Making Light

1
Making Light

• We can also make light by exciting atoms.
• From experiment, we see that different atoms emit
  different light. But each type of atom emits a
  very specific set of wavelengths called a
  spectrum.
• The hydrogen atoms emit three visible
  wavelengths one in the red, one in the
  blue-green, and one in the violet.

2
Making Light

• We need a model of the atom that will explain why
  atoms emit only certain wavelengths.
• First of all, what is the size of a typical atom?
  Lets take water (although that is a molecule,
  we know a lot about water its mass density 1
  gm / 1 cc, it is H2O so it has 18 grams/mole, and
  we know Avagadros number 6.02 x 1023
  molecules/mole.

3
Making Light from Atoms

• (1 cc / 1 gm) (18 gms / mole) (1 mole / 6.02
  x 1023 molecules)
• 18 x 10-6 m3 / 6 x 1023 molecules
• 3 x 10-29 m3 30 x 10-30 m3 .
• Therefore, the size is about (30 x 10-30 m3)1/3
• 3 x 10-10 m. Thus the size of an atom should
  roughly be about 0.1 nm .

4
Making Light from Atoms

• Now that we know the size of an atom, how much
  mass does the atom have?
• From the mass spectrograph, we know that the mass
  of an atom comes in integer values of 1 amu
  1.66 x 10-27 kg. (In fact, this is important in
  getting Avagadros number!)

5
Making Light from Atoms

• Now that we know the size and mass, what parts
  does an atom consist of?
• We know that the atom has electrons of very small
  mass (me 9.1 x 10-31 kg), about 2000 times
  smaller than one amu and a negative charge of
  -1.6 x 10-19 Coul.

6
Making Light from Atoms
We also know that the atom is neutral, so the
part of the atom that is not the electrons must
have essentially all the mass and a positive
charge to cancel that of the electrons. But what
is the structure of these electrons and this
other part of the atom?
7
Making Light from Atoms

• Two possibilities come to mind
• The planetary model, where the very light
  electron orbits the heavy central nucleus.
• The plum pudding model, where the very light and
  small electrons are embedded (like plums) in the
  much more massive pudding of the rest of the atom.

8
Making Light from Atoms

• The Planetary Model
• If the light electron does go around the
  central, heavy nucleus, then the electron is
  accelerating (changing the direction of its
  velocity). But an accelerating electron should
  emit electromagnetic radiation (its electric
  field is wobbling).

9
Making Light from Atoms

• If the electron is emitting EM radiation, it is
  emitting energy.
• If the electron is emitting energy, it should
  then fall closer to the nucleus.
• The process should continue until the electron
  falls into the nucleus and we have the plum
  pudding model

10
Making Light from Atoms

• In addition, the frequency of the EM radiation
  (light) emitted by the accelerating (orbiting)
  electron should continuously vary in frequency as
  the frequency of the electron continuously varies
  as it spirals into the nucleus. This does not
  agree with the experimental results the spectrum
  of hydrogen.

11
Making Light from Atoms

• The plum pudding model has no such problem with
  accelerating electrons, since the electrons are
  just sitting like plums in the pudding.

12
Rutherford Scattering

• To test the plum pudding model, Rutherford
  decided to shoot alpha particles
• (mass 4 amus charge 2e moving very fast)
  
• at a thin gold foil and see what happens to the
  alpha particles.
• (gold can be made very thin - only several atoms
  thick thus there should be very few multiple
  scatterings)

13
Rutherford Scattering

• If the plum pudding model was correct, then the
  alphas should pretty much go straight through -
  like shooting a cannon ball at a piece of tissue
  paper. The positive charge of the atom is
  supposed to be spread out, so by symmetry it
  should have little effect. The electrons are so
  light that they should deflect the massive alpha
  very little.

14
Rutherford Scattering

• Results
• Most of the alphas did indeed go straight through
  the foil.
• However, a few were deflected at significant
  angles.
• A very few even bounced back!
• (Once in a while a cannot ball bounced back off
  the tissue paper!

15
Rutherford Scattering

• The results of the scattering were consistent
  with the alphas scattering off a tiny positive
  massive nucleus rather than the diffuse positive
  pudding.
• The results indicated that the positive charge
  and heavy mass were located in a nucleus on the
  order of 10-14 m (recall the atom size is on the
  order of 10-10 m).

