Basic Physics III Lecture 21
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| Teil | 21 | |
| Anzahl der Teile | 27 | |
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| Identifikatoren | 10.5446/12957 (DOI) | |
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TeilchenWellenlängeDiffraktometerPhotonKaliber <Walzwerk>AmplitudenumtastungSchmalfilmHeftKugelschreiberLinearmotorElektronGammaquantPassfederTiefdruckgebietTrenntechnikAtombauGauß-BündelFestkörperBragg-ReflexionKristallwachstumHöchstfrequenztechnikTrägheitsnavigationSchnittmusterLadungstrennungWeißTeilchenWellenlängeWelle <Maschinenbau>DiffraktometerUrkilogrammKristallwachstumElektronErderAnstellwinkelMessschieberWerkzeugMark <Maßeinheit>RöntgendetektorGammaquantSpannungsänderungSchmucksteinProfilwalzenSpannungsabhängigkeitFahrzeugsitzKopfstützePhotonKugelschreiberWerkzeugverschleißTransmissionsgitterStundeComte AC-4 GentlemanTrenntechnikLichtstreuungKlangeffektLichtSchwingungsphaseAtomistikFeldeffekttransistorMagnetisierungPatrone <Munition>Angeregter ZustandSchienePaarerzeugungInterferenzerscheinungPhotoelektrizitätProof <Graphische Technik>KraftfahrzeugzulieferindustrieSpeckle-InterferometrieVollholzBeschleunigungRegelstreckeMasse <Physik>AmperemeterWiderstandserwärmungDruckgradientLambda-HyperonSpannungsmessung <Elektrizität>ColourMessschraubeMagnetspuleErsatzteilCompton-EffektBroglie, Louis deElektronenstrahlComputeranimation
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Vorlesung/KonferenzComputeranimation
Transkript: Englisch(automatisch erzeugt)
00:05
Hello. So you remember that two days ago we started our discussion of modern physics with the blackbody radiation and the photoelectric effect. Einstein had this theory that light is actually consisting out of particles, where he explained the photoelectric effect
00:31
by stating that the energy of a photon, a light particle, is the Planck quantum times the frequency.
00:40
Therefore, if we look at the liberated electron kinetic energy, then it is affected by the frequency of the light and not so much by the intensity, which would be the prediction of the wave theory. Therefore, if you plot the kinetic energy of the electrons as a function of the frequency of
01:01
the photons or of the light particles, then below a certain frequency no electrons are liberated at all. That is the equivalent amount of energy that is necessary to eject electrons from the potential in which they are sitting inside the atom. And then after that you get a linear relationship. This is in agreement with experiments and therefore the quantum mechanics took its start.
01:36
Let's look at an example and let's calculate the energy of a blue photon, meaning a photon
01:42
that corresponds to blue light or about a wavelength of 450 nanometers in air or in vacuum. Since we know that the energy of the photon is given by the Planck quantum times the frequency, we first need to find the frequency but we know that the frequency is the speed of light
02:02
divided by lambda and therefore the energy is the Planck quantum times the speed of light divided by lambda. We can calculate what is the product of Planck quantum and speed of light and we come up with this number but we can also directly look up the product of this.
02:23
I am going to use as the product of hc these numbers either in units of joule meter or in units of electron volts micrometer. One electron volt is 1.6 x 10-19 joule. It is the energy that an electron receives in a potential of 1 volt.
02:46
In other words it is nothing but the electron charge 1.6 x 10-19 coulombs multiplied with 1 volt. If we use this number in these units then we can get the energy of the photon simply by dividing this constant
03:06
by 450 nanometers and we come up with 4.44 x 10-19 joule for 1 photon of wavelength of 450 nanometers. Or in units of electron volts we use the other value the 1.240 electron volt micrometer
03:26
divided by the 0.45 micrometer of the wavelength and that gives us 2.76 electron volts. So if we have a visible light then the energy per photon is a few electron volts, something like 2 or 3 electron volts.
03:45
Let's also estimate how many photons are emitted in your typical 100 watt light bulb every second. For that calculation we assume that the bulb has an efficiency of about 3% meaning that 97% of the
04:05
energy goes into heat radiation just as we learned with the blackbody radiation and only 3% goes into visible light. We will also assume some average wavelength of 500 nanometers.
