Download Equipment o Ppt. o Gas discharge tubes and hand

Phys 233
Day 8, Q8: Spectra
1
Tues 10/4
Wed. 10/5
Thurs 10/6
Q8 Spectra S5, S8
L5: Photoel & Best Fit, LC5 LR 3.5, 8; (Q3) Summer Researcher
Q10 Schrod Eqn S1, 2, 5
RE-Q8: Lab Notebook & Theory section
PL5; Quiz 4:Q6,Q7
RE-Q10
Mon. 10/10
Tues. 10/11
Wed. 10/12
Study Day
Study Day
L6: Electron diff. & linearizing, LC6 LR 3.2, 9; (Q4)
Study Day
Study Day
PL6; Lab Notebook & Analysis sect, Quiz 5: Q8Q10
HW5: Q8: S.2, S6, S8; Q10: S.5, S.8, R.1
Equipment
o Ppt.
o Gas discharge tubes and hand-held spectrometers
o PhET spectrum for switching between emission and absorption.
Q8 Spectra
Q8.1 Energy-Level Diagrams
Q8.2 The Spontaneous Emission of Photons
Q8.3 Spectral Lines
Q8.4 Absorption Lines
Q8.5 The Physics of Spin
Q8.6 The Pauli Exclusion Principle
Questions?
Need to revise my discussion of spin, it‟s
not that the z projection is really just
1/sqrt(2) while the full magnitude is 1/2,
it‟s that the projection is ½ while the full
magnitude is sqrt(s(s+1)) so sqrt(3)/2. See
footnote on page 161. Note: Q8X.6
discusses that imposing the field does
cause a precession around the field‟s axis.
Should probably point folks to Q8.5 in
supplementary reading for Q5
From Phys 231. We‟d gone into much of this in Phys 231, so hopefully this is a little familiar.
Now we‟ll focus a little more closely on it, and we‟ve got it in the context of a more detailed
development of our quantum mechanical picture.
From last time
We‟d connected the boundary conditions in a square well – that the wavefunction must go to 0 at
the edges, with possible wavelengths, thus momenta, and thus energies.
See ppt.
We‟d made similar arguments for the harmonic potential and for the electric potential via the
Bohr model.
Q8.1 Energy-Level Diagrams
Dig-up notes from 231 discussion of energy-level diagrams
Different sketches, plots, and diagrams are useful visualization tools for addressing
different questions. Last time, we drew some potential energy curves, superimposed total
energy levels, and used that to discuss kinetic energy‟s dependence on position. This
time, we don‟t really care how much energy is potential and how much is kinetic, just
how much total there is. So that‟s all we‟ll focus on.
Phys 233
Day 8, Q8: Spectra
2
Good model for: gas particles in container low-level inter-atomic bonds hydrogen atom
Q8.2 The Spontaneous Emission of Photons
Classical Picture. From classical electricity and magnetism, we know that oscillations in
electric/magnetic fields, i.e. light, are produced when a charged particle oscillates back
and forth. Of course, there‟s energy invested in / carried away by light, so if a charged
particle oscillates and thus generates light, it gives up energy in the process.
Quantum Picture. Quantum mechanics doesn‟t contradict any of this, we just want to
translate it into the lingo of quantum.
o Radiation from varying prob. Density. In our quantum mechanical
representation of objects, to say something is oscillating back would be
tantamount to saying that the most probable location where you‟d find the object
2
oscillates back and forth through time, that is, it‟s probability density,
,
o Energy eigenstates have static prob density, no radiation. As you saw in the
previous chapter, the probability density of an object in an energy eigenstate does
not do this – it‟s static in time, like a frozen standing wave (also referred to as a
„stationary state‟ because it‟s not going anywhere.) That‟s how we understand the
stability of atoms – electrons occupying the energy eigenstates not radiating away
their energy and falling into the nucleus.
o Spontaneous emission and decay from all other states. Of course, any other
state could be described as a superposition of multiple energy eigenstates, and as
you saw in the last chapter, the probability density of such superpositions does
oscillate back and forth in time. Thus an electron in anything but an energy
eigenstate does generate electromagnetic waves does loose energy to the light it‟s
producing, and does decay until it lands in an energy eigenstate.
o Perturbation & de-excitation from ‘energy eigenstates’ to lower ones. That‟s
not the complete story though. Consider an electron in a hydrogen atom; say the
Phys 233
Day 8, Q8: Spectra
3
electron‟s not in the lowest-energy eigenstate, but a higher one. Why doesn‟t it
stay there? If we truly had an isolated system, it could, but we don‟t, and so it
doesn‟t.
 Classical Picture of Perturbation. Classically, an analogy would be a
doll house with a miniature ball on the stairs. If you left the dollhouse
alone, the ball would stay put; however, if a semitruck went rumbling
down the road, the ball might tumble down the to the bottom step. Sure,
the top step of the stairs would be an equilibrium position for the ball, but
it‟s not stable in the face of little perturbations.
 Electron in higher energy-eigen state. Something analogous happens
when an electron sits in a higher energy eigen state of the atom. If you
can‟t completely isolate the atom from the rest of the universe, then it‟s
not really stable, not really an energy eigenstate when you take into
consideration the small fluxuations in electric and magnetic field that it‟s
always subject to. So, given time, it will eventually get enough of a nudge
to shed its „excess‟ energy and fall into a lower-energy state. This would
go on until the electron „landed‟ in the lowest available state. Even that
one would still be subject to the rattling, but there‟d be no where else for
the electron to go.
Q8.3 Spectral Lines
Q8T.5 - Imagine that light is emitted by electrons that could be either in a “box” of a
certain unknown length L or in a harmonic oscillator potential energy function with a
certain unknown spring constant ks. Imagine that the electrons emit only a single
spectrum line in the visible range. Measuring the wavelength of this line is sufficient to
determine which kind of potential energy function is involved, true or false?
o False. Heck, some specific pairing of L and ks values could lead to the same
transition energy. However, if you saw two transitions from the system, you
could then check whether the energy‟s being emitted were consistent with
transitions between one or another well‟s energy levels.
Now, as we saw in the previous chapter, a bound state „fits‟ only specific energy
eigenstates, and thus there are only discrete energy „levels‟ for a quanton to transition
between. Visualized with the help of an energy-level diagram, the quanton moves from
having the energy of one rung to another. If the quanton can shift between having only
specific energies, then the energy that it gives off in the process, carried away by light,
also only has specific energies / frequencies / colors.
o Naturally, when a quanton produces a photon or adsorbs one, the energy
acquired/given by the photon is the energy lost/acquired by the quanton.
Equanton
Eq. final Eq.initial
Eq.initial Eq. final

