Modern Physics notes Paul Fendley fendley@virginia.edu Lecture 6 ...

Modern Physics notes
Paul Fendley fendley@virginia.edu
Lecture 6
• Size of the atom
• A digression on hand-waving arguments
• Spectral lines
• Feynman, 2.4-5
• Fowler, “Spectra”, “The Bohr atom”
The size of the atom
Let’s follow Feynman and use the uncertainty principle to make an argument about the size
of an atom. Classically, you can think of an atom as being a mini-solar system. There are
neutrons and protons in the middle (the nucleus). Orbiting the nucleus are the electrons, which
are much less massive. They don’t feel the force (the “strong” force) which holds the nucleus
together. Instead, they are bound by electric forces, just like the planets are bound by gravity.
Because electromagnetism is different from gravity, this picture ends up not working classically.
For example, an accelerating electron radiates classically, and so would emit all its energy and
crash into the nucleus. You need quantum mechanics to fix this up, and we’ll explain how.
Say the electrons moving in orbits of around radius a. If we do experiments, we’ll see that the
electron is in a different position each time we look. So its uncertainty in position ∆x is about
a. By the uncertainty principle, we then know that the uncertainty in momentum is about h/a.
(All of these numbers can be multiplied or divided by factors of 2 or 3 or π: what we are really
doing is dimensional analysis combined with quantum mechanics). This means the momentum
itself is probably around h/a. (If it were larger, we could then do scattering experiments with
light of momentum larger than h/a without disturbing the electron. We could then determine
its position to better accuracy then a, contradicting our original assumption that ∆x ∼ a.) If
its momentum is about h/a, then its kinetic energy is about (h/a) 2 /2M.
I just discovered that Feynman was a little sloppy. He writes h in the text, but when he
gets the numbers out, he actually doesn’t use h. You’ll remember that in our example, the
uncertainty was order h, but the careful derivation of the uncertainty principle says that the
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uncertainty is of order h/4π. For this reason, and others I’ll mention soon, physicists usually
use the constant which is merely Planck’s constant divided by 2π:
≡ h
2π = 1.055 × 10−34 J · s
The symbol ≡ means “is defined as”. (One reason is that it’s easier to say “h-bar” than “Planck’s
constant”.) So to get the same numerical answer as Feynman, let’s say the momentum is of order
/a instead of h/a. This is a hand-waving argument anyway!
So now let’s look at the energy of the hydrogen atom. The charge on the nucleus is just −e,
and the charge of the electron is e. (We define e to be negative; the answers will depend only on
e 2 , so this definition doesn’t affect the results, a useful check.) In addition to the kinetic energy,
there’s an electrostatic potential energy coming from the two charges being a distance a apart.
The total energy is therefore about
E =
2
2Ma −
e2
2 4πɛ 0 a
The minus sign in the second term is because the electron and the nucleus are attractive. Notice
that as a decreases, we lower the potential energy, but we also increase the kinetic energy (the
uncertainty principle means that cramming a particle into a smaller box increases its energy).
The value of a is determined by finding the minimum of this energy The reason is that in
the absence of any external sources of energy, it’s possible to get rid of energy by say emitting
light, but there’s no way of getting new energy. Thus the system will settle down into its lowest
energy state, and then be stable. To find this minimum value of energy, we find where dE/da = 0.
Precisely, at the minimum a = a 0 , we have
Solving for a 0 gives
a 0 = 4πɛ 0 2
m e e 2 =
0 = dE
da = − 2
+ e2
m e a 3 0 4πɛ 0 a 2 0
(1.055 × 10 −34 ) 2
(8.988 × 10 9 )(9.106 × 10 −31 )(1.602 × 10 −19 ) 2 m = .529 × 10−10 m
So the size of the hydrogen atom is about 10 −10 m, which is called an angstrom. (People today
tend to use nm = 10 −9 m instead of angstroms.) Using a simple hand-waving argument, we know
the approximate size of an atom! (note that this factor of 2π Feynman is sloppy on changes the
answer by a factor of 40, because it’s squared!)
To find the energy of this electron at this minimum, we can now plug our value of a 0 back
into the expression for E. After a little algebra, we get
e2
E = − = −13.6 eV
8πɛ 0 a 0
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The negative energy means that the electron has less energy when it is in the atom than it has if
it is not in the atom (E = 0 when a → ∞). This is good, because it means that electrons want
to stay in atoms, instead of moving away! This number 13.6 eV is called a Rydberg. It turns
out that we’ve been lucky. We’ve ignored factors of two and pi and so forth, so there’s no reason
this should be the exact answer. However, for the hydrogen atom this is the correct binding
energy: it takes one Rydberg of energy to pull the electron away from the hydrogen atom.
