known as Planck’s constant
$f$ is the frequency of the wave or photon

By using $v = f \lambda$ and rearranging for $f$:  $f = \frac{v}{\lambda}$ and subbing this into the the equation for the energy of a photon gives;

$E = h \frac{v}{\lambda}$

For waves that travel through a vacuum and hence at the speed of light (or in air since the speed is minimally different), this equation can be written as:

$E = \frac{hc}{\lambda}$

If the wavelength of EM radiation is known then $E = \frac{hc}{\lambda}$ is used, but if the frequency is known then you would use $E = hf$ .

Planck’s constant, $h$ is equal to $6.63 \times 10^{-34}$ . The units of this constant can be determine from one of the equations with it in, e.g. rearranging $E = hf$;

$h = \frac{E}{f}$

$h_{unit} = \frac{J}{Hz} = \frac{J}{s^{-1}} = Js$

Example question: How much energy does a photon of red light with a wavelength $685 nm$ have?

$E = ?$
$h = 6.63 \times 10^{-34} Js$  (given in the formula booklet)
$\lambda = 685 \times 10^{-9} m$
$\lambda = 3.00 \times 10^{8} ms^{-1}$

$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}3.00 \times 10^{8}}{685 \times 10^{-9}}$

$E = 2.90 \times 10^{-19} J$

If the same calculation if done for blue light with a wavelength of 475 nm, you can determine that the energy a photon would have is $E = 4.18 \times 10^{-19} J$ . This is larger than the energy a red photon has which is what you should expect.

These values for energy are tiny and so a more appropriate unit, known as the electronvolt is used – you should have come across it before, but below is a recap.

The Electron-Volt

If you read back through the notes on electromotive force and potential difference you will see that an equation was derived which contained $eV$ . This is a really useful term  in Physics and is labelled the electronvolt, it is a unit of energy that is used for particles that have a very small number of joules associated with it. For example, an electron travelling at 90% speed of light will have how much kinetic energy?

$E_{k} = \frac{1}{2} m_{e} v^{2}$

where
$m_{e} = 9.11 \times 10^{-31} kg$
$v = 0.90 \times 3.00 \times 10^{8} ms^{-1} = 2.70 \times 10^{8} ms^{1}$

$E_{k} = \frac{1}{2} 9.11 \times 10^{-31} (2.70 \times 10^{8})^{2}$

$E_{k} = 3.32 \times 10^{-14} J$

This is such a tiny amount of energy (and frequently smaller energies again are calculated for particles) that it is not always useful to refer to it in joules.

An electronvolt, $eV$, is an alternate unit for energy which is derived directly from the equation for the work done on charged particles, $W = QV$;

If say an electron is accelerated through a potential difference of 1 Volt;

$W = QV$

where
$Q = e$
$V = 1 V$

$W = 1 eV$

Hence the term electronvolt. Since work done is measured in joules, we can use this equation to determine how many joules are in 1 eV;

$e = 1.6 \times 10^{-19} C$
$V = 1 V$

$W = 1 \times 1.6 \times 10^{-19} CV = 1.6 \times 10^{-19} J$