Photoelectric Effect Light hits a metal plate, and electrons are ejected. These electrons are collected in the circuit and form a current. light ejected electron A + - V

Photoelectric Effect The following graphs illustrate what the wave theory predicts will happen: Icurrent Icurrent Ilight intensity Voltage Icurrent frequency of light

Photoelectric Effect We now show in green what actually happens: Icurrent Icurrent Ilight intensity V-stop Voltage Icurrent f-co frequency of light

Photoelectric Effect In addition, we see a connection between the stopping voltage and frequency above f-co: V-stop frequency f-co

Photoelectric Effect Einstein received the Nobel Prize for his explanation of this. (He did NOT receive the prize for his theory of relativity.)

Photoelectric Effect Einstein suggested that light consisted of discrete units of energy, E = hf. Electrons could either get hit with and absorb a whole photon, or they could not. There was no in-between (getting part of a photon). If the energy of the unit of light (photon) was not large enough to let the electron escape from the metal, no electrons would be ejected. (Hence, the existence of f-cutoff.)

Photoelectric Effect If the photon energy were large enough to eject the electron from the metal (here, W is the energy necessary to eject the electron), then the following equation would apply: energy of the photon absorbed (hf) goes into ejecting the electron (W) plus any extra energy left over which would show up as kinetic energy (KE).

Photoelectric Effect This extra kinetic energy (KE) would allow the electron to climb up a “hill”, but the size of the hill that the electron could climb up would be limited to the extra kinetic energy the electron had. By measuring the steepest hill, we could arrive at the extra energy of the electron. Hill sizes in electrical terms are in VOLTS: KE = PE = qVstop.

Photoelectric Effect Put into a nice equation: hf = W + e*Vstop where f is the frequency of the light W is the “WORK FUNCTION”, or the amount of energy needed to get the electron out of the metal Vstop is the stopping potential When Vstop = 0, f = fcutoff , and hfcutoff = W.

Photoelectric Effect - Example Most metals have a work function on the order of several electron volts. Copper has a work function of 4.5 eV. Therefore, the cut-off frequency for light ejecting electrons from copper is: hfcutoff = 4.5 eV, or fcutoff = 4.5 x (1.6 x 10-19 C) x (1 V) / 6.63 x 10-34 J-sec = 1.09 x 1015 Hz,

Photoelectric Effect - Example or cutoff = c/ fcutoff , or cutoff = (3 x 108 m/s) / (1.09 x 1015 cycles/sec) = 276 nm (in the UV range) Any frequency lower than the cut-off (or any wavelength greater than the cut-off value) will NOT eject electrons from the metal.

Photoelectric Effect From Einstein’s equation: hf = W + e*Vstop , we can see that the straight line of the Vstop vs f graph should have a slope of (h/e) . This gives a second way of determining the value of h. [The first was from fitting the blackbody curve.] When we do this, we get the same value for h that Planck did: 6.63 x 10-34 Joule*sec .

Compton Scattering When light encounters charged particles, the particles will interact with the light and cause some of the light to be scattered. incident photon scattered photon light wave electron motion of electron after hit electron motion of electron

Compton Scattering From the wave theory, we can understand that charged particles would interact with the light since the light is an electromagnetic wave!

Compton Scattering But the actual predictions of how the light scatters from the charged particles does not fit our simple wave model. If we consider the photon idea of light, some of the photons would “hit” the charged particles and “bounce off”. The laws of conservation of energy and momentum should then predict the scattering.

Compton Scattering As we will see in part five of the course, photons DO HAVE MOMENTUM as well as energy. The scattered photons will have less energy and less momentum after collision with electrons, and so should have a larger wavelength according to the formula:  = scattered - incident = (h/mc)[1-cos()]

Compton Scattering  = scattered - incident = (h/mc)[1-cos()] Note that Planck’s constant is in this relation as well, and gives a further experimental way of getting this value. Again, the photon theory provides a nice explanation of a phenomenon involving light.

Compton Scattering  = scattered - incident = (h/mc)[1-cos()] Note that the maximum change in wavelength is (for scattering from an electron) 2h/mc = 2(6.63 x 10-34 J-s) / (9.1 x 10-31 kg * 3 x 108 m/s) = 4.86 x 10-12 m =  which would be insignificant for visible light (with l of 10-7m) but NOT for x-ray and -ray light (with l of 10-10 m or smaller) .

Making Light: Exciting atoms 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 discrete spectrum. The hydrogen atoms emit three visible wavelengths: one in the red, one in the blue-green, and one in the violet.

Atoms and Light Hydrogen’s spectrum (in the visible) consists of just three lines: purple, blue-green, and red. Helium has quite a bit different set of lines in its spectrum.

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.

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 .

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!)

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.

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?

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.

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).

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

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.

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.

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)

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.

Rutherford Scattering Results: Most of the alphas did indeed go almost 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!

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).

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 (this is done in PHYS 380).

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”!

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.

Rutherford Scattering If the mass takes up only 1 trillionth of the space, why can’t I walk right through the wall?

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.

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.

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.]

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)

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.)

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).

The Bohr Theory Actually, what he needed was this: L = n*(h/2n* where  = h/2 ( is 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.

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

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)] .

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).

ALL THREE MATCH THE ACTUAL SPECTRUM OF HYDROGEN! 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!

Spectrum of Hydrogen Hydrogen’s spectrum (in the visible) consists of just three lines: purple, blue-green, and red. Helium has quite a bit different set of lines in its spectrum.

The Bohr Theory This matching of theory with experiment is the reason Bohr made his assumption that L = n (instead of L = nh).

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.

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?)