By the way, note that Feynman didn’t forget the 4πɛ 0 : he’s working in units of charge where
this is 1. In these units (called stat-coulombs), the charge of the electron is 1.6 × 10 −19√ 9 × 10 9
stat-coulombs, so you of course get the same answer at the end.
So quantum mechanics explains why atoms are stable. If the electron were to spiral into the
nucleus, ∆x would be smaller, and the uncertainty principle would require that the momentum
be larger. This increase in kinetic energy balances with the decrease in potential energy right at
a 0 , as we’ve shown.
Quantum mechanics also explains why we don’t fall through the floor. The atoms in our shoes
push against the atoms in the floor. When the atoms are squashed into a smaller space, their
momentum and hence energy must increase, because of the uncertainty principle. Remember that
energy wants to be minimized, so the force resists the compression. Classically, the squashing
would lower the potential energy, but without the uncertainty principle, there would be no
corresponding increase in kinetic energy.
A digression
Probably you’re not yet used to hearing “hand-waving” arguments like the one I just gave, so
let me digress and say what I mean by that. These are arguments where you’re not being very
precise, but rather are more along the lines of a plausibility argument. In some sciences, that’s
pretty much all you can ever do (biology is in an interesting transitional period). In physics, we
often can go much farther: we can write down precise mathematical equations which describe
the experimental reality. And in probably all of your physics classes so far were done in this
fashion. But all scientists, including physicists, give plausibility arguments all the time.
One reason is that when you’re doing research, it’s usually a good idea to try to know the
answer before you find it. Of course, the best researchers are prepared for when they don’t
get the answer they were looking for – often the greatest discoveries (e.g. the X-ray, and we’ll
see, the cosmic microwave background) were made this way. But knowing what to expect is
important in these cases – previously-unknown physics is necessary to explain why you didn’t
see what you thought you would see.
A second reason for giving plausibility arguments is that before you understand the math
describing what’s going on, you need to decided what’s going on. (Even good mathematicians
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do it this way, by the way.) So you have to take a plausibility argument the way it is intended:
to give you a short-cut to the right answer. As long as you don’t take it too seriously (i.e. get
upset if it ends up giving the wrong answer), it’s a useful thing.
In this case we know the hand-waving argument is essentially correct because we know how
to do the computations correctly, and will explain this later in the course. Trouble starts in
research when people don’t know the correct answer, but write papers anyway containing only
the hand-waving argument! Then you get to have fun yelling at your colleagues whose handwaving
arguments give a different answer than yours. You then all go to conferences in nice
parts of the world to continue the arguments, so we all win.
Spectral lines
The existence of photons was our first reason for the word “quantum” in quantum mechanics.
Light at a given frequency occurs in discrete lumps, “quanta”. These lumps can be seen easily
experimentally, although not quite by the naked eye (although your eye apparently would need
to be only about 10 times more sensitive to be able to see individual photons).
A second interesting thing also is apparent in the photoelectric effect experiment we did.
You’ll notice something about the light which was coming from the mercury vapor. Instead
of just giving out light of all colors, there were noticeable lines of distinct frequency (you can
check experimentally that these are of distinct frequency, and measure the wavelength, by doing
interference experiments). Since we now know that light frequency is proportional to photon
energy, this tell us that the mercury atoms are not emitting energy of arbitrary sizes. Instead,
the energy seems to be coming out in chunks. Thus again, where classically you might expect
something continuous, here you get something quantized.
Since the entire subject is named “quantum mechanics”, obviously the word quantum is
rather important. To give a first shot at understanding quanta, let’s first understand one way
the wavelength of light can be quantized classically. Say we put light into a box of length L, so
that the light cannot get out. This is called a standing wave: you can think of it as the light
just bouncing back and forth off of mirrors.
Let’s just do a one-dimensional box. The spatial dependence of light of a given wavelength
λ is either
e −2πix/λ or e +2πix/λ
Remember, we can always add two solutions of Maxwell’s equations in a vacuum and get another,
so we can add these two together with arbitrary coefficients ⃗ A and ⃗ B. In a equation:
⃗E = ⃗ Ae −2πix/λ + ⃗ Be +2πix/λ
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However, here we need to take into account the presence of the box. This says that A, ⃗ B ⃗ and λ
are not arbitrary. Since no light is escaping, there can be no light at the edges of the box, which
we put at x = 0 and x = L. This means E ⃗ = 0 at the edges. To satisfy this, we need to make
( πn
)
⃗E ∝ sin
L x
where n is some integer. Because sin nπ vanishes for any integer n, this ⃗ E behaves appropriately
at the edges. So note that we have ⃗ A = − ⃗ B, and more importantly
λ = 2L n
for some integer n. The wavelength is quantized in units of the size of the box.