ph
o Q: The book gives an expression for the possible energies of photons emitted by a
quanton transistioning between energy states in a hydrogen atom, what would
they be for a quanton transitioning between energy states in a square well?
2 2
h 2 ni2 h n f
h2
E
E
ni2 n 2f

ph
q .initial
q . final
8mL2 8mL2 8mL2
o A photon can be characterized in terms of its energy, frequency, or wavelength,
but most devices for actually determining the energy of a photon make use of the
fact that, according to its wavelength, light gets refracted differently as it passes
across an air-glass boundary or diffracted differently as it goes through a
Phys 233
Day 8, Q8: Spectra
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diffraction grating. So these „spectrometers‟ are often calibrated in terms of
photon wavelength.
hf ph hc / ph
hc / ph
ph
ph

hc
ph
Eq.initial
Eq. final
Demo: Gas discharge tubes and spectrometers.
o Q: The book gives some examples for the case of the „quanton‟ being an electron
in the electric potential energy well of an atom. What‟s the expression for possible
wavelengths emitted by a massive quanton in a box?
hc
hc
ph
2
h
ph
ni2 n 2f
2
8
mL

8mL2 c
8 L2 mc 2
1
ph
2
2
2
hc
h ni n f
ni n 2f
Q8S.3 – Imagine that an electron is trapped in a box whose length is 0.80 nm. Draw a
spectrum chart (like figure Q8.2) showing all its visible (700nm< <400nm) emission
lines.
8L2 mc 2
1
2
hc
ni n 2f
ph
2
8 0.8nm 0.511 106 eV
1
2
1240eVnm
ni n 2f
2,110nm
1
2
i
n
n 2f
2
And


ph
h2
hc
ni2 n 2f
ni2 n 2f
2
8mL
8mc 2 L2
1) 1.76eV
ph ( 2
boarder line
1) 703nm
ph ( 2
ph
(3
2) 2.94eV
ph
(3
2) 422nm
0.588 eV ni2
n 2f