04:22
We first observe that the energy that is used up every second or that is coming from the wall plug is 100 watts times 1 second or 100 joule. And since we have a 97% inefficiency or a 3% efficiency that means only 3 joules out of those 100 joules actually goes into visible light.
04:47
So now we make the assumption that the light has a wavelength of roughly 500 nanometers and therefore one photon has the energy of 1.986 times 10 to the minus 25 joule meters divided by 500 times 10 to the minus 9 meters.
05:05
And that gives us about 4 times 10 to the minus 19 joules. And if we divide the 3 joule by the 4 times 10 to the minus 19 joule we get about 7.5 times 10 to the 18 photons.
05:21
So quite a few photons in a single 100 watt light bulb. There were some technological applications of the photoelectric effect. These days most of them are somewhat obsolete. For example the so called electric eye.
05:42
Or when you still had celluloid films then you could encode the soundtrack of a movie next to the images actually. And use a small light source to read the brightness or darkness of the soundtrack and then measure the resulting current in a photocell.
06:02
Of course semiconductor photo diodes work better than a high voltage tube. And by now everything is encoded digitally anyway. I have chosen in the layout of this class to not cover special relativity but we cannot avoid it completely.
06:23
So in two slides the difference is between classical mechanics and special relativity. So in classical mechanics the mass of a particle is a constant and the momentum is this constant times the velocity.
06:40
And we can get the kinetic energy for that particle by forming the product of one half the mass and the velocity square. Using the momentum, the momentum equals mass times velocity. This also means that the kinetic energy is the momentum squared divided by the mass.
07:03
We simply multiply both numerator and denominator with two times the mass. The energy is just kinetic energy when there is no potential around. And particles can have any velocity and any mass that is bigger or equal zero.
07:22
In special relativity the most famous of physics formulas is that the energy equals the mass times the speed of light squared. However in this case the mass m is a dynamic mass meaning that it is not a constant, it depends on the speed of the particle.
07:43
So we define a rest mass m0 which is the mass when there is no motion. The more you speed up the particle the bigger the mass will become. And then there is the Einstein energy momentum relation which says that the energy squared is the rest
08:04
mass times the speed of light squared squared plus the momentum squared times the speed of light squared. And of course because of this relation this must be equal to the dynamic mass times the speed of light squared and then the whole thing squared because the energy is squared.
08:25
The momentum is defined just as in classical mechanics only that it is defined with the dynamic mass. So the momentum is the dynamic mass times the velocity. And if we plug that into this equation then we get m0c squared plus mv squared and
08:47
I've divided out the speed of light squared here equals mc squared and the whole thing squared. Or the dynamic mass equals the rest mass times the speed of light divided
09:03
by the square root of the speed of light squared minus the velocity squared. And usually it is put in this form that the dynamic mass is the rest mass divided by the square root of this factor. The kinetic energy is the energy minus the rest energy meaning the energy that we get in the rest frame of the particle.
09:28
And that has now the following consequences that if you have a rest mass that is bigger than zero then particles cannot reach the speed of light. They have to be slower than the speed of light and the reason for that is
09:43
that if v becomes c then the denominator becomes zero and the dynamic mass becomes infinite. So it requires an infinite amount of energy to accelerate bigger than zero rest mass particles to the speed of light. If this rest mass is zero then particles can't be slower than the speed of light.
10:05
They always have to travel forever at the speed of light and we can see that in the following way. In the Einstein relation between energy and momentum this term falls out because there is no
10:20
rest mass and that means that the energy equals the momentum times the speed of light. And if we plug in that the momentum equals mass times velocity we arrive at the conclusion that the velocity has to be the speed of light. So such particles you cannot slow them down and one particle we already know that has that quality and that is the photon.
10:46
So photons of course by definition move at the speed of light. So we have the following photon properties. It has no rest mass since it moves at the speed of light and its energy is given by the Planck quantum times
11:07
the frequency or we can also say Planck quantum divided by two pi which is usually called h bar times the angular frequency omega. Kinetic energy and energy are the same thing because there is no rest mass
11:24
and the momentum is just given by the energy divided by the speed of light. And therefore since we know that the energy is h times f or h bar omega divided by c we get that the momentum of the photon is the Planck quantum divided by the wavelength or h bar times the wave number k.
11:50
And photons have all the usual particle interactions including collisions. So you can collide photons with other particles.
