Episode 506: Particles as waves
This episode introduces an important phenomenon: wave  particle duality.
In studying the photoelectric effect, students have learned that light, which we think of as waves, can sometimes behave as particles. Here they learn that electrons, which we think of as particles, can sometimes behave as waves.
Summary
Demonstration: Diffraction of electrons. (30 minutes)
Discussion: de Broglie equation. (15 minutes)
Worked examples: Using the equation. (15 minutes)
Student questions: Using the equation. (30 minutes)
Discussion: Summing up. (10 minutes)
Student questions: Practice calculations. (30 minutes)
Demonstration:
Diffraction of electrons
The diffraction of electrons was first shown by Davisson and Germer in the USA and G P Thomson in the UK, in 1927 and it can now be observed easily in schools with the correct apparatus.
S how electron diffraction. It will help if students have previously seen an electronbeam tube in use (e.g. the fine beam tube, or e/m tube).
TAP 5061: Diffraction of electrons
Before giving an explanation, ask them to contemplate what they are seeing. It is not obvious that this is diffraction/interference, since students may not have seen diffraction through a polycrystalline material. (Note that the rigorous theory of crystal diffraction is not trivial  waves scattered off successive planes of atoms in the graphite give constructive interference if the path difference is a multiple of a wavelength, according to the Bragg equation. The scattered waves then appear to form a wave that appears to “reflect” off the planes of atoms, with the angle of incidence being equal to the angle of reflection. In the following we adopt a simplification to a 2D case  see TAP 5061)
Qualitatively it can help to show a laser beam diffracted by two ‘crossed’ diffraction gratings. Rotate the grating, and the pattern rotates. If you could rotate it fast enough, so that all orientations are present, you would see the array of spots trace out rings.
TAP5062: Diffraction of light
From their knowledge of diffraction, what can they say about the wavelength of the electrons? (It must be comparable to the separation of the carbon atoms in the graphite.) How does wavelength change as the accelerating voltage is increased? (The rings get bigger; wavelength must be getting smaller as the electrons move faster.)
Discussion:
de Broglie equation
In 1923 Louis de Broglie proposed that a particle of momentum p would have a wavelength given by the equation:
wavelength of particle = h/p
where h is the Planck constant,
or = h/mv for a particle of momentum mv.
The formula allows us to calculate the wavelength associated with a moving particle.
Worked examples:
Using the equation
1. Find the wavelength of an electron of mass 9.00 10^{31} kg moving at 3.00 10^{7 }m s^{1}.
= h/p = [6.63 10^{34}] / [9.00 10^{31} 3.00 10^{7}] = 6.63 10^{34 }/ 2.7010^{23}
= 2.4610^{11} m = 0.025 nm
This is comparable to atomic spacing, and explains why electrons can be diffracted by graphite.
2. Find the wavelength of a cricket ball of mass 0.15 kg moving at 30 m s^{1}.
= h/p = [6.63 10^{34}] / [0.15 30] = 1.4710^{34} m = 1.5 10^{34} J s (to 2 s.f.)
This is a very small number, and explains why a cricket ball is not diffracted as it passes near to the stumps.
3. It is also desirable to be able to calculate the wavelength associated with an electron when the accelerating voltage is known. There are 3 steps in the calculation.
Calculate the wavelength of an electron accelerated through a potential difference of 10 kV.
Step 1: Kinetic energy E_{k} = eV = 1.6 10^{19} 10000 = 1.6 10^{15} J
Step 2: E_{K} = ½ mv^{2} = ½m (mv)^{ 2} = p^{2 }/ 2m, so momentum
p = √2mE_{k} = √2 9.1 10^{31} 1.6 10^{15} = 5.4 10^{23} kg m s^{1}
Step 3: Wavelength = h / p = 6.63 10^{34} / 5.4 10^{23} = 1.2 10^{11} m = 0.012 nm.
Student questions:
Using the equations
A useful set of questions can be found at Resourceful Physics on the web
This can also be accessed from
http://resourcefulphysics.org/download/361/waves_and_particles.doc
and is available to subscribers.
Discussion:
Summing up
You may come across a number of ways of trying to resolve the waveparticle dilemma. For example, some authors talk of ‘wavicles’. This is not very helpful.
Summarise by saying that particles and waves are phenomena that we observe in our macroscopic world. We cannot assume that they are appropriate at other scales.
Sometimes light behaves as waves (diffraction, interference effects), sometimes as particles (absorption and emission by atoms, photoelectric effect).
Sometimes electrons (and other matter) behave as particles (beta radiation etc), and sometimes as waves (electron diffraction).
It’s a matter of learning which description gives the right answer in a given situation.
The two situations are mutually exclusive. The wave model is use for ‘radiation’ (i.e. anything transporting energy and momentum, e.g. a beam of light, a beam of electrons) getting from emission to absorption. The particle (or quantum) model is used to describe the actual processes of emission or absorption.
Student question:
Interpreting electron diffraction patterns
TAP 5063: Interpreting electron diffraction patterns
TAP 5064: Electron diffraction question
TAP 506 1: Diffraction of electrons
Start by setting up the electron diffraction demonstration tube according to the manufacturer’s instructions. Check the connections before switching on as it is easy to burn out the graphite grid unless manufacturer’s instructions are correctly followed.