Q8T.2 – (work through on whiteboard). Imagine that the energy of the longestwavelength photon emitted by a quanton in a box of length L is Eph. What is the energy
of the photon with the fourth-longest wavelength emitted by a quanton in this box?
A. 3Eph
B. 4Eph
C. 5Eph
D. 7Eph
E. 8Eph
F. 9Eph
G. Some other energy (specify)
Just considering ph hc / ph , longest wavelength means smallest energy, from
ph
h2
ni2
8mL
n 2f or looking at the quanton-in-a-box energy diagram, that‟s when
h2 2 2
h2
2i 1 f 3
ni=2 and nf=1. So we‟re calling E ph
. Now the photon with
8mL
8mL
the 4th longest wavelength is that with the 4th smallest energy, looking at the diagram,
Phys 233
Day 8, Q8: Spectra
5
that would correspond to a transition from n=5 to n=4, or
h2 2
h2
h2
E ph
5i 4 2f
9
3*3
3E ph .
8mL
8mL
8mL
o Q: What‟s the expression for possible photon energies emitted by a quanton in a
simple harmonic potential?
k
E q.initial Eq. final  ni 12  n f 12  ni n f
 s ni n f

ph
m
hc
ph
ph
hc
1
 k s / m ni n f
2 c
1
k s / m ni n f
2 c
m
1
k s ni n f
Q8B.4 – The longest wavelength associated with vibrational transitions of the diatomic
H2 molecule is 4540 nm. What is the value of ћω for this molecule?
hc 1240 eVnm
0.273 eV
o First, ph
4540 nm
ph
o Next, we‟re told that this is for the longest wavelength, thus the smallest energy,
thus the energy of transition between two adjacent states, n=1.
 ni n f
 1 so, this is the value of 
o
0.273eV .
ph
Returning to the question of determining the energy structure from the light emitted,
Q8R.1 – When excited, electrons in a certain substance give off photons having a variety of
wavelengths. The longest two wavelengths emitted are 830 nm (in the near infrared) and 496 nm
(blue-green). Are the electrons in this substance bound by a spring-like interaction, or are they
trapped in a box? Carefully defend your response.
How are the longest two wave wavelengths related for one or the other type of energy well?
For the spring-like simple harmonic oscillator, the longest two wavelengths would correspond to
transitioning between adjacent energy levels and between next-nearest neighbors.
1 1
m1
m 1
/
2 In
2 c
, next longest ph..next.longest 2 c
, so ph.longest / ph.longest
ph.longest
1 2
ks 1
ks 2
the problem given to us, we have 830nm/496nm=1.67, nope, that‟s not what emitted this light!
How about for the particle in a box, then the longest wavelength comes from the n=2 to n=1
transition, and the next longest comes from the n=3 to the n=2 transition.
ph.longest
So
ph.longest
8L2 mc 2
1
2
hc
2i 12f
/
ph.longest
8L2 mc 2 1
,
hc
3
ph..next.longest
8L2 mc 2
1
2
hc
3i 2 2f
8L2 mc 2 1
hc
5
1 1
/
5 / 3 167 , that‟s a match!
3 5
Q8.4 Absorption Lines
Demo: PhET spectrum – emission to absorption.
Absorption as Resonance. Moore makes a great point that simply applying conservation of
energy isn‟t sufficient to understand why electrons don‟t often accept more energy than they
need and then simply re-emit a weaker photon to shed the excess energy and get them into
energy states. The photon constitutes an oscillating electric and magnetic field and that would
Phys 233
Day 8, Q8: Spectra
6
contribute to the electron‟s potential, making it oscillate around the unperturbed configuration at
the frequency of the photon. That means that the wavefunction oscillates slightly at the light‟s
frequency and so does the amplitude squared – the probability density, this must have the form
2
( x)
f ( x) g ( x) cos light t .
Now, if the frequency of the light corresponds to the energy difference between the electron‟s
original energy state and another one,
En Em
,
light

