Making Light How do we make light Making

Making Light How do we make light?

Making Light How do we make light? – Heat and Light: Incandescent Lighting (10 -20% efficient) – Atoms and Light: Fluorescent Lighting (40 -60% efficient) – Semiconductors and Light: Light Emitting Diodes (LEDs) (60%-80% efficient) We’ll review Heat and Light first. Later in this part we will consider Atoms and Light.

Blackbody Radiation: Heat loss by radiation was discussed in PHYS 201 in the Thermo part. The light from a blackbody is light that comes solely from the object itself rather than being reflected from some other source. The sun makes its own light. The moon we see by the reflected light that comes to it from the sun. The sun is thus a very good blackbody. The moon is not, at least in the visible.

Blackbody Radiation: Recall that a good absorber is also a good emitter, and a poor absorber is a poor emitter. We use the symbol to indicate the blackness (absorbs and emits light well, =1) or the whiteness (reflects light well but doesn’t emit light well, =0) of an object.

Blackbody Radiation: What are the parameters associated with the making of light from warm objects? – Temperature of the object, T. – Surface area of the object, A. – Color (black versus white) of the object,

Blackbody Radiation: Experimental Results At 310 Kelvin, only get IR emitted. Intensity UV blue yellow wavelength red IR

Blackbody Radiation: Experimental Results At much higher temperatures, get visible emitted - look at blue/red ratio to get temperature Intensity log scale UV blue yellow wavelength red IR

Experimental Results The preceding graph is a little misleading in that the scales for the Intensity axis are not the same for the different temperatures. For example, comparing peaks: the peak of the T=300 K curve is about 160, 000 times smaller than the peak of the T=6, 000 K curve. At 9. 7 microns (in the IR) which is the peak of the 300 K curve, the 300 K curve is about 120 times lower than the 6, 000 K curve at that same wavelength.

Blackbody Radiation: Experimental Results Ptotal = AT 4 where = 5. 67 x 10 -8 W/m 2 *K 4 peak = b/T where b = 2. 9 x 10 -3 m*K Intensity log scale UV blue yellow wavelength red IR

Example If you eat 2, 000 calories per day, that is equivalent to about 100 joules per second or about 100 Watts - which must be emitted. Let’s see how much radiation you emit when the temperature is comfortable, say 75 o. F=24 o. C=297 K, and pick a surface area, say 1. 5 m 2, that is at a temperature of 93 o. F=34 o. C=307 K: Pemitted = AT 4 = (5. 67 x 10 -8 W/m 2 K 4)*(. 97)*(1. 5 m 2)*(307 K)4 = 733 Watts emitted!

Example continued But this is not the whole story: besides emitting radiation, we receive radiation from the outside: assuming the outside temperature is 72 o. F = 22 o. C = 297 K Pabsorbed = AT 4 = (5. 67 x 10 -8 W/m 2 K 4)*(. 97)*(1. 5 m 2)*(297 K)4 = 642 Watts absorbed! Hence, the net power emitted by the body via radiation is: Pnet = 733 Watts - 642 Watts = 91 Watts. The peak of this radiation is at: peak = b/T = 2. 9 x 10 -3 m*K / 307 K = 9. 5 m which is in the infrared (as expected).

Blackbody Radiation: Wave Theory We’ll now try to understand why radiation works this way – via the wave theory. 1. Certain waves resonate in an object (due to standing wave), such that n( /2) = L. From this it follows that there are more small wavelengths that fit than long wavelengths. (see next slide for example)

Blackbody Radiation: Wave Theory Resonance: n( /2) = L Example: for L = 1 meter, we have the following wavelengths that “fit”: 1 = 2 m; 2 = 1 m; 3 =. 67 m; 4 =. 50 m; 5 =. 40 m; 6 =. 33 m; 7 =. 29 m; 8 =. 25 m; etc. For the range of ’s, we have permitted 1 - 1. 99 m; 1. 50 -. 99 m (half the range size), 2. 25 -. 49 m (half again the range size), 4 etc.

