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? Let’s 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 Avagadro’s 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 Avagadro’s 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 E&M 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 E&M 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 amu’s; 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 can’t 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 can’t 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. We’ll 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 Let’s start to consider the planetary model for the simplest atom: the hydrogen atom. Use Newton’s 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 Planck’s constant, h, had the units of angular momentum: L = mvr (kg*m2/sec = Joule*sec) 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 n=1, 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 > 2, we only get IR photons.

33. The Bohr Theory Problems with the Bohr Theory: • WHY is angular momentum quantized (WHY does L=n 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 theelectron 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 n=1 and n=2. What we now have is a wave/particle duality for light (E&M 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 electronsDO 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 > /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 > /2 . There is a third part: * L > /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 Let’s 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 .