Wire carefully, no bare conductor above 40 V
School EHT supplied are limited to a maximum output current of less than 5 mA so they are safe, but can still make the user jump. The lead connecting the supply to the anode should have a female connector so that no metal is exposed.

You will need two supply voltages:

a lowvoltage supply to provide the current to heat the cathode;

a highvoltage supply to provide the potential difference between the cathode and the anode, to accelerate the electrons.
These are both usually supplied by the EHT unit.
Qualitative experiments
Check that you can identify the following:

the electron gun, which includes the cathode and the anode;

the graphite target;

the fluorescent screen, which will glow when the electrons strike it.
When the cathode has heated up (you will see it glowing), increase the accelerating voltage V.
An invisible beam of electrons emerges from the electron gun and passes through the graphite film. To show that there is an electron beam present, bring up a bar magnet close to the tube and observe the pattern shifting. Magnetic fields deflect moving electrons. So this qualitative demo shows both the wave aspects (the diffraction pattern) and the particle aspects (deflection by a magnetic field) in the same equipment.
Look for a diffraction pattern on the screen at the end of the tube. What shape is the pattern? Where can you see constructive interference? And destructive interference?
Now predict: If you increase the accelerating voltage, how will the energy and speed of the electrons change? How will this affect the diffraction pattern?
Test your ideas.
Quantitative experiments
Choose a feature of the diffraction pattern that is easy to measure.
It might be the diameter of the first bright ring, or of the first dark ring. We will call this chosen quantity d.
Investigate how d depends on V. Make several measurements.
Determine the mathematical relationship between d and V.
If the relationship is of the form d proportional to V^{n}, you can find the value of n by plotting
a loglog graph. The gradient of the graph is equal to the value of n.
What other information would you need if you were to use this experiment to determine the atomic spacing in graphite?
Practical advice
It is difficult to present a convincing discussion of wave–particle duality in a few lines. We have adopted the approach that it is an observed phenomenon, witness the diffraction of electrons.
The de Broglie equation wavelength = h/p simply allows us to translate between the wave and particle pictures.
Increasing the accelerating voltage increase the electrons’ energy and momentum; greater momentum means shorter wavelength, which in turn means that the electrons are diffracted through a smaller angle. Hence the diameters of the diffraction rings get smaller. Students could make measurements of the rings and relate these to the accelerating voltage. A logl graph will reveal the relationship between them. It is reasonable to say that ring diameter is proportional to wavelength; a loglog graph of diameter against voltage will thus be a straight line with
gradient (1/2).
Apparatus