Blackbody Radiation: Wave Theory 2. From thermodynamics, we have the equipartition of energy: each mode on average has an energy proportional to the Temperature of the object: Emode = k. BT, where k. B = 1. 38 x 10 -23 Joules/Kelvin.

Blackbody Radiation: Wave Theory The standing wave theory and the equipartition of energy theory together predict that the intensity of light should increase with decreasing wavelength: This works very well at long wavelengths, but fails at short wavelengths. This failure at short wavelengths is called the ultraviolet catastrophe.

Blackbody Radiation: Wave Theory wave theory: UV catastrophe Intensity log scale experiment wavelength

Blackbody Radiation: Planck’s idea Need to turn the Intensity curve down when gets small (or frequency gets large). Planck’s idea was to: • Keep the standing wave idea and number of modes. • Look at the equipartition theory and how the energy per mode got to be k. BT (where k. B is Boltzmann’s constant: k. B = 1. 38 x 10 -23 J/K).

Blackbody Radiation: Planck’s idea Eavg ≡ Ei / 1 = P(E)*E / P(E) where P(E) is the probability of having energy, E. From probability theory we have the Boltzmann probability distribution function: P(E) = Ae-E/k. T (this is demonstrated in the next few slides). If we assume that energy is continuous, then the summation can become an integral, and we can get a value for the average energy (which we do after discussing the Boltzmann probability distribution).

BOLTZMANN DISTRIBUTION The Boltzmann Probability of one atom having n units of energy, P(En) = P(n*E 1), is based on equal likelihood of any atom having any of the n units. Following is a listing of all possible cases for 4 atoms (called A, B, C and D) having 3 units of energy. After that we will state the results of 4 atoms having 5 units of energy.

BOLTZMANN DISTRIBUTION CASE I: four atoms having three units of energy: ABCD ABCD 3000 2100 1200 1020 1002 0300 2010 0210 0120 0102 0030 2001 0201 0021 0012 0003 (3000) 4 (2100) 12 1110 1101 1011 0111 (1110) 4

BOLTZMANN DISTRIBUTION Case I: Probability of atom A having n units: P(n): PA(3) = 1/20 =. 05 PA(2) = 3/20 =. 15 PA(1) = 6/20 =. 30 PA(0) =10/20 =. 50 Note that the sum of probabilities equals 1 as it should: PA(n) = 20/20 = 1. 00

BOLTZMANN DISTRIBUTION Note that this distribution is different than the distribution based on straight probability. For example, it may seem that the probability of an atom getting a unit of energy would be ¼ (since there are 4 atoms) for each of the three units available, so the probability that one atom would get a unit of energy in all three distributions, PA(3), would be PA(3) = (1/4)3 =. 0156 (compared to the Boltzmann probability – see previous slide - of PA(3) =. 0500). But we can’t have all 4 atoms having 3 units of energy since that would mean we would need 12 units of energy when we only have 3 units total!

BOLTZMANN DISTRIBUTION Likewise, the straight probability of one atom not getting the unit of energy in each distribution would be (3/4) – three of the atoms would not get the unit, only one would - so the straight probability of one atom getting zero after three distributions would be PA(0) = (3/4)3 =. 422 (compared to the Boltzmann’s probability of PA(0) =. 500), but we can’t have all four atoms having zero units since we do have 3 units of energy that must be distributed.

BOLTZMANN DISTRIBUTION CASE II: four atoms Case I: Probability of atom A having: five units of energy: three units of energy P(5) = 1/56 =. 018 P(4) = 3/56 =. 054 P(3) = 6/56 =. 107 P(2) =10/56 =. 179 P(1) =15/56 =. 268 P(0) =21/56 =. 375 P(5) =. 000 P(4) =. 000 P(3) =. 050 P(2) =. 150 P(1) =. 300 P(0) =. 500

Plot of P(E) vs E 3 units 5 units For more units of energy, the probability of having lower energy is less and the probability of having higher energy is greater.

