Basic Physics III Lecture 22
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Basic Physics III22 / 27
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Transcript: English(auto-generated)
00:04
Hello, we will start today our discussion with talking about the nature of electrons. So how does an electron look like? Well, if you look it up in Google Images,
00:21
you find something like this. It's a blue ball with E minus printed on it. But that's, of course, not an electron. One can't really see an electron At the end of the last lecture, we actually found out that it is quite easy to make electron waves that
00:41
are much shorter in wavelength than most of electromagnetic waves. So one can't really see an electron. One can't just measure the effects of an electron. And it has a definite mass of 9.1 times 10 to the minus 31 kilograms, or in terms of energy
01:01
over c squared, 511 keV. Then it has a charge, a negative elementary charge of minus 1.6 times 10 to the minus 19 coulombs. And indeed, it has some intrinsic angular momentum, meaning that it's almost like as
01:21
if it was spinning of this angular momentum. And it is both a particle and it is a wave. And finally, if one wants to take the question literally, what is the nature of electron? Electron is the weak word for amber.
01:41
And that's what it is. So the reason why we use electron for that particle is that the first electric experiments were carried out using pieces of amber and cat fur and such things. Since we know that electrons are waves, among other things, that means that it ought to be possible to construct
02:05
optical instruments with electrons. There was an idea from this fellow that it ought to be possible to construct magnetic lenses, meaning magnetic coils that produce magnetic fields,
02:22
so that electrons that are affected by those magnetic fields behave in very similar ways that photons behave when they pass through lenses. And therefore, this was a theoretical idea. But because the electron wavelength is too,
02:41
even for very modest acceleration voltages, is far smaller than that of light, of optical light, definitely, one can theoretically at least achieve much greater magnification with an electron microscope than with a light microscope.
03:01
And so this idea was eventually developed by this fellow, who eventually got the Nobel Prize for it. Modern instruments are up to 1,000 times better than light microscopes. And they're still limited by aberration effects rather than Rayleigh resolution criteria.
03:20
So that is a huge success story. Not only can you make a transmission electron microscope like this one, which is modeled after the light microscope. So we have a source, an electron source. We make a parallel beam out of it using a condensing lens.
03:43
Then we have the objective lens. And we finally have the eyepiece. And we project the image on some screen, film, or semiconductor detector. But we can also construct a different type of microscope where we bounce the electron beam off the object.
04:04
So in this case, we steal the electron beam and scan it across the specimen. And then once you hit the specimen, it will emit a spray of electrons, which you detect. And then you use a computer to produce an image of that.
04:21
And the advantage of that procedure is although it has somewhat less resolution typically than the transmission microscope, it gives you three dimensional images. Giancoli gives you two beautiful images of virus attacking bacteria.
04:42
So this one is a transmission picture. And this one is coming from a scanning microscope. And you can see the incredible detail you get from the viruses as well as from the cell in both two or three dimensions. So that is a nice application of the wave nature of electrons into technology.
05:05
So now let us discuss something very fundamental, the structure of atoms. So since about the beginning of the 20th century, we know that all matter that we know of is made out of atoms.
05:21
Those atoms were not like originally thought really indivisible and sort of fundamental as they are still used in chemistry, but they have a structure themselves. The first hint of such a structure came from experiments in the 19th century
05:43
when Thompson discovered the electrons. So small, negatively charged particles. And so he made an atom model based on that, sometimes referred to as the plum-putting model. And so he assumed that we have small negative charges
06:02
that are embedded in some extended positively charged material. And such an entity forms an atom and it is held together by the electrostatic force between the negatively charged electrons and the positively charged material.
06:20
Rutherford, a New Zealand fellow, he put that to the test and he was this model and he was shooting alpha particles, those are positively charged particles, on a metal foil. And if this model is correct, then what you would expect is that an alpha particle comes in,
06:40
it sort of just tears through the atom and goes through pretty much undisturbed because the electrons will just get pushed out of the way. The alpha particle is much heavier than an electron. And to his great surprise, he saw a large angle scattering of alpha particles and even reflections where the alpha particles
07:02
was coming back towards the emitter. He was very surprised by that. I believe he used an analogy. It's like shooting with a big cannon on a piece of paper and then see the cannonball reflect back at you. And so what he reasoned is that this model cannot be correct in this form.