16
Rutherford Scattering

• If the electric repulsion of the gold nucleus is
  the only force acting on the alpha
• (remember both alpha and the nucleus are
  positively charged)
• then the deflection of the alpha can be
  predicted.

17
Rutherford Scattering

• The faster we fire the alpha, the closer the
  alpha should come to the gold nucleus.
• 1/2 m v2 q?(kqgold/r)
• We will know that we have hit the nucleus (and
  hence know its size) when the scattering differs
  from that due to the purely electric repulsion.
  This also means that there must be a nuclear
  force!

18
Rutherford Scattering

• Note how small the nucleus is in relation to the
  atom the nuclear radius is 10-14 m versus the
  atomic radius of 10-10 m - a difference in size
  of 10,000 and a difference in volume of 1012 (a
  trillion!).
• The electron is even smaller. It is so small
  that we cant yet say how small, but it is
• less than 10-17 meters in radius.

19
Rutherford Scattering

• If the mass takes up only 1 trillionth of the
  space, why cant I walk right through the wall?

20
Rutherford Scattering

• The electric repulsion between the orbiting
  electrons of the wall and the orbiting electrons
  of me - and the electric repulsion between the
  nuclei of the atoms in the wall and the nuclei of
  my atoms, these repulsions keep me and the wall
  separate.
• The nuclear force does not come into play. Well
  say more about the nuclear force in part V of
  this course.

21
Making Light from Atoms

• We now know that the atom seems to have a very
  tiny nucleus with the electrons somehow filling
  out the size of the atom - just what the
  planetary model of the atom would suggest.
• However, we still have the problem of how the
  electrons stay in those orbits, and how the atom
  emits its characteristic spectrum of light.

22
The Bohr Theory

• Lets start to consider the planetary model for
  the simplest atom the hydrogen atom.
• Use Newtons Second Law
• Fel macircular , or ke2/r2 mv2/r
• (one equation, but two unknowns v,r)
• Note that the theory should predict both v and
  r.

23
The Bohr Theory

• We need more information, so try the law of
  Conservation of Energy
• E KE PE (1/2)mv2 -ke2/r E
• (a second equation, but introduce a third
  unknown, E total unknowns v, r, E)

24
The Bohr Theory

• Need more information, so consider
• Conservation of Angular Momentum
• L mvr
• (a third equation, but introduce a fourth
  unknown, L unknowns v, r, E and L.)

25
The Bohr Theory

• We have three equations and four unknowns.
• Need some other piece of information or some
  other relation.
• Bohr noted that Plancks constant, h, had the
  units of angular momentum L mvr
• (kgm2/sec Joulesec)
• so he tried this L nh
• (quantize angular momentum).

26
The Bohr Theory

• Actually, what he needed was this
• L n(h/2??????n?
• ??where ? h/2? (called h-bar)??
• This gave him four equations for four unknowns
  (treating the integer, n, as a known). From
  these he could get expressions for v, r, E and L.

27
The Bohr Theory

• In particular, he got
• r n2?2/(meke2) (5.3 x 10-11 m) n2
• (for n1, this is just the right size radius for
  the atom) and
• E -mek2e4/2?2(1/n2) -13.6 eV / n2
• (where 1 eV 1.6 x 10-19 Joules).
• This says the electron energy is QUANTIZED

28
The Bohr Theory

• In particular, when the electron changes its
  energy state (value of n), it can do so only from
  one allowed state (value of ninitial) to another
  allowed state (value of nfinal).
• ?E hf -13.6 eV(1/ni2) - (1/nf2) .

29
The Bohr Theory

• ?E hf -13.6 eV(1/nf2) - (1/ni2)
• In the case of ni 3, and nf 2,
• ?E (-13.6 eV)(1/4 - 1/9) 1.89 eV
• ?E hf hc/? , so in this case,
• ?????emitted hc/?E
• (6.63x10-34 J-sec)(3x108 m/s)/(1.89 x 1.6x10-19
  J)
• 658 nm (red light).

30
The Bohr Theory

• Similarly, when ni 4 and nf 2, we get
• ?E 2.55 eV, and???emitted 488 nm
• (blue-green) and
• when ni 5 and nf 2, we get
• ?E 3.01 eV, and???emitted 413 nm
• (violet).
• ALL THREE MATCH THE ACTUAL SPECTRUM OF HYDROGEN!

