AQA AS Physics – Module 1 revision notes
AS Module 1: Particles, Radiation, and Quantum Phenomena
General Notes:
Main headings in blue, number refers to AQA syllabus reference.
Red text is not implicitly required for this syllabus, but included for the sake of
completeness.
Tasks are in italics.
Refraction at plane surfaces (10.2.1)
When light (or any electromagnetic radiation) travels from one medium to another, it’s
velocity changes. This also results in a change in the direction of travel (unless the light
hits the boundary between the two surfaces at exactly 90o).
This effect is called refraction. The amount of refraction depends upon how much the
velocity of the wave is altered.
The refractive index for light passing from medium 1, to medium 2 is expressed as 1n2.
Where,
n2 = v1/v2
1
Also, the change in direction can be found using Snell’s law of refraction;
n2 = sin 1 / sin 2
1
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AQA AS Physics – Module 1 revision notes
Practical Exercise
Find the refractive index of perspex to 3 sig. fig. using a ray box and a rectangular
perspex block.
Record between 8 and 12 data sets of 1 and 2 , and use these to find the refractive
index of perspex by plotting a suitable graph.
Hint, use the equation of a straight line y=m x + c method.
Then use the refractive index to calculate the velocity of light in Perspex if the velocity of
light in air is 3.00 x 108m/s.
Relationship between refractive Indices of different Media
Total Internal Reflection
When light travels from a more dense to a less dense medium (e.g. from glass to air) it’s
path is bent away from the normal (fig a). If you increase the angle of incidence, angle i,
then the angle r also increases. However, as angle r is always greater than angle i, it will
become equal to 90 before angle i does (fig b). This value for the angle of incidence is
called the critical angle, angle c. At this point, and for all angles greater than the critical
angle, since the light can no longer be reflected, it is all reflected, hence Total Internal
Reflection.
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AQA AS Physics – Module 1 revision notes
The relationship between the critical angle and the refractive index is:
NB This gives the refractive index for light going from the less dense to the more dense
medium.
[Proof:
1n2 = sin 1 / sin 2
1n2 = sin 90 / sin c
but sin 90 = 1, therefore;
n2 = 1 / sin c
1
]
Fibre Optics
AQA syllabus requires simple treatment of fibre optics including function of cladding
with lower refractive index around central core limited to step index only; candidates
should be familiar with modern applications of fibre optics, e.g. endoscopy,
communications, etc.
The optical fibre (in its simplest form) consists of a core of glass of one refractive index,
and a cladding of a slightly lower refractive index. A protective plastic coating protects
the fibre from damage.
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AQA AS Physics – Module 1 revision notes
A light source coupled to the optical fibre emits a cone of light into the core of the fibre.
Only rays such as the blue and the magenta ones which meet the requirement for total
internal reflection to occur will be guided in the fibre along a zig-zag path.
However, other rays such as the red one will be refracted in the cladding and will not be
propagated along the fibre.
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AQA AS Physics – Module 1 revision notes
The Photoelectric Effect (10.2.2)
Description
The photoelectric effect is when light is incident on a metal surface, electrons are
liberated from that metal surface. The light has to be of a sufficiently high frequency (the
threshold frequency), for this effect to occur. The threshold frequency is different for
different metals, but is usually in the UV region of the E-M spectrum.
The photoelectic effect is evidence for the particle nature of light.
Observation 1:
Light intensity does not increase the KE of the liberated electrons, but it does increase the
number that are emitted.
Conclusion:
Light is made up of particles (photons), and increasing the intensity, increases the number
of photons incident on the surface. Each photon can release one electron.
Observation 2:
Below a certain frequency, no electrons are emitted from the surface.
Conclusion:
The energy of a photon is proportional to it’s frequency, and below the threshold
frequency, the energy of the photon is not sufficient for the electron to overcome the
forces binding it to the surface of the metal.
Energy in the Photoelectric effect
When a photon is incident on the metal surface and releases an electron, the photon
disappears and it’s energy is used in two ways. Some energy is used to overcome the
forces holding the electron to the surface of the metal (the work function), and the
remainder is the maximum KE that the electron may acquire as it leaves the surface.
hf =  +Ek
hf = energy of photon of frequency f
h = Planck’s constant
 = work function of metal
Ek = maximum kinetic energy of emitted photon
The work function is the minimum energy required to eject a (photo-)electron from the
metal surface.
Common examination errors:
Saying that a photon is emitted instead of a (photo-)electron.
Confusing this effect with ionisation, the metal atom is not ionised.
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AQA AS Physics – Module 1 revision notes
Energy Levels and Spectra (10.2.3)
The electronvolt
Because the energies involved are very small, the electronvolt (eV), a much smaller unit
than the joule, is often used.
1 eV = 1.6 x 10 -19 J
Remember to convert all energies from eV to Joules before doing any calculations.
Ionisation and excitation
If an electron gains energy it may rise to a higher energy level within the atom. To do this
it can absorb a photon, as long as the energy of the photon is exactly the energy
difference between two energy levels, e.g. hf=E2 - E1. In this case the photon disappears
although a new photon will be emitted when the electron de-excites a short time later.
An alternative mechanism for excitation is that the electron can gain energy from a
collision with another particle. In this case the electron gains some but not all of the KE
of the colliding particle, which will continue on it’s way with the remainder. Again the
amount of energy transferred will be equivalent to the difference between two energy
levels.
Ionisation occurs when an electron gains enough energy to escape from an atom. If the
energy gained by the escaping electron is greater than the energy it needed to escape (the
ionization energy) then the difference will be carried off as KE.
Photon emission
After excitation, the electron very quickly falls back down to a lower energy level. In
doing so, a photon is released with an energy equivalent to the difference between two
energy levels, e.g. hf=E4 - E1. It may drop down directly to the lowest ground state, or, if
it is at level 3 or higher, it may drop down in two or more stages, in which case it will
release a photon at each stage, e.g. E4 -> E1 or E4 -> E3 then E3 -> E1.
Line Spectra
When excited electrons de-excite (drop down to a lower energy level) and release a
photon, these photons can be observed in the form of a line spectrum. Each of the bright
lines in the spectrum represent one specific energy level transition. As different elements
will have different electron configurations, each element will have a unique line
spectrum. The simplest atom is hydrogen with only one electron so not surprisingly it’s
line spectrum is the simplest and is most commonly used in examination questions.
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AQA AS Physics – Module 1 revision notes
Wave-particle duality (10.2.4)
State what is meant by the wave-particle duality of electromagnetic radiation:
The properties of electromagnetic radiation can be explained by considering it behave as
either a wave or as a particle.
Evidence for Electrons behaving as waves:
Electrons can be diffracted.
Evidence for Electro-magnetic radiation behaving as particles:
The photo-electric effect.
Evidence for Electro-magnetic radiation behaving as waves:
Diffraction, interference and refraction.
Calculation of the de Broglie wavelength of electrons:
Wavelength = h/mv
h = Planck’s constant
(Note: mv = momentum)
Sometimes you will be asked to calculate what frequency electromagnetic radiation of
this wavelength would posses, in which case use wave speed = frequency x wavelength.
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