07:21
Rather, the positive charge has to be concentrated in an atomic nucleus like so that is much, much smaller than an atom. And only then does it have enough charge and enough mass to actually interact
07:40
with the alpha particle in this fashion. So that was the Rutherford model that you have an atomic nucleus on the order of 10 to the minus 15 meters and atoms have a diameter on the order of 10 to the minus 10 meters. So about five orders of magnitude smaller. And then electrons,
08:01
they are in a cloud surrounding this nucleus and make the atom neutral. And this was then further refined by Bohr and Sommerfeldt who considered the electrons orbiting around the nucleus in a similar way as the planets orbit around the sun.
08:25
So kind of like a miniature solar system. This is how this would work. We have on a charged nucleus, it is charged by the elementary charge times a capital Z depending on which element it is. And then if you consider just one electron,
08:43
there is a attractive force between the electron and the nucleus, the electric force, which is this one. And the magnitude of it is given by Z times the elementary charge, the charge of the nucleus times the charge of the electron which is also the elementary charge
09:01
divided by the distance square and divided by one over four pi epsilon naught. If you assume a circular orbit and eventually Sommerfeldt assumed elliptical orbits just like in the solar system, but if you make the simplifying assumption that the orbit is circular,
09:22
then we can actually solve this problem classically. So we have a centripetal force and then if the electron has a velocity, it will remain stable in a circle around the nucleus. And we need a balance of this electric force with the required centripetal force.
09:43
And we can solve that for radius R and get some solution. However, for the solution to make sense with experimental data, one needs to require that only some radii are allowed
10:01
or possible in an atom and not all of them. And the question is why? And another theoretical problem is with Maxwell's equation. If you have an electron orbiting around the nucleus, then looked at it from the side, it looks like the electron is sort of
10:20
in the planetary plane. The electron is sort of just oscillating back and forth. So we have an oscillating charge and it ought to emit radiation just like any oscillating charge emits radiation. And when it does emit radiation, it loses energy and therefore it will move closer to the nucleus,
10:41
which makes it go even faster, oscillate even faster. And that means that it emits more radiation. So basically it should just spiral into the nucleus and then be absorbed by the nucleus. And such a thing does not happen. Atoms don't destroy themselves in a radiation catastrophe.
11:02
And the question is again, why? So we can use the concept of electrons as waves rather as particles to rescue this scenario. So we observe that if you have electrons, an electron wave propagating,
11:21
then it has some wavelength lambda, which is given by the Planck constant divided by the momentum of the electron or the Planck constant divided by the mass of the electron times its velocity. And if the propagation is in a straight line,
11:40
then we sort of have a traveling wave like that. And if we fix the line on both ends, then we get a standing wave and only certain wavelengths are possible just like with the waves on a string. In this case, we have a circular orbit. So we have a different kind of boundary condition.
12:00
If we go once around the orbit, then the phase of the wave has to match itself. And therefore, since we know that the circumference of a circle is two pi times its radius, this distance must be an integral number times the wavelength.
12:21
Otherwise, we get destructive interference, like in this case, for example. And since it keeps running around and around and around, the destructive interference will be complete if any of it is apparent. So we get only a stable solution for a standing wave on a circle
12:42
if we have this condition fulfilled, meaning that the distance along the circumference of the circle is an integral number of n times lambda. And if we now use the expression for the wavelength of the electron, then lambda equals H over momentum,
13:05
where momentum is mass times velocity. Then we get the condition that the radius times the electron mass times its velocity must be an integral number times the Planck constant divided by two pi, which this constant H over two pi
13:22
is usually referred to as H bar. One can look at this condition in a different way, and we observe that mass times velocity is the momentum, R is the distance to the nucleus,
13:40
and so R times momentum, or R times mass times velocity is the angular momentum of the electron with respect to the nucleus. So that angular momentum, usually called L in general, L is the radius crossed into the linear momentum.
14:08
So in this case, we just look at one component and we get this. So we find that from our requirement
14:20
that we have a standing electron wave, we find that only certain quantized values of angular momentum are allowed. So in a very similar fashion, as we had quantized the energy of the electromagnetic field or of the vibrations of molecules
14:43
in the black body radiation. So if we now use this condition and solve it for the velocity, so we simply solve this equation for the velocity, which means we divide by the radius and the electron mass, then we get for the velocity
15:01
that it's NH bar divided by radius and electron mass. And we can now plug that into our previous equation. So if you wanna know what is MeV square R, which is what we had from the Bohr Sommerfeld criterion,
15:21
for that criterion, we needed MeV square R. And if you now use the equation for the velocity coming from the standing electron wave, then MeV square R is MeR times NH bar divided by MeR square.