31
The Bohr Theory

• This matching of theory with experiment is the
  reason Bohr made his assumption that
• L n? (instead of L nh).

32
The Bohr Theory

• Note that we have quantized energy states for the
  orbiting electron.
• Note that for all nfinal 1, we only get UV
  photons.
• Note that for all nfinal gt 2, we only get IR
  photons.

33
The Bohr Theory

• Problems with the Bohr Theory
• WHY is angular momentum quantized
• (WHY does Ln? need to be true.)
• What do we do with atoms that have more than one
  electron? (The Bohr theory does work for singly
  ionzed Helium, but what about normal Helium with
  2 electrons?)

34
DeBroglie Hypothesis

• Problem with Bohr Theory WHY L n? ?
• have integers with standing waves
• n(?/2) L
• consider circular path for standing wave
• n? 2?r and so from Bohr theory
• L mvr n? nh/2????get 2?r nh/mv n??
• which means ? h/mv h/p .

35
DeBroglie Hypothesis

• ?????DeBroglie h/mv h/p
• In this case, we are considering the electron to
  be a WAVE, and the electron wave will fit
  around the orbit if the momentum is just right
  (as in the above relation). But this will happen
  only for specific cases - and those are the
  specific allowed orbits and energies that are
  allowed in the Bohr Theory!

36
DeBroglie Hypothesis

• The Introduction to Computer Homework on the
  Hydrogen Atom (Vol. 5, number 5) shows this
  electron wave fitting around the orbit for n1
  and n2.
• What we now have is a wave/particle duality for
  light (EM vs photon), AND a wave/particle
  duality for electrons!

37
DeBroglie Hypothesis

• If the electron behaves as a wave, with
• ? h/mv, then we should be able to test this
  wave behavior via interference and diffraction.
• In fact, experiments show that electrons DO
  EXHIBIT INTERFERENCE when they go through
  multiple slits, just as the DeBroglie Hypothesis
  indicates.

38
DeBroglie Hypothesis

• Even neutrons have shown interference phenomena
  when they are diffracted from a crystal structure
  according to the DeBroglie Hypothesis ? h/p .
• Note that h is very small, so that normally ?
  will also be very small (unless the mv is also
  very small). A small ? means very little
  diffraction effects 1.22 ? D sin(?).

39
Quantum Theory

• What we are now dealing with is the Quantum
  Theory
• atoms are quantized (you can have 2 or 3, but not
  2.5 atoms)
• light is quantized (you can have 2 or 3 photons,
  but not 2.5)
• in addition, we have quantum numbers
• (L n? , where n is an integer)

40
Heisenberg Uncertainty Principle

• There is a major problem with the wave/particle
  duality
• a) a wave with a definite frequency and
  wavelength (a nice sine wave) does not have a
  definite location. At a definite location at a
  specific time the wave would have a definite
  phase, but the wave would not be said to be
  located there.
• a nice travelling sine wave A sin(kx-?t)

41
Heisenberg Uncertainty Principle

• b) A particle does have a definite location at a
  specific time, but it does not have a frequency
  or wavelength.
• c) Inbetween case a group of sine waves can
  add together (via Fourier analysis) to give a
  semi-definite location a result of Fourier
  analysis is this the more the group shows up as
  a spike, the more waves it takes to make the
  group.

42
Heisenberg Uncertainty Principle

• A rough drawing of a sample inbetween case, where
  the wave is somewhat localized, and made up of
  several frequencies.

43
Heisenberg Uncertainty Principle

• A formal statement of this (from Fourier
  analysis) is ?x ?k????????
• (where k 2?/?, and ? indicates the
  uncertainty in the value)
• But from the DeBroglie Hypothesis, ? h/p, this
  uncertainty relation becomes
• ?x ?(2?/?) ?x ?(2?p/h) 1/2 , or
• ?x ?p ?/2.

44
Heisenberg Uncertainty Principle

• ?x ?p ?/2
• The above is the BEST we can do, since there is
  always some experimental uncertainty. Thus the
  Heisenberg Uncertainty Principle says ?x ?p gt
  ?/2 .