then the electron‟s wavefunction must be expressible as a superposition of those two states,
( x) a En ( x)e Ent /  b Em ( x)e Emt /  (we‟re essentially running backwards through the logic
of one of the RE problems you did for Q7)
and there‟s a reasonable probability that the electron will absorb the photon‟s energy and settle in
the higher of the two states. Alternatively, driving it at a frequency that doesn’t correspond to a
transition of energy states makes the electron‟s wavefunction a superposition of many more
energy eigenstates,
( x) a En ( x)e Ent /  b Em ( x)e Emt /  c El ( x)e Elt /  d Ek ( x)e Ekt /  ...
each more weakly represented, and there‟s less chance that it will settle anywhere but its
dominant original state.
Stimulated emission. If you think about it, an almost identical argument could be made for
getting an electron to transition down from one energy eigen state to a lower (if vacant) state.
That‟s known as Stimulated emission and is key to how lasers work.
Q8.5 The Physics of Spin
Finally in this chapter, the text gives a nice discussion of Spin and how it interacts with the
magnetic field in the Stern-Gerlach machine. Reading over it points out to me that I‟d
misinformed you on one point – while it is true that the spin never fully aligns with any measured
axis, and so the true magnitude is greater than the projection on an axis that one would measure,
1
the magnitude of the spin‟s projection on an axis is S z s where s
2 , and the full
magnitude is S
 s(s
1)
 3 / 2 . So, I‟d remembered that there was a root, and that the
full magnitude was larger than the projection, but I‟d gone the wrong way.
That the projection of the spin angular momentum would be half integer corresponds to saying
that the probability density of the electron‟s wavefunction must be symmetric upon 360°
rotation. This seems plausible but since it‟s the wavefunction, and not just its square (probability
density) that must match itself upon 360° rotation for orbital angular momentum leaves
something lacking in the argument.
Q8.6 The Pauli Exclusion Principle
What follows is more of a logical proof than a satisfying explanation for why quantons with halfinteger spin, fermions, can‟t share a state, but quantons with integer spin, bosons, can. If you‟re
interested, read on, if not, feel free to skip to the problems at the end, where we‟re simply using
this fact.
The argument goes something like this, we‟ll first consider swapping two quantons between
otherwise identical states in two different potential wells, and then we‟ll be ready to ask about
Phys 233
Day 8, Q8: Spectra
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their being in the exact same state in the same well. We‟ll see that integer-spin quantons can, but
we get a logical inconsistency when we try to have half-integer-spin quantons cohabitate in the
same state. Here goes:
Say you have two quantons, A and B, in each in a different potential energy well, 1 and 2, then
the wavefunction that describes your combined knowledge of both is
( A @1 & B @ 2)
A (r1 )
B (r2 )
Now, if they‟re the same kind of fundamental particle, there‟s no measurable difference between
having A in well 1 and B in well 2, or A in well 2 and B in well 1. The wavefunction for the
latter would be (not the swap of A and B subscripts)
( B @1 & A @ 2)
B (r1 )
A (r2 )
Now, as you‟ve noticed with position, measurements depend upon the probability density of a
wavefunction, not the wavefunction itself, so it isn‟t actually necessary that
( A @1 & B @ 2)
(b @1 & A @ 2)
A ( r1 )
B ( r2 )
B ( r1 )
A ( r2 )
For the two situations to be measurably indistinguishable, just as long as
( A@1 & B @ 2)
2
(b @1 & A@ 2)
2
2
2
(r1 )
B (r2 )
B (r1 )
A (r2 )
Of course, that means you just need them to be equal to within a + or – sign:
( A@1 & B @ 2)
(b @1 & A@ 2)
A ( r1 )
B ( r2 )
B ( r1 )
A (r2 )
As I‟ll now argue, whether it‟s the + sign or the – sign depends upon whether the quantons have
integer or half-integer spin. Now, let‟s think about a mechanism for actually swapping quantons
A and B between wells 1 and 2 that involves rotations; I‟ll illustrate them as balls with axes so
we can keep track of orientation, alternatively, visualize coffee mugs with their handle‟s sticking
off on one side. I‟ll also keep track of it in my notation with a second entry to indicate
orientation.
A
A
(r1 ,0 )
B
(r2 ,0 )
A
B
Well 1
Well 2
Pick them both up and rotate to place them in each other‟s well
B
A
(r1 ,180 )
A (r2 ,180 )
Swapped and 180° rotated
B
Well 1
Well 2
Phys 233
Day 8, Q8: Spectra
8
Now, if you rotate them between the two wells, you end up with them being rotated 180°. So I‟ll
add in a final twist to my operation, and spin them where they sit by another 180°.
B
A
Well 1
Well 2
(r1 ,360 )
A (r2 ,360 )
Swapped and 360° rotated
B