15:41
And therefore we get one of the MeR cancels and MeV square R is equal to N square H bar square divided by MeR. And we can now set that in relation to the electric force. So this is the term from the electric force.
16:02
Once we do that, we have an expression that tells us what radius is possible and what radius is not possible. Because of that standing wave condition, not all radii are possible anymore in the electron because many radii are eliminated by destructive interference.
16:22
And that sort of is an explanation for why only certain radii are allowed. So we have basically a quantized radius. So the radius has to be N square times H bar square multiplied by four pi epsilon naught over Me Z e square.
16:43
Or the Nth radius is given by the first radius times N square over Z. Where the first radius we simply use N equals one in this expression. And the integer N is called a quantum number
17:01
because it describes the state of the system. So the value for this first radius which is also called the Bohr radius is 52.92 picometers or 0.529 times 10 to the minus 10 meters.
17:20
And now you remember that I told you that an atom has roughly the size of 10 to the minus 10 meters which is simply twice this Bohr radius. We can also look at this expression for the Bohr radius in a different way. And we see it as the product of two factors.
17:41
The first one is this one here. And this is the second one. And the second one we recognize as the Compton wavelength of the electron. So lambda e from the last lecture. And the second constant
18:00
that's actually a new constant. And it is usually called alpha or fine structure constant. And it has no units. So it's relatively easy to remember one over alpha is roughly 137. It is only calculated with constants of nature.
18:22
So it is a constant of nature itself also. And in fact, it characterizes the strength of the electromagnetic force I mean. So in other words, we get that the Bohr radius is the Compton wavelength of the electron
18:41
divided by two pi times the fine structure constant. We can also look at the electron energy that results from that. Now that we have found which radii are possible in atoms, we can also find what energies are possible in atoms.
19:01
And we have two types of energy, kinetic energy and potential energy. And for simplification, we are going to use non-relativistic approximations rather than doing a fully relativistic calculation as Sommerfeld did.
19:21
So we simply say the kinetic energy is one half electron mass times the velocity square. And from the expression that we have for the velocity, this is one half times the quantum number square times H bar square divided by Me R square. So we simply have plugged in to this equation,
19:43
the expression for V that we had derived earlier. So if we now use the condition which radii are possible for this R square, so we get the Bohr radius square divided by Z square times the quantum number to the fourth power.
20:03
And that means that some of the quantum number powers and H bar powers are canceled and we are left with one half Z square H bar square divided by Me N square, the quantum number square times the Bohr radius square. And if we now find what is the electron mass
20:22
times Bohr radius square, then we come up with this. And finally, plugging it all in, we get this expression for the kinetic energy, the charge of the nucleus square times the mass of the electron times the elementary charge to the fourth power.
20:41
Then in the denominator H bar square, the quantum number square and some pre-factor from the electromagnetic force. Then for the potential energy, it's actually much simpler. For the potential energy, we observe that the potential energy of U of an electron
21:05
which has charge minus E in a voltage V is given by this simple expression. And we know that the voltage produced by the nucleus
21:21
is the charge of the nucleus multiplied with the elementary charge divided by four pi epsilon naught and the radius. So if we use that,
21:42
then the potential energy is given by this expression. And if you plug in the number for the Bohr radius, we find out after some complicated math that it is exactly minus twice the kinetic energy. Therefore, if we add up the potential
22:02
and the kinetic energy, so we get minus twice the kinetic energy plus the kinetic energy is just minus once the kinetic energy. So we simply have to take this expression and put a minus sign in front of it and we get the total energy. And we can express that as the charge of the nucleus
22:21
square divided by the quantum number square times another constant, so-called Rydberg energy, where the Rydberg energy is given by this expression or in units of the fine structure constant, it is minus alpha square times the rest energy
22:40
of the electron divided by two. Or if you prefer a numerical value, the Rydberg energy is minus 13.61 electron volts. It is the ionization energy of the hydrogen atom, meaning if you pump this much energy into the electron,
23:01
it will get ejected out of the hydrogen atom. So in summary, using standing electron waves, we have quantized the radius and we find only those radii are possible where the lowest one is the Bohr radius with roughly 53 picometer.
23:23
We have quantized the energy with the ionization energy being about minus 13.6 electron volts. And we have quantized the angular momentum and because it is a standing wave, we have also solved the second problem of the Bohr Sommerfeld model.