45
Heisenberg Uncertainty Principle

• A similar relation from Fourier analysis for time
  and frequency ?t ?? 1/2 leads to another
  part of the Uncertainty Principle (using E hf)
  ?t ?E gt ?/2 .
• There is a third part ??? ?L gt ?/2 (where L
  is the angular momentum value).
• All of this is a direct result of the
  wave/particle duality of light and matter.

46
Heisenberg Uncertainty Principle

• Lets look at how this works in practice.
• Consider trying to locate an electron somewhere
  in space. You might try to see the electron by
  hitting it with a photon. The next slide will
  show an idealized diagram, that is, it will show
  a diagram assuming a definite position for the
  electron.

47
Heisenberg Uncertainty Principle

• We fire an incoming photon at the electron, have
  the photon hit and bounce, then trace the path of
  the outgoing photon back to see where the
  electron was.

incoming photon
electron
48
Heisenberg Uncertainty Principle

screen
slit so we can determine direction of the
outgoing photon
outgoing photon
electron
49
Heisenberg Uncertainty Principle

• Here the wave-particle duality creates a problem
  in determining where the electron was.

photon hits here
slit so we can determine direction of the
outgoing photon
electron
50
Heisenberg Uncertainty Principle

• If we make the slit narrower to better determine
  the direction of the photon (and hence the
  location of the electron, the wave nature of
  light will cause the light to be diffracted.
  This diffraction pattern will cause some
  uncertainty in where the photon actually came
  from, and hence some uncertainty in where the
  electron was .

51
Heisenberg Uncertainty Principle

• We can reduce the diffraction angle if we reduce
  the wavelength (and hence increase the frequency
  and the energy of the photon).
• But if we do increase the energy of the photon,
  the photon will hit the electron harder and make
  it move more from its location, which will
  increase the uncertainty in the momentum of the
  electron.

52
Heisenberg Uncertainty Principle

• Thus, we can decrease the ?x of the electron only
  at the expense of increasing the uncertainty in
  ?p of the electron.

53
Heisenberg Uncertainty Principle

• Lets consider a second example trying to
  locate an electrons y position by making it go
  through a narrow slit only electrons that make
  it through the narrow slit will have the y value
  determined within the uncertainty of the slit
  width.

54
Heisenberg Uncertainty Principle

• But the more we narrow the slit (decrease ?y),
  the more the diffraction effects (wave aspect),
  and the more we are uncertain of the y motion
  (increase ?py) of the electron.

55
Heisenberg Uncertainty Principle

• Lets take a look at how much uncertainty there
  is ?x ?p gt ?/2 .
• Note that ?/2 is a very small number
• (5.3 x 10-35 J-sec).

56
Heisenberg Uncertainty Principle

• If we were to apply this to a steel ball of mass
  .002 kg /- .00002 kg, rolling at a speed of 2
  m/s /- .02 m/s, the uncertainty in momentum
  would be 4 x 10-7 kgm/s .
• From the H.U.P, then, the best we could be sure
  of the position of the steel ball would be ?x
  5.3 x 10-35 Js / 4 x 10-7 kgm/s
• 1.3 x 10-28 m !

57
Heisenberg Uncertainty Principle

• As we have just demonstrated, the H.U.P. comes
  into play only when we are dealing with very
  small particles (like individual electrons or
  photons), not when we are dealing with normal
  size objects!

58
Heisenberg Uncertainty Principle

• If we apply this principle to the electron going
  around the atom, then we know the electron is
  somewhere near the atom,
• (?x 2r 1 x 10-10 m)
• then there should be at least some uncertainty in
  the momentum of the atom
• ?px gt 5 x 10-35 Js / 1 x 10-10 m 5 x 10-25 m/s
  

59
Heisenberg Uncertainty Principle

• Solving for p mv from the Bohr theory
• KE PE Etotal, (1/2)mv2 - ke2/r -13.6 eV
• gives v 2.2 x 106 m/s gives
• p (9.1 x 10-31 kg) (2.2 x 106 m/s)
• 2 x 10-24 kgm/s
• this means px is between -2 x 10-24 kgm/s and 2
  x 10-24 kgm/s, with the minimum ?px being 5 x
  10-25 kgm/s, or 25 of p.

60
Heisenberg Uncertainty Principle

• Thus the H.U.P. says that we cannot really know
  exactly where and how fast the electron is going
  around the atom at any particular time.
• This is consistent with the idea that the
  electron is actually a wave as it moves around
  the electron.

