Download Physics 228 Today: April 4, 2013 Do we fully

Physics 228 Today: April 4, 2013
Do we fully understand the hydrogen atom? No.
Ch 41: 4-7:
Zeeman effect, electron (and proton) spin, manyelectron atoms, ...
Notes for next week: regular classes, except
Thursday April 11: no lecture, exam review
session (optional). Exam 10:00 - 11:20 PM in Hill
114 and ARC103.
Website: Sakai 01:750:228 or www.physics.rutgers.edu/ugrad/228
Thursday, April 4, 2013
Hydrogen Atom Review: 3 Quantum Numbers
Principle QN n = 1, 2, ...: states of same n are degenerate.
Angular momentum QN l: L =
[l(l+1)]½ħ,
l = 0, ... n-1
magnetic" or "z-component" QN ml = 0, 1, ... l
Shell
K
L
"
M
"
"
N
etc...
Thursday, April 4, 2013
n
1
2
2
3
3
3
4
l
0
0
1
0
1
2
0
ml
0
0
0, ±1
0
0, ±1
0, ±1, ±2
0
notation
1s
2s
2p
3s
3p
3d
4s
semiclassical
picture
What happens when you put the
Hydrogen atom in a magnetic field?
Here is the classical E&M physics analysis:
If the hydrogen atom has a magnetic moment, then there is an
interaction energy U = -μ.B. And a torque τ = μxB and precession.
From our study of E&M, if the electron moves around the proton,
there is a current, and thus a magnetic moment: μ = IA.
Classically, if we have an orbit of radius r and speed v, then the
period of the orbit is T = 2πr/v, and the current, the charge per unit
time passing some point, is I = e/T = ev/2πr. Thus, the magnitude of
the magnetic moment is μ = (ev/2πr)(πr2) = evr/2.
Remembering that the angular momentum is L = |rxp| = mvr, we
obtain μ = (e/2m)L. The ratio μ/L is called the gyromagnetic ratio (g).
Thursday, April 4, 2013
The Bohr atom in a magnetic field
The Bohr atom has circular orbits with the constraint that
2πrn = nλn. This leads to pn = h/λn = nh/2πrn = nħ/rn and
Ln = mvnrn = pnrn = nħ.
Thus μn = (-e/2m) Ln = (-e/2m) nħ. So Un = -μn.B = (eB/2m) nħ.
The energy shift of the nth level is proportional to n.
But this is not what is found. The spectrum is more
complicated, because in hydrogen one has multiple L values
for fixed n (except n=1).
Thursday, April 4, 2013
The Schrödinger Eq. atom in a magnetic field
How do we make the connection between the
S.E. atom and the magnetic moment? It is not
so obvious in QM how the electron actually
moves around in the atom and what to do
about a current and area.
We use the classical equations: μn = (-e/2m)
Ln, and U = (eB/2m) nħ. The combination
eħ/2m is common, and we define the "Bohr
magneton" as μB = eħ/2m = 5.79x10-5 eV/T.
Now the s states have L=0, so μ = 0, and
there is no effect from the magnetic field.
For the d state shown to the right, μ is
aligned to L, is quantized, and has several
possible projections, so U = -μ.B is quantized,
and the level is split into (2l+1) sub-levels.
Thursday, April 4, 2013
The Schrödinger Eq. atom in a magnetic field II
We have U = -μ.B = -μzB, with μz = (-e/2m)Lz =
-ml(eħ/2m) = -mlμB.
Thus U = mlμBB. The (2l+1) sub-levels have even
separations / splittings of μBB. The magnetic
field removes the degeneracy.
Pieter Zeeman discovered this in 1896.
But so far, the m=0 states of the same n have
equal energy.
Thursday, April 4, 2013
iClicker
You put an n=3 Hydrogen atom in a magnetic field. You
see photons corresponding to n=3 to n=1 transitions.
How many lines do you see?
a) 1
b) 3
c) 5
d) 8
e) 9
Thursday, April 4, 2013
At this point since the splittings are
all the same it might seem there are
5 distinct energies, but we will see in
the next few slides that 2 of these
are not allowed.
Selection Rules
Consider the splitting of a d state into 5
levels as the B field is increased.
Now consider a transition from, e.g., a 3d
to a 2p state, which has split into 3 levels.
How many different photon energies are
possible? It seems we have (Ed ± (0,1,2)μBB)
- (Ep ± (0,1)μBB) = (Ed - Ep) ± (0,1,2,3)μBB.
But there are not 7
possible energies. The
photon can only carry
of 1 unit of angular
momentum, so we are
limited to Eg = (Ed Ep) ± (0,1)μBB. Only
certain transitions are
possible.