electron diffraction tube

power supply (6.3 V) for cathode

e.h.t. supply (0 – 5000 V dc) with voltmeter

connecting leads
Great care must be taken to set the tube up correctly. The graphite can be damaged by incorrect connections. Notice that the positive e.h.t supply terminal is used without the protective resistor in some set ups. Take care.
As the discussion in this section is almost entirely in terms of electron diffraction, students might get the idea that only electrons exhibit wave–particle duality whereas in fact it applies to all particles. You might like to mention that neutron diffraction is also used as a tool for probing material structure. Large biomolecules have now been diffracted.
The chosen method is to introduce wave particle duality via the experiment however a little mathematics may be of use. This mathematics is given so an experimental or theoretical route can be chosen or the mathematics later used to back up the experiment.
E=hf and since f c then E =hc/
Taking E = mc^{2} then mc^{2} = hc/and so mc = h/
momentum = mass x velocity = p =mc
And this gives the de Broglie relation p = h/
Do notice that the switch from mc to mv has not been justified other than it works. (Note, this derivation really only refers to photons and not to electrons).
For the diffraction tube where V is the anode cathode pd and v the electron speed then
½ mv^{2} = eV so v = (2eV/m)^{ 1/2} giving mv = m (2eV/m)^{ 1/2} = (2meV)^{ 1/2}
p = h/ = (2meV)^{ 1/2}
Simplifying the Bragg formula into a two dimensional treatment for a transmission grating, for diffraction θ = /b (approximately) where b is the separation between the planes of carbon atoms. D, if required, can be obtained from manufacturers details.
θ = d/D from geometry so d/D = /b or proportional to d as diffracting gap size b and target screen distance D are constants.
So using the momentum and accelerating pd equation above then
p = h/ = (2meV)^{ 1/2}
or
1/ is proportional to V^{1/2 }
or
is proportional to V^{1/2 }
External references
This activity is taken from Salters Horners Advanced Physics, section PRO, activity 18, with extra material added in the last two sections, adapted from Revised Nuffield Advanced Science Physics teachers guide, book 2, section L2.
TAP 5062: Diffraction of light
Apparatus

torch bulb in holder with power supply

diffraction gratings, 2 each of various spacing, e.g. 100 lines mm^{1}, 300 lines mm^{1}

red, blue and green filters (one of each colour)

2 glass slides (e.g. microscope slides)

lycopodium dust
Safety
Lycopodium is dried pollen. Some people are allergic to pollen so, before use, it is wise to check for hayfever sufferer. (See Safeguards in the School Lab 4.4). Lycopodium powder in air will explode so extinguish all flames before using it.
Technique
Observe what happens when light is diffracted by:

a grating of parallel lines;

a regular grid of crossed lines;

a random array of fine dust particles.
Explore the effects of changing the separation of the lines and the wavelength of the light.
Set up a small (6 V) lamp. You are going to observe how light from the lamp is diffracted in different situations.
Write down or sketch what you observe.
Hold a diffraction grating and a colour filter close to your eye.
Look through it at the lamp. Rotate the grating.
(A diffraction grating consists of many parallel, equally spaced lines, as much as several thousands in each centimetre.)
Repeat with a finer or a coarser grating (using the same colour filter as before).
Now use two diffraction gratings. Place one on top of the other, so that their lines are at
90 degrees to each other. You have made a grid of lines through which the light is diffracted. Look through the grid at the lamp.
Try holding the gratings so that their lines cross at a different angle – say, 60° or 45°. Try combining gratings with different line spacing.
Finally, sprinkle fine dust (e.g. lycopodium powder) on to a glass slide. Blow off any excess dust. Look through the slide at the lamp.
(Now you are looking at light diffracted by a random array of tiny particles.)
Repeat the above, using a different colour filter.
Sum up your observations by saying how the diffraction patterns depend on:
Practical advice
This activity is intended to remind students of diffraction effects they may have previously observed. Points to bring out are:

the size of the pattern increases with increasing wavelength (red light is diffracted more than blue)

the size of the pattern depends inversely on the spacing of the diffracting objects (the finer the grating, the more widely spaced the diffraction images)