61
Quantum Theory

• But if an electron acts as a wave when it is
  moving, WHAT IS WAVING?
• When light acts as a wave when it is moving, we
  have identified the ELECTROMAGNETIC FIELD as
  waving.
• But try to recall what is the electric field?
  Can we directly measure it?

62
Quantum Theory

• Recall that by definition, E F/q. We can only
  determine that a field exists by measuring an
  electric force! We have become so used to
  working with the electric and magnetic fields,
  that we tend to take their existence for granted.
  They certainly are a useful construct even if
  they dont exist.

63
Quantum Theory

• We have four LAWS governing the electric and
  magnetic fields MAXWELLS EQUATIONS. By
  combining these laws we can get a WAVE EQUATION
  for EM fields, and from this wave equation we
  can get the speed of the EM wave and even more
  (such as reflection coefficients, etc.).

64
Quantum Theory

• But what do we do for electron waves?
• What laws or new law can we find that will work
  to give us the wealth of predictive power that
  MAXWELLS EQUATIONS have given us?

65
Quantum Theory

• The way you get laws is try to explain something
  you already know about, and then see if you can
  generalize. A successful law will explain what
  you already know about, and predict things to
  look for that you may not know about. This is
  where the validity (or at least usefulness) of
  the law can be confirmed.

66
Quantum Theory

• Schrodinger started with the idea of Conservation
  of Energy KE PE Etotal .
• He noted that
• KE (1/2)mv2 p2/2m, and that ?h/p, so that p
  h/? (h/2?)(2?/?) ?k p, so
• KE ?2k2/2m
• Etotal hf (h/2?)(2?f) ??.

67
Quantum Theory

• He then took a nice sine wave, and called
  whatever was waving, ?
• ?(x,t) A sin(kx-?t) Aei(kx-?t) .
• He noted that both k and ? were in the exponent,
  and could be gotten down by differentiating. So
  he tried operators

68
Quantum Theory

• ?(x,t) A sin(kx-?t) Aei(kx-?t) .
• pop? i?d?/dx i?-ikAe-i(kx-?t) ?k?
• (h/2?)(2?/?)? (h/??? p? .
• similary
• Eop?? i?d?/dt i?-i?Aei(kx-?t) ???
• ((h/2?)(2?f)? (hf?? E? .

69
Quantum Theory

• Conservation of Energy KE PE Etotal
• becomes with the momentum and energy operators
• -(?2/2m)(d2?/dx2) PE? i?(d?/dt)
• which is called SCHRODINGERS EQUATION. If it
  works for more than the free electron, then we
  can call it a LAW.

70
Quantum Theory

• What is waving here? ?
• What do we call ?? the wavefunction
• Schrodingers Equation allows us to solve for the
  wavefunction. The operators then allow us to
  find out information about the electron, such as
  its energy and its momentum.

71
Quantum Theory

• To get a better handle on ?, lets consider
  light how did the EM wave relate to the
  photon?

72
Quantum Theory

• The photon was the basic unit of energy for the
  light. The energy in the wave depended on the
  field strength squared.
• Recall energy in capacitor, Energy (1/2)CV2,
  where for parallel plates, Efield V/d and
• C// K?oA/d, so that
• Energy (1/2)(K?oA/d)(Efieldd)2
• K?oEfield2 Vol, or Energy
  ??Efield2.

73
Quantum Theory

• Since Energy is proportional to field strength
  squared, AND energy is proportional to the number
  of photons, THEN that implies that the number of
  photons is proportional to the square of the
  field strength.
• This then can be interpreted to mean that the
  square of the field strength is related to the
  probability of finding a photon.

74
Quantum Theory

• In the same way, the square of the wavefunction
  is related to the probability of find the
  electron!
• Since the wavefunction is a function of both x
  and t, the the probability of finding the
  electron is also a function of x and t!
• Prob(x,t) ?(x,t)2

75
Quantum Theory

• Different situations for the electron, like being
  in the hydrogen atom, will show up in
  Schrodingers Equation in the PE part.
• Different PE functions (like PE -ke2/r for the
  hydrogen atom) will cause the solution to
  Schrodingers equation to be different, just like
  different PE functions in the normal Conservation
  of Energy will cause different speeds to result
  for the particles.