WEBVTT

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PROFESSOR: Hydrogen atom,
the first thing to do

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is to describe the
potential V or r.

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And it would be in the units
that we'd like to use minus e

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squared over r.

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e is the charge of the electron,
and the electron and proton

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with the same charge.

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The potential
energy is negative.

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And that sign you should
be comfortable with.

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It's suggesting that you
go closer and closer.

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You're going down and
energy is favored.

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The two particles to go
on top of each other.

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Immediately we want to
generalize it a little so

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that it's hydrogen like atoms.

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And we'll put a Z here, assuming
that the proton, instead

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of the program, there's
a nucleus with z protons.

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And there probably would
be about z electrons,

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but we're worried now
about just one electron.

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And in this case, a
charge in the proton

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is ze multiplied by the charge
of the electron gives you

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this as the potential energy.

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And this is advantageous.

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You sometimes have
an alpha particle

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that captures an electron,
and then there's two protons.

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And you don't want
to solve this, again,

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from the beginning.

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So just put the z, and
that's what we will do.

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So a few numbers.

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We've done some of
these numbers before,

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but the Bohr radius and one way
of calculating the Bohr radius

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is to just think of units
and think of energy.

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Energy goes like h
squared over ma squared.

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This has units of energy.

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You remember p squared over 2m.

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And p is h over distance.

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So if you have a
Bohr radius a 0,

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this quantity has
units of energy.

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But a potential
has units of energy

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and we just put e
squared over a0.

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So this is a consistent equation
between two quantities that

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have units of energy
from which you

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can get the unique
length that has

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units of energy, which is a0.

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And a0 is h squared
over me squared.

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It's a very simple and nice
constant is the Bohr radius.

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Intuitively one thing
that should remember,

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e is appearing in
the right place.

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And you could imagine if
the strength of electricity

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was weaker and weaker,
like setting e going to 0,

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the atom would become
bigger and bigger.

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It would just not
be able to hold it.

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So it's reasonable to
expect this to happen.

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So at this moment,
we can calculate what

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this is, at least, estimate it.

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And for that, we multiply by
a c squared e squared over m--

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mc squared.

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And then we recall
that e is squared

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over hc is about 1 over 137.

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So we write this as
hc over e squared

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over hc times mc squared.

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Now in here what mass is
the mass that we should put.

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I will not be all that careful.

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It should really be
the reduced mass.

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But in differs by a factor
of one part in 1,000 or less

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even from the mass
of the electrons.

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So I'll put just the
mass of the electron.

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This it's about
197 mev for Fermi.

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This is 1 over 137.

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And for the electron is 0.5
times 10 to the 6 ev 0.5 mev.

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I won't run the numbers.

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The answer is about
0.529 angstroms,

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which is about 53 picometers.

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Angstrom is 10 to
the minus 10 meters.

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Picometer 10 to the
minus 12 meters.

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So that's a length scale
you've seen several times.

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There's an energy scale
that is famous to.

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And that's e squared over a0.

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Because energy comes
here, the energy scale

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is e squared over e0.

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So you can substitute what a0
is because you know it already.

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And you get e to the
fourth over m over h

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squared, which is e to
the fourth over h squared

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c squared times Mc squared.

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And you see it's kind of nice to
see these quanities appearing,

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because here you have e
squared over hc squared times

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mc squared.

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So the typical energy
of the hydrogen atom

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is the fine-structure constant,
sometimes called alpha squared

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times an energy.

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And what energy's
available in the problem?

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The rest energy of the electron.

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So if the bound state
energy should be something,

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it should be a number
proportional to the energy

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that the problem already has.

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And the problem has one
energy, the rest energy

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of the electron.

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So it's not surprising.

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So it's one over 137
squared times 511,000 ev.

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And that's about 27.2 ev.

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And the reason this
may sound familiar

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is because the true ground state
energy of the hydrogen atom

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is this number divided
by 2, which is 13.6 ev.

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So of course, you would
not know at this stage,

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because you're
just doing numbers.

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I may remind you of things did
a long time ago in this course.

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You calculated a couple
of other constants.

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And you showed that alpha--

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again, the
fine-structure constant

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comes a0 was the so-called
Compton wavelength

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of the electron, with a bar.

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So it's h bar over mc.

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Remember, that the
Broglie wavelength

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is h over the momentum.

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The Compton wavelength
this h over mc.

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And the Bard can wave
at the h bar over mc.

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And that's what alpha
times a zero is.

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And that quantity
is about 400 Fermi.

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It's already much smaller
than the Bohr radius.

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It's smaller by 137
from the Bohr radius.

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And then if you do
alpha squared a0,

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that actually was the
classical electron radius.

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So you must divide
this by 137, again,

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and it gives you
about 2.8 Fermi.

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And what you can remember
is that the size of a proton

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is about a Fermi.

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So that gives you a
little bit of intuition.

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So those are the basic
numbers that we begin with,

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with the hydrogen atom.

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It gives you a scale of what's
going on, the size of an atom,

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and the energies that
we're supposed to get.