12:01
However unlike other particles photons can't be slowed down if they lose energy they just shift in frequency but they still move at the speed of light. So we can look at a little simulation of collisions of photons with electrons.
12:25
So here we have a photon coming in and we have an electron sitting at rest initially. And the photon goes in and it collides with the electron it becomes a photon with a slightly longer wavelength or smaller frequency and then the electron is moving.
12:45
We can now look what happens when we change the angle of the scattering of the photon. So if I make the angle very small that the photon goes almost in the same direction as when it started then it has very little change in wavelength.
13:09
And you can see that on the bottom graph here which shows this green line. Here you can see how the wavelength changes.
13:22
So this is a scale in nanometers and here we have the original photon wavelength and here at this position we have the shifted photon. And of course the wavelength gets longer since the frequency becomes smaller since the photon loses energy in the collision. You can also see what happens if you go to shorter wavelength and it
13:45
doesn't affect the shift in how many nanometers you change when you undergo the scattering. Let's actually calculate this collision process. Well before we do that let's first look again at our example that we calculated with a light bulb and see how much
14:07
momentum is actually absorbed by a piece of paper when if you focus all photons from a 100 watt light bulb onto it. And we know that the momentum of one photon, again we assume 500 nanometers, is H divided by lambda.
14:28
So this is the Planck constant divided by 500 nanometers. Then we get 1.33 times 10 to the minus 27 kilogram meters per second. And that means since we have 7.5 times 10 to the 18 photons that
14:46
the total momentum transfer every second is 10 to the minus 8 kilogram meters per second. So that every second that means that the force which is kilogram meters per second squared is 10 to the minus 8 newtons.
15:02
So in other words a light bulb exerts a force of 10 to the minus 8 newtons. If you focus all of its photons on some object. And we can also look at the efficiency of photosynthesis. In photosynthesis molecules like the chlorophyll molecule in plants capture the energy of the sunlight and
15:27
changes CO2 to useful carbohydrates that the plant can then make use to grow for example. And it requires about 9 photons for this transformation to take place.
15:40
And the reverse process where we simply burn wood for example releases 4.9 electron volts per molecule consumed. So if we assume 670 nanometers as a typical wavelength since chlorophyll absorption is somewhere between 650 nanometers and 700 nanometers.
16:03
Then we can calculate how efficient the photosynthesis is. So again we calculate the energy of one photon. So it's 1.240 electron volt micrometers divided by the wavelength which is 0.67 micrometers.
16:23
So this gives us 1.85 electron volts for the energy of one photon. Since we know that photosynthesis requires 9 photons we multiply that by 9 and we get the total energy, solar energy that is needed of 16.7 electron volts.
16:41
And if we now form the ratio of the 4.9 electron volts that we gain when we burn the product molecule. Over the 16.7 electron volts that was required to manufacture it then the efficiency is about 30%. So photosynthesis is about 30% efficient.
17:03
So let's now calculate the scattering of photons with electrons like it was shown in the Java simulation. The Compton effect which is what this is called is an elastic collision. And that means that both momentum and energy is conserved in such a collision.
17:25
And for the momentum we have two components. The one in the direction of the incident photon and the one that is perpendicular to the incident photon. So let's assume that the scattered photon makes an angle of pi with the incident direction.
17:43
And that the electron which is initially addressed makes an angle theta. So if we look at the x component of the momentum then initially of course we simply have the momentum of the incident photon, all of it.
18:00
And afterwards we have the electron momentum pe prime after the collision times the cosine theta. That's the component that is pointing along the x direction. Plus the scattered photon times cosine phi. So we can solve this for pe prime cosine theta and we get this equation.
18:27
Then we look at the y component. There was no y component before the collision so there should be no y component after the collision. So that means that pe prime times the sine of theta which is the electron y component.
18:47
Must be equal and opposite to the y component of the scattered photon which is gamma prime sine phi.
19:00
So since we have those two equations pe prime cosine theta and pe prime sine theta we can square both equations and add them up. And we then get pe prime square because sine square theta plus cosine square theta equals 1. So this is the y component, this part square and this is the x component, comes from the x component.
19:26
And if we multiply this out, this term, then we get a term with cosine square phi, this one. And a term with p gamma square, this one, and a mixed term, which is this one.