23:42
Since the wave is a standing wave, it means that on average, the electron doesn't move at all. Or in other words, there's a wave going around this way and there's a counter wave going around that way. And so the sum of both waves results in no motion. And since the charge doesn't move on average,
24:01
it doesn't emit any radiation either. As long as those conditions are fulfilled. So this model isn't perfect. And we will talk about its inadequacies later, but it captures many of what is really happening.
24:23
You can also look at the UCI demo website and you find an applet there where you can get the whole thing very easily. Let's switch to emission here. So here we have hydrogen and if you want helium,
24:42
this one is helium or a neon, but it was mentioned sodium looks like this. Let's look at magnesium or iron. And the further we move down in the periodic table,
25:00
the more complex the structure of those spectra becomes. And you can also measure those lines in some sense. If you click on one of them, it will show you in units of angstrom or 10 to the minus 10 meters, what the wavelength of this particular wave line is.
25:25
So this one is 452 nanometers. So let's see whether we can actually understand those atomic spectra. So the way we make those lamps is we have
25:42
rarefied gases and we operate them in a high voltage discharge tube. And then the accelerated charges in there, they excite the atoms to radiation and they radiate with the characteristic frequencies
26:01
that they possess. And from the spectrometer, which is basically with the diffraction grading, what we just did, we can extract a spectrum characteristic to the element. One can use a continuous spectrum like the solar black body spectrum and pass it through gas, like gas surrounding the sun.
26:23
And you can then find an inverted image or an absorption lines coming from the various elements. And the little applet you can switch around between absorption spectra and emission spectra. So in other words, when the atom absorbs radiation,
26:40
it absorbs it at the same frequencies as it emits it. One can now go through the enormous task of making a catalog of the various lines spectra, emission spectra, for example. And here it is just for hydrogen.
27:01
So there are several so-called series, the Lyman series, the Baumer series, the passion series in various parts of the electromagnetic spectrum. And each one of them consist out of a whole bunch of lines. So if you, for example, zoom in on the Baumer series, then you can see it in more detail.
27:21
Eventually by, almost by trial and error, it was figured out the underlying rule behind it. So the difference, one can fit this to this following law, that the inverse of the wavelength is the constant R times one over N prime square minus one over N square.
27:44
For N equal one, we recover the Lyman series for N equals two, we recover the Baumer series. And for N equals three, we recover the passion series. So this constant R is called the Rydberg constant,
28:01
and you can determine it experimentally and it has this value. So with this law, we can explain the emission lines of hydrogen. So if we look at this Rydberg constant, here again is its value, and if we multiply it
28:22
with the Planck constant and the speed of light, then we get 13.61 electron volts. And so that is the same constant that we got for the ionization energy of the hydrogen atom. So in other words, the Rydberg energy is nothing but minus the Rydberg constant times H times C.
28:45
So the Rydberg constant is the ionization energy of hydrogen, and that means that we can sort of picture how this rule came into being. So we know that from when we calculated
29:04
the hydrogen atom, we know that the energy levels of the hydrogen atom, or of any other atoms, are given by Z squared times this Rydberg energy
29:22
divided by N squared. So we can sort of plot as a function of, we can plot the various energies, and we find the spacing like that,
29:42
and we can now understand the various spectral lines as the transition between those energy levels. And that is, because it's this transition, that is why we get the difference between those numbers N and N prime squared, because those refer
30:03
to the energy levels in the atom. And so if you make a transition from one energy level to another energy level, then simply by setting the energy equal to H times the light frequency, you can calculate the wavelength. And this is why the Rydberg constant
30:21
and the ionization energy are related by HC. That is simply the conversion from energy to wavelength. So let's look at that in an example. So let's determine the wavelength of the first Lyman line.
30:41
So that's the transition of level two to level one. And the question is also in what region of the electromagnetic spectrum is this located? To solve this, we set the energy difference to H times the frequency, where H is the Planck constant. So the energy difference is this difference
31:02
between second and first energy level, that is minus 3.4 electron volts, which we read off this graph here for N equals two, minus 13.6 electron volts, which is the lowest level or the ionization energy. And we get therefore 10.2 electron volts.