One can actually couch these operations in mathematical terms but the point is that I can swap
the two quantons in this manner. Now, for quantons with integer spin, their wavefunctions are
symmetric upon rotation of 360°, that is
for integer spin - bosons
B (r1 ,360 )
A (r2 ,360 )
B (r1 ,0 )
A (r2 ,0 )
So, for such particles, this way of swapping has no other effect than just swapping their
positions. And if they are truly indistinguishable quantons, then swapping them in this way
means not only can you not measure a difference when you swap quantons this way, but even the
wavefunctions are unchanged (not just the square of the wavefunctions)
A (r1 )
B (r2 )
B (r1 )
A (r2 )
(1)
for integer spin - bosons
A @1 & B @ 2
B @1 & A @ 2
However, as the book notes, quantons with half integer spin have wavefunctions that are
antisymetric upon 360° rotation (only their magnitude squared is symmetric upon 360° rotation).
So,
B (r1 ,360 )
A (r2 ,360 )
B (r1 ,0 )
A (r2 ,0 ) for half-integer spin - fermions
So, for such particles, this way of swapping the quantons also flips the signs of their
wavefunctions. And if they are truly indistinguishable quantons, then swapping them in this way
means that while there‟s no measurable difference, the wavefunctions do have flipped signs.
A (r1 )
B (r2 )
B (r1 )
A (r2 )
(2) for half-integer spin - fermions
A @1 & B @ 2
B @1 & A @ 2
Again, our condition is that swapping the quantons has no measurable effect, and that‟s true
since the change in sign of the wavefunctions gets squared away when we take magnitude square
which is a step in predicting a measurement.
Now here‟s the kicker. All this talk about having our quantons in the same state but for two
different wells was just to get us to this point; what we‟re really interested in is putting two
quantons in the same state in the same well. Now we‟re ready to consider that. Say the two
quantons both reside in the same well, say well 1:
Phys 233
Day 8, Q8: Spectra
9
So, now equation (1), which applies to integer-spin quantons reads
A & B @1
B & A@1
(r1 )
B (r1 )
B (r1 )
Which is an obviously true equality.
A
A
(r1 )
Now equation (2), which applies to half-integer spin quantons reads
A & B @1
B & A@1
(r1 )
B (r1 )
A (r1 )
B (r1 )
But comparing the left and right hand sides, we‟ve got something equals its own negative!
That‟s only true for 0. So we must conclude that
A & B @1
B & A @1 0
That means the probability density for two quantons with half-integer spin to be in the exact
same state is universally 0, i.e., there‟s 0 chance of it‟s ever being observed - it ain‟t gonna
happen.
Two quantons with half-integer spin can’t occupy the exact same state.
A
Q8T.8 -We can model cyanine dye molecules as if they contained N electrons trapped in a
“box,” like the potential energy function, where N is some even integer. The energy of a single
quanton in a box is given by En = E1n2, where n is an integer and E1 is the energy of the ground
(n = 1) energy level. If for a given molecule there are four electrons trapped in the molecule‟s
“box,” what is the lowest possible total energy of these electrons?
A. Etot = E1
B. Etot = 4E1
C. Etot = 8E1
D. Etot = 10E1
E. Etot = 30E1
With the Pauli exclusion principle, we can have 2 electrons in the ground state and 2 in
the first excited state. So, Etot = 2E1 + 2E2 but E2 = 22E1, so we‟ve got Etot = 2E1+8E1 =
10E1.
Q8S.9 – The potential energy function for a proton or neutron in an atomic nucleus can be
crudely modeled as being a “box” that is about 4 fm = 4 x 10-15 m wide. Imagine that we put 12
neutrons into such a box. Find the minimum total energy of the 12 neutrons in this case. (Hint:
Only two neutrons can occupy each energy level. Explain why.)
Approximately, how much lower would this minimum total energy become if we changed one
half of the neutrons to protons?
For the first question, we‟ve got 2 neutrons in each energy level thanks to their being fermions
(having half-integer spin) and the Pauli Exclusion Principle. So, with 12 neutrons, we‟re filling
up through the 6th level:
Etotal 2 E1 2 E2 2 E3 2 E4 2 E5 2 E6
Etotal
2 E1 2 E1 2 2
2 E1 32
Etotal
2 E1 1 4 9 16 25 36
2
As for E1, E1
h
8m p L2
2 E1 4 2
2
hc
8 m p c 2 L2
2 E1 5 2
2 E1 6 2
182E1
1240eVnm
2
12.8MeV
2
8 939.57 106 eV 4 10 6 nm
So, the total amount of energy (measured relative to the bottom of the well) is 2,330MeV.
Phys 233
Day 8, Q8: Spectra
10
If half of these were protons, since they have roughly the same mass, and neglecting the
electric repulsion (which isn‟t so bad since we‟d probably pack these so protons weren‟t right by
each other), we‟d essentially have two boxes, one for protons and one for neutrons, and so we‟d
only have 6 quantons in each – 2 boxes of six would have
Etotal 2 2 E1 2 E2 2 E3
2 E1 2 2
Etotal
2 2 E1
Etotal
4 E1 1 4 9
so only 4/13ths as much energy.
2 E1 32
56E1
716.8MeV