19:40
And since we have from the other component already one with sine square phi, we collect those terms here and recognize that this is the momentum of the scattered photon square. So we get that the momentum square of the electron after the collision is the sum of the scattered photon momentum square plus the incident momentum of the photon square minus this mixed term.
20:11
So we remember that result and we just copy it onto the next page. So we can make this a little bit nicer by expressing this sum of those two momentum squares in a slightly different way.
20:28
We observe that the binomial formulas give us this expression that p gamma prime minus p gamma square equals p gamma prime square plus p gamma square minus the mixed term. And we can therefore express this sum by the difference in photon momentum square plus two times the product of the photon momenta.
20:54
If you plug this into here, then we get the momentum of the electron after the collision as
21:01
the difference square of the photon momenta plus this mixed term which depends on the scattering angle phi. All of this is coming just from momentum conservation. Now we are using energy conservation. So first of all the photon energy before the collision is e gamma and if
21:23
we divide that by the speed of light then we get the initial momentum p gamma. And the electron energy is initially, since the electron is at rest, just the rest mass of the electron times the speed of light square. If I divide by the speed of light then I get just the rest mass of the electron times the speed of light.
21:45
Then after the collision I get e gamma prime and dividing by the speed of light gives me p gamma prime. And the electron energy I have to now use the Einstein relation. So this is the Einstein relation which I get after the collision.
22:12
Gamma is just my way of denoting that it's the photon. So gamma has no particular value, it's just a symbol for the photon.
22:23
Since I'm a particle physicist I think of photons as gamma rays typically. It's just a notation, I could have written photon there. So since the energy is conserved that means that initial energy of photon and electron must be equal to final energy of photon and electron.
22:48
And if I solve this for e prime and divide by the speed of light then I get this and I recover here the momentum of the initial photon and the momentum of the scattered photon.
23:04
And here the rest mass of the electron. So now if I square the whole thing I can use the Einstein relation on the left hand side which is this one. And on the right hand side I just multiply it out with the binomial formulas.
23:21
So the difference in momenta squared plus a mixed term plus the rest momentum squared. And I get this relation for the momentum of the electron after the collision. But I already have a relation for that coming from momentum conservation which is this one which we had derived on the previous page.
23:43
So I can set those two things equal and the first term is identical so that cancels out and the factor of two cancels out. So I'm just left with this term equaling that term. And now if I multiply with the Planck constant and divide by the two momenta of the
24:11
photon and the rest mass of the electron and the speed of light then this term goes away.
24:20
Here I get h over p gamma prime since I divide by both p gamma and p gamma prime. But this is just the wavelength of the scattered photon. And then for the next term I get h over p gamma but that is just the initial photon wavelength.
24:40
And then on this side both of the p gamma and p gamma prime go away and I'm just left with h over m e c times one minus cosine phi. So in other words the result is that the wavelength of the photon shifts by a constant amount because all of these are constants of nature.
25:03
This is the rest mass of the electron, the speed of light and the Planck constant. So the shift in the wavelength is just controlled by the scattering angle and by nothing else. So therefore if you now look at various angles at zero degrees there is no shift at all, a zero scattering angle.
25:25
Then at 45 degrees there is some shift, at 90 degrees there is a bigger shift and the biggest shift you will observe at 180 degrees which is sometimes called backscattering. Furthermore we can give this expression of constants of nature a name and we can
25:45
call that the Compton wavelength of the electron because it has units of a wavelength. So the Compton wavelength of the electron is h over m e c and if you plug in numbers meaning 511 kilo electron volt per speed of light square for the electron mass and for Planck's constant I'm using this expression.
26:11
Then I get 2.427 picometers for the Compton wavelength of the electron. Therefore since we have this expression we have a Compton shift or a wavelength shift
26:25
in lambda that is the Compton wavelength of the electron times 1 minus cosine 5. Compton scattering has some application if you use gamma rays instead of x-rays. Gamma rays have even shorter wavelengths than x-rays and you can produce gamma rays with radioactive sources.
26:46
The number of scattered photons will be proportional to the number of electrons that are around, sort of intuitive. Therefore it is proportional to the density of electrons in a particular material.
27:04
You can use the Compton scattering to scan the density of bones since the density of electrons ought to be proportional to the density of bone material. You can do that to diagnose bone diseases. Sometimes the onset of such diseases is accompanied by change in the bone density.
27:26
So it has some diagnostic purposes. Let's calculate an example. Let's assume we have x-rays of wavelength 0.14 nanometers or 140 picometers.