Thursday, April 4, 2013
Allowed transitions
md
mp
Δm = md-mp = mγ
2
1
1
2
0
2
2
-1
3
1
1
0
1
0
1
1
-1
2
0
1
-1
0
0
0
-1
0
1
The 6 md < 0 transitions are left as
an exercise for those interested.
Thursday, April 4, 2013
The table shows the change
in the z-component of the
orbital angular momentum.
Since angular momentum is
conserved, the photon must
carry at least an angular
momentum l = Δm, since its
z-component of angular
momentum must equal the
change in the atomic z
component. (Angular
momentum, like momentum,
is conserved in each
component.) But the photon
is spin 1, so the red-colored
lines are forbidden. Only 3
of the 7 possible energies
are allowed.
The Stern-Gerlach Experiment
In the magnet the energy of an atom with a magnetic moment is
U = -μzB = mlμBB. But if the magnetic field is not constant,
classically their is a force F = ∇μ.B = mlμB(dB/dz). So a bunch of
atoms coming from the oven will separate into 2l+1 separate
bunches if there is a B field with enough spatial variation.
Thursday, April 4, 2013
When you run the experiment, you can get an even number of blobs.
Thursday, April 4, 2013
The Stern-Gerlach Experiment and
More on Line Spectra - iclicker!
For odd Z nuclei, not only do we get an even number of blobs, but
we also can get extra splittings of spectral lines in magnetic fields
not explained by the normal Zeeman effect - the book has some
confusing pictures on this. With the angular momentum, we have
an integral magnitude l, and the degenerate states are split into
the 2l+1 levels by a magnetic field. But 2l+1 is always an odd
number, so how can we get an even number of blobs in the Stern
Gerlach experiment?
a) The data are wrong - there is an invisible line off to the side.
b) You don't see the m=0 line, since there is no force on those
atoms.
c) 2l+1 can be even for integral l - what's the problem?
d) l can be half-integral, not just integral.
e) the electron has an intrinsic spin of 1/2.
Thursday, April 4, 2013
Electron spin
Subatomic particles have intrinsic spins, that can be integral
(0 for the pi meson, 1 for the photon, rho meson, ...) or halfintegral (1/2 for the electron or neutrino or proton or
neutron, 3/2 for the Delta resonance, ...)
The classical analogy almost always made is to a top or the
earth, which have rotational angular momentum, in addition
to, in the case of the each, orbital angular momentum of
revolution around the sun. This is a bad analogy as the
particles are not actually rotating.
The algebra of spin is:
sz = ±½ħ = msħ,
S = √((1/2)(3/2)) ħ = √(¾) ħ, and
μz = -2.00232 (e/2m) sz.
Thursday, April 4, 2013
The algebra of spin is:
Huh? 2.00232?
sz = ±½ħ = msħ,
S = √((1/2)(3/2)) ħ = √(¾) ħ, and
μz = 2.00232 (-e/2m) sz = g (-e/2m) sz. = gμB/2.
In non-relativistic quantum mechanics, there is not natural
explanation for spin; it is added in an ad hoc manner.
Two different ways of combining relativity with quantum mechanics
were developed in the 1920s (now combined into QFT). One (Klein
and Gordon) describes particles with integral spin. The other (Dirac)
describes particles with half-integral spin. In a first approximation,
the electron has a g factor of 2. You might say spin contributes
twice as much to the magnetic moment as orbital angular
momentum. In modern QED theory higher order corrections can be
calculated, these make the g factor of point particles (electrons)
slightly different from 2, and predict it to about 1/1012.
Thursday, April 4, 2013
Anomalous Zeeman Effect
We now have a spin-½ electron. As a result, each of our atomic
energy levels can have two electrons, one "spin up" and the
other "spin down". Thus, even for s orbits, the spin of the
electron can give rise to a splitting between the spin-up and
spin-down states in the atom, and we can get doublets of lines.
Note that with spin 1/2, but g = 2, the coefficient of B above is
the same as for orbital angular momentum.
Thursday, April 4, 2013
Spin-Orbit Coupling & fine structure
In the rest frame of the electron - thinking semi-classically the proton is revolving about the electron. Since the proton is a
charged particle, it generates a magnetic field at the position of
the electron. (In the view with the proton at rest the electron
generates a B field at the position of the proton.)
As a result, there is a magnetic correction to the energy U = μ.B where B is proportional to L. Since for the electron μ is
proportional to S, we can rewrite the energy as proportional to
L.S; this is preferred since we know L and S of the orbit, but not
the B field. This is then referred to as the spin-orbit force. It
leads to a splitting of spin-up and spin-down electron states even
in the absence of a magnetic field!
These splittings related to the magnetic interactions are referred
to as "fine structure".
