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Photon energy
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.
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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.
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Incandescence
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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.
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Emission spectra
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Explaining the origin of line spectra
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Dropping energy levels
Each possible drop between atomic energy levels in an
atom corresponds to the emission of one specific frequency
of photon.
n=4
n=3
n=2
n=1
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.
ground
hydrogen
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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.
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Understanding thermal emission
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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.
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H emission spectrum
H absorption spectrum
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Analysing light from a source
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Energy of photons in light
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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).
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:
1 eV = 1.6 × 10-19 J
How many electronvolts to one joule?
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.
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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
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Wave and photon calculations
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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.
1 2 3 4 5 6
The difference between two
-0.85 eV
energy levels gives the
-1.5 eV
energy of the photon
corresponding to that jump,
and this can be used to find
-3.4 eV
frequency and wavelength:
hf = E1 – E2
hc/λ = E1 – E2
-13.6 eV
What wavelengths of light are emitted in transitions 1–6?
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The discovery of photoelectricity
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.
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Measuring photoelectricity
In the late 19th century, scientists
used apparatus like this
photocell to analyse and
measure photoelectricity.
photocell
When light is shone on the cathode
of the photocell, electrons are
released. They are attracted to the
anode, causing a current to flow.
anode
cathode
vacuum
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The gold leaf electroscope
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Einstein’s photoelectric equation
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The photoelectric effect equation
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Wave–particle duality
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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.
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De Broglie wavelength calculations
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Waves and particles
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Glossary
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What’s the keyword?
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Multiple-choice quiz
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