27:40
We scatter them from a very thin slice of carbon. The question is what will be the wavelength of the x-rays after scattering when they scatter at 0 degrees, at 90 degrees or at 180 degrees. For the Compton formula we first just look at the scattering angle and 1 minus cosine 0 degrees is 1 minus 1 since cosine of 0 degrees is 1.
28:07
So this gives us 0 for part a and for part b it is 1 minus cosine of 90 degrees and cosine of 90 degrees is 0. So for part b we get 1 and for part c we actually get 1 minus minus 1 or 2.
28:26
So now we can calculate the Compton shift which is just the Compton wavelength times these expressions. So we get no Compton shift at all for part a. We get the Compton wavelength or 2.43 picometers for part b or we get twice the Compton wavelength of the electron or 4.85 picometers for part c.
28:51
So since we have originally 140 picometers as the wavelength, therefore the scattered wavelength is in part a again 140 picometers since there is no shift.
29:04
And in part b it is 142 picometers and in part c it is 145 picometers. And now you see why you want to use gamma rays for this kind of measurement rather than x-rays because all those wavelengths are relatively close by each other.
29:23
So you need fairly high precision in measuring your wavelengths. Well if you have say a wavelength of just 10 picometers then the Compton shift is far more noticeable. So the shorter the wavelength the easier it is to detect the Compton shift.
29:43
Since the Compton shift doesn't change in its size it is independent of the initial wavelength. We therefore have now a small collection of photon interactions. So we have the Compton scattering and in this case photon survives, it just has a longer wavelength.
30:05
Then there is the photoelectric effect and in this case the photon is completely absorbed but an electron is emitted. And if we don't have enough frequency or enough energy to actually emit the electron we can still excite the atom.
30:22
So in this case the photon is also vanished, it is absorbed and no electron is emitted. However the excited atom can de-excite after some time and emit another photon. And finally there is yet another effect which is called pair production which is shown here.
30:44
So we have a photon coming in and out of nothing it creates an electron E- and a positron E+. And the positron and the electron are called a pair. Positron is an electron that has all the same properties as an electron but the opposite charge.
31:05
And that's why this process is called pair production. Pair production can only happen in some material because you actually need a nucleus to participate in the interaction.
31:23
Because otherwise it is impossible to conserve both energy and momentum of the process in pair production. So you need the nucleus to absorb some of the momentum of the photon otherwise this process cannot happen. The opposite of pair production is also possible.
31:44
So you can have an electron and a positron colliding with each other. And when they do two photons are emitted and both electron and positron is destroyed.
32:10
So that is called annihilation. If the momentum of the electron and the positron is small enough then before they annihilate they can form a bound state which is called positronium.
32:22
Which sort of looks like an atom where the nucleus of the atom is replaced with the positron. Because of this process, this annihilation process, positrons don't last very long in nature unless you isolate them from all the other electrons.
32:43
Which is somewhat tricky to do. Just one example about pair production. So we can calculate what the minimum energy of a photon has to be so that it can actually produce such an electron and positron pair. In this case we are not concerned with momentum conservation.
33:03
We just assume that the nucleus takes care of it and we just look at energy conservation. So we know that the electron mass is 511 keV per C square. And I just told you that the positron has all the properties of an electron except that it has the opposite charge.
33:22
So the positron has also a rest mass of 511 keV per C square. And therefore we need twice the rest energy of an electron or 1.022 MeV to produce both electron and positron. And that is the minimum energy that the photon has to carry because the photon needs to supply that energy.
33:46
So we can calculate what wavelength that corresponds to by taking the Planck constant and divide the energy. We get the resulting wavelength and it's 1.21 times 10 to the minus 6 micrometers or 1.21 picometers.
34:11
So this is in the gamma ray range of the electromagnetic spectrum. So it requires hard gamma rays to produce electron-positron pairs.
34:22
So x-rays are not energetic enough to do it. We have seen in the last few weeks a lot of experimental evidence that electromagnetic waves and light are a wave. For example we looked at interference and diffraction and refraction and all those effects that depend on wavelength and on phase.
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And it seems to prove beyond any doubt that electromagnetism or that light is a wave phenomena. However now we have collected also an impressive list of experiments that show the particle nature of light.