This quantization shouldn’t be so shocking if you’ve ever played a stringed instrument. When
you attach the ends so that they do not move, they can only vibrate with certain wavelengths.
But now, let’s remember that electrons are waves, too. The wave function/probability amplitude
squared gives the probability an electron is in a given place. If we’re trapping the electron in
a box, then the wave function must vanish at the edges of the box. This yields the same
quantization condition that we saw for the electric field of light. This means that the momentum
(and hence the kinetic energy) of an electron is quantized! Thus the velocity of the electron can
only take on certain values, if the electron is trapped in a box. On this homework, you’ll check
what values of velocity this is, for typical box sizes.
Note that this is a little easier to write out if instead of using wavelength, we use the wavenumber
k = 2π/λ. Then our wave is ∝ sin(kx), and the boundary conditions require that
k = πn
L .
The two ways of writing this are obviously equivalent; you’ll notice that Feynman tends to use
k instead of λ. To get rid of the factors of 2π this introduces, people introduced the constant .
In terms of these new variables,
Once we start using , then it’s easier to to use
for frequency so that for photons
p = h λ = k
ω ≡ 2πν
E = ω
You should have seen ω before, when you were doing rotational motion. There ω is the angular
frequency as measured in radians. The point of all of these definitions is to absorb factors of 2π:
in terms of k, ω and , you don’t see any factors of 2π in E = ω, p = k and a wave looks like
e ikx−iωt .
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The fact that the wavelength and hence the momentum and energy of a particle in a box
are quantized is quite important. Let’s study the atom by making some more hand-waving
arguments. First think about the radial direction. The arguments above indicated that to
minimize the energy and hence be stable, the electron should remain within some radius a. This
doesn’t violate the uncertainty principle: we can’t predict precisely where it appears in this
region (although later we will be able to compute the probability amplitude). Thus this is like
confining a particle to a box of radius a, and so the radial momentum is quantized. The angular
component of the momentum is quantized for a similar reason. Think of an orbit at fixed radius
and momentum. Since we are now in quantum mechanics, this is a probability wave going around
the atom. Instead of demanding that the probability vanish at the edges, we demand that the
probability be periodic: if you go around once, the probability must be the same. Put another
way, one must have an integer number of wavelengths n around the orbit. Thus
where a is the radius of electron’s orbit.
nλ ∼ 2πa
Don’t take this argument too seriously, but the result is very important. The momentum
and energy of the electron in an atom are quantized. By similar reasoning, the angular
momentum of the electron is quantized as well. (The angular component of the ordinary
momentum is related to the angular momentum, but the two are not the same. How do they
differ?)
This now explains why we saw the distinct spectral lines in the mercury vapor. The vapor
in the lamp is fairly dilute, so the atoms are far enough apart so that they can be treated
individually. This means that the electrons in the atoms have distinct energy levels. To illustrate
this, physicists often draw the diagrams of the type you see in Fig. 2-9 of Feynman.
So how do discrete energy levels translate into what we see? When you heat a gas, or run
an electrical current through the gas, it puts energy into the system. This “excites” the atom.
This means that the energy you put in lifts the electrons out of the lowest energy levels and into
higher ones. But as we said before, the electron won’t stay there, if it can. It will fall down
into the levels with lower energy. So what happens to the now-excess energy. When falling
down levels, a photon is emitted! The energy of this photon is not arbitrary: when an electron
falls from energy E 1 to the lowest-energy state (the “ground” state) with energy E 0 , the photon
emitted must have energy E 1 − E 0 . But now quantum mechanics comes in again: we know that
this photon of fixed energy must have a fixed frequency. Thus this transition will emit a photon
of frequency
ν = E 1 − E 0
.
h
The discrete lines we saw in mercury arise from the various transitions in mercury.
Quantum mechanics therefore explains something completely obscure classically: the existence
of spectral lines. The existence of spectral lines with known frequencies was of vital
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importance in understanding relativity experimentally, and is still of vital importance in understanding
cosmology.
Typically, the spacing between these levels in an atom is on the order of an eV . This is
because the values of these energies are related to the size of the atom, not the size of the box
the mercury vapor is in. If they were related to the size of the system, you wouldn’t see the
quantization. Remember that k ∝ 1/L for a system in a box of size L, while k ∝ 1/a for the
atom. Since a/L is around 10 −10 , the levels due to the size of the box would be spaced much
closer together than the levels due to the size of the atom.
Another thing to note is that if the atoms are close enough together so that different atoms
interact with each other, the behavior can change dramatically. For example, in a metal, there
are electrons which are not bound to a given atom. This is why metals conduct current!
We’ll derive all this much more carefully later. The important point to take home is that we
now have seen two kinds of quantization unsuspected classically. Light comes in bundles called
photons, and electrons in atoms have discrete energy levels.
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