the geometry of the diffraction pattern depends on the geometry of the diffracting objects

a random arrangement of small obstacles gives rise to a diffraction pattern that is a set of concentric rings.
External reference
This activity is taken from Salters Horners Advanced Physics, section PRO, activity 17
TAP 5063: Interpreting electron diffraction patterns
For this question to make sense, students need to be aware that an obstacle (e.g. a nucleus) is equivalent to a ‘hole’ of the same size. So far they have only met diffraction by ‘gaps’.
Use the electron diffraction patterns shown in Figures (a) and (b) to determine the diameters of carbon and oxygen nuclei.
The diffraction patterns represented by the two graphs in Figures (a) and (b) were obtained using electrons accelerated to 420 MV.
By following the steps below, you will get a flavour of how such diffraction data are interpreted.
You will also see some of the equations that must be used when electrons are moving at relativistic speeds (that is, close to the speed of light).
Calculate the kinetic energy E_{k} of an electron accelerated through 420 MV.
To calculate the momentum of the electron, you cannot use p=mv and E_{K} = ½ mv^{2}, because these equations don’t apply to relativistic particles.
Instead, it turns out that the electron’s momentum p is related to its kinetic energy by
E_{k} ~ pc
where c is the speed of light.
Calculate the momentum of the electron.
The de Broglie relationship
p = h
applies to relativistic and nonrelativistic electrons because Relativity Theory is used in its derivation.
Calculate the electron’s de Broglie wavelength.
The angle of the first minimum θ_{min} on the graph is a clearly identifiable point, and this is used to calculate the diameter of the nuclei.
From the graphs in Figure (a) and (b), determine the angle of the first minimum for carbon nuclei, and for oxygen nuclei.
The smaller the nucleus, the more it diffracts the electrons.
From the angles you found, which are smaller, carbon nuclei or oxygen nuclei? Is this what you would expect from their atomic numbers?
The diameter d of the nucleus is related to the electron wavelength and the diffraction angle θ_{min} by
Use this relationship to obtain estimates of the diameters of carbon and oxygen nuclei.
Explain why your answers are only estimates.
External references
This activity is taken from Salters Horners Advanced Physics, section PHM, activity 19 which was an adaptation of Revised Nuffield Advanced Physics section L question 37(L).
Answers and worked solutions
E_{k }= 420 x 10^{6}_{ }x 1.60 x 10^{19} = 6.72 x 10^{11} J
E_{k} ~ pc
so p = 6.72 x 10^{11} / 3 x 10^{8} = 2.24 x 10^{19} N s
= h/p = 6.63 X 10 ^{34} / 2.24 x 10^{19}. = 2.96 x 10^{15} m
(a) carbon nuclei, 50°
(b) oxygen nuclei. 42°
d = 1.22
θ_{ min}
where d is the diameter of the nucleus
Remember, the angle should be in radians or use sin θ.
(a) carbon d = (1.22 x 2.96 x 10^{15}) / 0.766 = 4.7 x 10^{15} m
(b) oxygen d = (1.22 x 2.96 x 10^{15}) / 0.669 = 5.3 x 10^{15} m
TAP 506 4: Electron diffraction questions
1 In an electron diffraction experiment using graphite the larger ring formed by rows of carbon atoms 1.23 x 10^{10} m apart was formed at an angle of 0.167 radian.
(a) What is the wavelength? [θ in radians = / b where b is the diffracting object size]
(b) Write an expression for the kinetic energy of an electron (½ mv^{2}) in terms of its charge, and accelerating voltage V
(c) Obtain an expression for momentum, p, in terms of e, V and m.
The accelerating voltage was 5000 V
(d) Work out h in mv = h/
Answers
(a) 0.205 x 10^{10} m
(b) KE = 0.5mv^{2} = Ve
(c) p = mv = (2meV)^{ 1/2}
(d) h = 7.78 x 10^{34} J s (the accepted value is 6.6 x 10^{34} J s)
External reference
This activity is taken from Revised Nuffield Advanced Physics Unit L question 32.