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Like the black body radiation, the photoelectric effect, the Compton scattering and the pair production actually is also difficult to explain with the wave theory. So how does this fit together?
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It doesn't really fit together but we simply state that it does. So we say light has a dual nature, it has some aspects of waves and it also has some aspects of particles. And that is called the principle of complementarity that if you want to understand light then you
35:42
have to take into consideration both the particle as well as the wave nature of the light. But it gets even weirder because in quantum mechanics matter also has a dual nature. So in other words every particle also has a wave nature and every wave has a particle nature.
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So this principle of complementarity refers to everything. For example we know that a photon of wavelength lambda carries a momentum of h over lambda.
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We can turn this around and say a particle of momentum p has a wavelength of h over p. Simply the same formula solving it for lambda. And Louis de Broglie thought of this first. And so this is called the de Broglie wavelength of a particle of momentum p.
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And what it means is that you can actually observe a diffraction of particles like electrons for example when they hit some crystal just like you observed a diffraction pattern of x-rays when they hit a crystal.
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So let's say we have some regular spaced atoms or scattering centers. Then we have an incident electron beam. Then we can formulate diffraction criterion or interference criterion where we observe maxima and minima based on what angle we observe them.
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And we can do that experimentally and create a diffraction image that looks very very similar to the diffraction images of x-rays. So as long as the particles are coherent they have a wave nature as well. So let's look at an example.
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Let's say we have a 200 gram or 0.2 kilogram ball that is moving with a speed of 50 meters per second. What is the wavelength of that ball? And we use the relationship between wavelength and momentum.
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And this is the Planck constant this time again in joule seconds. And we divide by 0.2 kilograms times 50 meters per second. And we get that the wavelength is 2.21 times 10 to the minus 34 meters. And this is such a small number that there is not even a prefix invented for it.
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Prefixes only go to 10 to the minus 18. So it's a very very very small wavelength and we will not be able to observe very easily the diffraction pattern of 200 gram balls. On the other hand if we look at microscopic particles I just showed you that you can observe diffraction in electrons.
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Let's assume that we have electrons that have been accelerated through an electric potential difference of 100 kilovolts or 100 volts and calculate that wavelength.
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To do that we again have to use relativistic formulas. So the rest mass of the electron is 511 keV per C squared. That means the rest energy is 511 keV. And we add the potential energy which is 100 kilovolts times the charge of the electron or 100 keV.
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So we get 611 keV as the energy of the electron after it goes through that potential difference. And we can calculate the momentum using the Einstein relation. So this is the momentum squared which gives us 112,000 keV per C squared or 335 keV per C as the momentum of those electrons.
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And now we can look at the corresponding wavelength simply h over p or hc over pc.
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In other words I can divide 1.240 electron volt micrometers by 335 keV and I get 3.7 picometers if I do that. So this wavelength is small but not so different from that of gamma rays for example.
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So in the part B if I only have 100 volt rather than 100 kilovolt then I can proceed in the same way. Now the total energy is 511.1 keV.
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So I could do that but I can also use a classical formula. So this is the corresponding classical formula where simply the momentum squared is twice times the kinetic energy times the mass of the electron.
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And in both cases I get 10.1 keV per C as the momentum of the electron. Using the same relation this gives me 123 epicometers. So that is in the x-ray range of wavelengths.
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So in other words if you accelerate electrons by reasonable potential differences like between 100 volt and 100 kilovolts you get wavelengths that correspond to gamma rays to x-rays. For photons. As a further example we can actually calculate diffraction pattern.
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So let's say we have such a 100 volt potential and accelerate electrons. And let's say we observe a maximum and the first maximum occurs at 24 degrees. What is the separation d between atoms?
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And to solve that since the energy of the electrons is low they don't penetrate into the crystal like x-rays would do. So we simply look at the surface atoms and therefore we observe that there is a path length difference of the spacing times sine theta.
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For a maximum that needs to be an integral number of wavelength. And therefore we get that d sine theta has to be equal to lambda. And since we know that theta is 24 degrees we get that lambda has to be...
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The lambda we know is 123 picometers from the previous example. And therefore the d is 123 picometers divided by the sine of the 24 degrees or 0.407. And that means that the spacing of the atoms is 302 picometers.
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So we can calculate diffraction patterns from electrons just like we did for x-rays or for diffraction gradings for light. Let me conclude with this today and talk about the nature of the electron next time. Thank you very much.