31:23
If we now take those 10.2 electron volts and form HC over this 10.2 electron volts, where HC is 1.240 electron volt micrometers, we get 122 nanometers. And that is in the ultraviolet portion
31:43
of the electromagnetic spectrum. Or we can directly go to the Rydberg rule and simply calculate what is one over lambda using the Rydberg constant, and then times one over one square minus one over four
32:01
or one over two square. And this gives us again, 121 or 122 nanometers. So let's do another example and determine the wavelength of the fifth Balmer line. That's the transition from level six to level two. And again, the question is,
32:21
what part of the electromagnetic spectrum is this in? So this time I'm just going to use the Rydberg constant. So this is the Rydberg constant. And I have to subtract one over four from this two square minus one over 36 from level six.
32:41
And I get 410 nanometer once I form one over this expression. And 410 nanometers is violet light. The fifth Balmer line is in the visible part of the spectrum. And let's also look at the maximum wavelength
33:01
that hydrogen in its ground state can absorb. We said that the rule is that the electromagnetic energy that the atoms absorb is at the same line as it is emitting it. So in other words, the longest wavelength is the smallest amount of energy. And since it's sitting in the ground state,
33:22
the smallest amount of energy is the one that will lift it from the slowest level to the first one higher up. So it's the one two transition again. And we already know what energy that corresponds to and what wavelength it corresponds to from the previous example. So it's 122 nanometers.
33:42
The question is, what would be the next smaller wavelength after that? And that would be the one three transition. And in this case, it's 13.6 electron volts minus 1.5 electron volts, which we read off this graph. And so that gives us an energy difference
34:01
of 12.1 electron volts, or wavelength of 102 nanometers. And finally, let's calculate the ionization energy of helium plus rather than hydrogen. And calculate also the longest wavelength to ionize helium plus.
34:22
So in this case, since we know that the energy levels are given by the z squared times the Rydberg energy divided by the quantum number squared. And we assume that for the ionization energy, we look at the ground state.
34:40
So n equals one, z equals two. And because of z equals two, this gives us four times the 13.6 electron volts of the Rydberg energy, or minus 54.4 electron volts is the energy level of the ground state. And therefore the ionization energy of helium plus
35:02
is 54.4 electron volts. And if you wanna know what wavelength is required to carry such an energy, we simply use the Planck constant for that. And we find that it's 22.8 nanometers. At the very end, I'd like to show you a simulation
35:26
of the hydrogen atom here using the Bohr-Sommerfeld model. And so right now, you're looking at the first orbit,
35:41
the Bohr radius, and you see the electron running around here in the particle picture. Or if we can look at the wave picture, and the wave looks a little bit crummy in this case because it's very hard to plot with just one wavelength superimposed on a circle.
36:04
However, if we go to a higher level, let's say we go to a level number six, so now we see that the electron wave oscillates around the orbit and it finally meets itself with the perfect phase condition.
36:24
And so this is how it looks like in the wave model. And if you look at the particle model, then since the electron is further out, it has a smaller velocity and it's moving slowly around the nucleus like that. So I told you that this model isn't quite the entire truth
36:46
and in part, this is because it's two-dimensional. Or everything happens in a plane and in reality, there's spherical symmetry around the nucleus. So we can look at a full simulation of the hydrogen atom,
37:07
which is a different Java app. And let's look at the sixth orbit again. And it is actually a standing wave in three dimensions now.
37:22
Oh, let me increase the position here. So the color in the picture shows you the phase of the wave. And so you see the three-dimensional wave
37:41
oscillating back and forth, and it is a standing wave because there are no surfaces which do not oscillate at all. This is the same level that we looked at before, but in the case of three dimensions, there are actually more quantum numbers than we discussed.
38:02
So we just discussed one quantum number, which gives us the main energy level. This is the quantum number N. In three dimensions, there are also two quantum numbers associated with angular momentum,
38:20
which are called L and M. So let's look at an increasing one for L. So the first one is usually called S, the spherically symmetric one. The second one is called P.
38:41
And now you see a more complex structure than we had for the simple spherical one. They are now preferred directions. And let me rotate the coordinate system a little bit. We can go to the next one and look at the next higher quantum number,
39:01
and it gets more complex. So it's called D. And finally, F. Those are actually quite pretty pictures, and one can use such calculations to understand the chemistry of various elements as kind of like a symphony of a choir of atoms.
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And that is a fairly neat analogy, I think, how atoms really look like in nature. So resonances and standing waves are at the very core of understanding atoms,
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even when you use the full calculation, not just the simplified assumptions. Thank you for your attention.