Thursday, April 4, 2013
Proton Magnetic Moment & hyperfine
structure
Note also the proton has an intrinsic magnetic moment. Like the
electron, the proton is a spin-1/2 particle, and its magnetic
moment is given by μ = g (-e/2m) sz. But here the mass is the
proton mass, which is about 1800 times the electron mass, and
the magnetic moment is much smaller. As a result, we use instead
μ = gμN/2, where μN is the "nuclear magneton". Finally the
proton is a particle with structure, so g = 5.586, it is not close to
2. This is because the proton has a complex internal structure.
Nuclei, made up of protons and neutrons, tends to have magnetic
moments of similar size to the protons. The splittings resulting
from interactions of the nuclear magnetic moments with the B
field from the electron lead to small splitting, referred to as
"hyperfine structure" since they are much smaller than the fine
structure splittings resulting form the electron magnetic moment
interactions.
Thursday, April 4, 2013
Adding Angular Momenta
Classically, if we add angular momentum vectors, depending on
their relative orientation, we can get any result from the sum of
the two vectors to the difference.
This idea holds in quantum mechanics. For a single electron for
example, we would write J = L + S, the total angular momentum
is the orbital angular momentum plus the spin angular
momentum. But the angular momenta are still quantized. Thus j =
|l ± ½|. (We use the magnitude, since otherwise for l = 0, we
could get j = -½.) Or J = √(j(j+1)) ħ.
Thursday, April 4, 2013
Summary
In the simple picture, we have levels with quantum numbers n, l,
ml (and ms) that are degenerate.
We find that an external magnetic field removes the degeneracy,
splitting a level with angular momentum l into (2l+1) levels.
With the spin of the electron, we find an external field leads to
doublets for (spin up vs spin down) even for s states.
Even without an external field we have magnetic effects:
The magnetic interaction between magnetic fields from protons
and electron magnetic moment leads to an L.S force and "fine
structure".
The magnetic interaction between the smaller proton magnetic
moment and the magnetic field from the electron leads to a
similar "hyperfine structure".
Thursday, April 4, 2013
Many Electron Atoms
So far we have considered mainly
the hydrogen atom. When we go
to heavier atoms, we can largely
treat the nucleus as being a point
object with some mass and some
charge Ze. But the electrons in
orbit about the nucleus interact
with each other. You can see that
if one electron is generally outside
another, then in effect the binding
charge of the nucleus is reduced
by 1.
Thus we cannot simply say that
the energy of an orbit n becomes
En = -13.6Z2/n2 eV.
Na: Z=11, En = -1646/n2 eV
But E(3s) = -5.14 eV
E(3p) = -3.035 eV
E(3d) = -1.521 eV
E(4s) = -1.947 eV
Thursday, April 4, 2013
Pauli Exclusion Principle
For many electron atoms, we cannot simply use the 1/r Coulomb
potential, but the potential is mainly U(r), independent of θ and φ.
As a result, the characterization of the angular momentum states
still holds. We still have n>0, 0≤l<n, |ml|≤l, and ms=±½.
The Pauli exclusion principle is important for understanding atomic
structure: for electrons (and other spin-½ particles, or more
generally Fermions) each state can hold only 1 particle. No two
electrons can have the same 4 quantum numbers.
For example: the more bound orbits fill first, and when a major shell
is filled there is a large energy gap up to the next orbit. This leads
to the non-reactive "noble" gases. There are techniques to do
actual calculations; they are too complicated for this class.
Also: atoms with similar configuration of outer electrons tend to be
similar in their chemical properties - this leads to the Periodic
Table.
Thursday, April 4, 2013
Closing iClicker
You have one electron in the l = 1 2p state, and a
second electron in the l=2 3d state.
What are the possible total angular momenta of the
two electrons? (The answers are j in units of ħ, not
J = √(j(j+1)) in units of ħ.)
a) 1, 2, 3
b) 0, 1, 2, 3, 4
c) 3
d) when you add the angular momenta of two
electrons, it can be anything between the limits - it
is continuous, not quantized
e) none of the others are right, since the electron is
spin ½. The answer has ½-integral numbers.
Thursday, April 4, 2013
Closing iClicker
You have one electron in the l = 1 2p state, and a
second electron in the l=2 3d state.
What are the possible total angular momenta of the
two electrons? (The answers are j in units of ħ, not
J = √(j(j+1)) in units of ħ.)
a) 1, 2, 3
b) 0, 1, 2, 3, 4
We add l=1 to l=2 to s=½ to s=½; they
can add to numbers from 0 to 4.
c) 3
d) when you add the angular momenta of two
electrons, it can be anything between the limits - it
is continuous, not quantized
e) none of the others are right, since the electron is
spin ½. The answer has ½-integral numbers.
Thursday, April 4, 2013