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In 1905, Albert Einstein published a paper that described his photon theory of light. This explained experimental evidence, such as the photoelectric effect, that did not fit with the classical wave model of light.

Einstein proposed that, although light exhibited wave-like properties, it travelled in particles called photons.

He explained that photons contain discrete ‘energy packets’ called quanta, and that the energy of an individual quantum depends on the frequency of the light.

Photon energy


Thermal radiation

All objects with a temperature above absolute zero emit thermal radiation.

Why do objects emit this radiation?

All charged particles emit radiation when they accelerate.

If an object has a temperature above absolute zero, then its electrons and protons will vibrate and emit radiation.

Absolute zero is the temperature (-273°C, 0K) at which particles theoretically lose all their energy and stop vibrating.


Incandescence


The incandescent light bulb

How could you use incandescence to explain how a conventional light bulb works?

1. A current flows through the filament.

2. The filament heats up.

3. The thermal emission of the filament moves into the visible spectrum.

Why is the incandescent light bulb not an efficient device for producing light?

The filament’s emission spectrum remains mostly in the infrared, even when it is at its hottest. Most of the bulb’s energy input is therefore wasted as heat.


Emission spectra


Explaining the origin of line spectra


Dropping energy levels

Each possible drop between atomic energy levels in an atom corresponds to the emission of one specific frequency of photon.

This results in one line in an element’s emission spectrum. Three lines in hydrogen’s spectrum are shown here, along with the energy jumps that they correspond to.

hydrogen

n = 1

n = 2n = 3n = 4

ground


The fluorescent lamp

The incandescent light bulb is an inefficient device for producing light (typically 1–2%). The fluorescent lamp is an alternative, with an efficiency of around 10%.

Fluorescent lamps use electricity to cause excitation of mercury vapour. When the mercury atoms relax, they emit ultraviolet photons.

Ultraviolet light is not visible, but can be converted into visible light using a phosphor. This coats the inside of the bulb and fluoresces when bombarded with the ultraviolet light from the mercury vapour.


Understanding thermal emission



Quantised absorption

Just as an electron can drop between energy levels in an atom, releasing a single photon, it can also jump up one or more energy levels if it absorbs a photon of the right energy.

Only a single photon of the relevant energy can cause this. It is not possible for an electron to ‘store up’ energy from smaller quanta until it has enough to make the jump.

One result of this is that shining a continuous spectrum of light at a transparent material leads to a few discrete frequencies being absorbed, while the rest are transmitted. This forms an absorption spectrum.

H emission spectrum

H absorption spectrum


Analysing light from a source


Energy of photons in light


The electronvolt

A joule (J) is a large unit of energy when dealing with tiny atoms. Scientists often use an alternative unit for small amounts of energy, called an electronvolt (eV).

How many electronvolts to one joule?

1 eV = 1.6 × 10-19 J

1 J = 1/(1.6 × 10-19) eV = 6.25 × 1018 eV

Use these two conversion rates to change between the two. Be careful to use joules in calculations with other SI units.

An electronvolt is equal to the amount of energy transferred to a single electron if it is accelerated through a potential difference of 1 V:


Wave speed, wavelength and frequency

The speed, wavelength and frequency of a wave are related by the following equation:

wave speed = frequency × wavelength

c = fλ

What is the photon energy, in electronvolts, of red light of wavelength 685 nm?

E = hf = hc/λ = 6.63 × 10-34 × 3 × 108 / (685 × 10-9)

= 2.9 × 10-19 J

= 2.9 × 10-19 × 6.25 × 1018 eV

= 1.8 eV


Wave and photon calculations


Wavelength of emitted radiation

Given a set of values for atomic energy levels in a particular element, it is possible to calculate the wavelengths of radiation it can emit or absorb.

hf = E1 – E2

hc/λ = E1 – E2

The difference between two energy levels gives the energy of the photon corresponding to that jump, and this can be used to find frequency and wavelength:

-0.85 eV

-1.5 eV

-3.4 eV

-13.6 eV

1 2 3 4 5 6

What wavelengths of light are emitted in transitions 1–6?



Photoelectricity was first discovered in 1887 by Heinrich Hertz during investigations into radio waves using a ‘spark gap’.

Radio waves are produced when a high voltage is supplied across two electrodes causing a spark in the gap.

Hertz found that if ultraviolet light was shone on the electrodes, the sparks were much stronger and thicker.

The discovery of photoelectricity


In the late 19th century, scientists used apparatus like this photocell to analyse and measure photoelectricity.

When light is shone on the cathode of the photocell, electrons are released. They are attracted to the anode, causing a current to flow.

Measuring photoelectricity

anode cathode

vacuum

photocell


The gold leaf electroscope


Einstein’s photoelectric equation


The photoelectric effect equation



Wave–particle duality


Wave properties of particles

In 1924, Lois De Broglie came up with a radical new way of looking at the relationship between waves and particles. He suggested that all particles could behave as waves.

De Broglie deduced that a particle had a wavelength, and it was dependent on only one thing – the momentum of that particle:

λ = h / p

Three years later, this hypothesis was confirmed for electrons with the first observations of electron diffraction.


De Broglie wavelength calculations


Waves and particles



Glossary


What’s the keyword?


Multiple-choice quiz

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