P(E) vs E • Note that both of these look approximately like dying exponentials. • Note that P(E) for large values of E gets bigger and P(E) for small values of E gets smaller if there is more energy to share. • Note that since the sum of P(E) over all E gives 1, the value of P(0) decreases as E gets bigger. The function P(E) = Ae-E/k. T is consistent with all of these. What is important in calculating the average energy is the product of E and P(E). We look at this on the next slide.

Plot of E*P(E) vs E P(E) E*P(E)

Blackbody Radiation: Sum goes to Integral Eavg = LIM E→ 0 [ P(E) / P(E)] = = = Area under the curve / 1 = k. BT.

Blackbody Radiation: Planck’s idea Planck recalled that the SUM only becomes the INTEGRAL if you let E go to zero. Planck’s idea was NOT to let E go to zero. This is strange, since normally we think that energy, like most things, is continuous. Recall that vavg = Δx/Δt, and vinstantaneous = LIMΔt→ 0 = dx/dt. Not letting ΔE→ 0 is “thinking outside the box”, but it is following the math. If you require P(E) to be evaluated at the end of each E, then the SUM will decrease as E increases!

Blackbody Radiation: Planck’s idea As E gets bigger, Eavg gets smaller: E*P(E) = A*E*e-E/k. T. Area under red curve is more than area under blue is more than area under green. E*P(E) E

Blackbody Radiation: Planck’s idea It’s easy to see on the leading edge that as E gets bigger, the total Energy under the curve and hence the average energy gets smaller. This is in fact confirmed by an actual summation. Eavg ↓ as E ↑

Blackbody Radiation: Planck’s idea To get the intensity versus wavelength curve to fall at small wavelengths we need to decrease the average energy per mode more as the wavelengths get smaller - and the frequency gets bigger: Eavg ↓ as E ↑ ; he wanted Eavg ↓ as ↓ From f=c, ↓ as f ↑ , so Planck thought: E ↑ as f ↑. Planck tried the simplest relation: E = (constant) * f

Blackbody Radiation: Planck’s idea Planck found that he could match the curve and DERIVE the empirical relations, including the values of the constants and b: – P = AT 4 where = 5. 67 x 10 -8 m 2 *K 4 – -3 max = b/T where b = 2. 9 x 10 m*K with the simplest relation: E = (constant) * f if the constant = 6. 63 x 10 -34 J*sec = h. The constant, h, is called Planck’s constant.

How to Make Light The wave theory combined with the equipartition of energy theory failed to explain blackbody radiation. Planck kept the wave idea of standing waves but introduced E = hf. This was later identified with the idea of light coming in discrete packets (or photons) rather than continuously as the wave theory predicted.

How to Make Light From this theory we now have a way of relating the photon idea to the wave idea for color and type: E = hf. – Note that high frequency (small wavelength) light has high photon energy, and that low frequency (large wavelength) light has low photon energy.

How to Make Light E = hf High frequency light tends to be more dangerous than low frequency light (UV versus IR, x-ray versus radio). The photon theory gives a good account of why the frequency of the light makes a difference in the danger. Individual photons cannot break bonds if their energy is too low while big photons can!

Photons and Colors Electron volts are useful size units of energy 1 e. V = 1. 6 x 10 -19 Coul * 1 V = 1. 6 x 10 -19 J. • radio photon: hf = 6. 63 x 10 -34 J*s * 1 x 106 /s = 6. 63 x 10 -28 J * [1 e. V/ 1. 6 x 10 -19 J] = 4. 15 x 10 -9 e. V • red photon: f = c/ 3 x 108 m/s / 7 x 10 -7 m = 4. 3 x 1014 Hz, red photon energy = 1. 78 e. V • blue: = 400 nm; photon energy = 3. 11 e. V. • x-ray photon: hf = 6. 63 x 10 -34 J*s * 1 x 1018 /s = 6. 63 x 10 -16 J = 4, 150 e. V = 4. 15 Ke. V

Power and photons Example How many photons are emitted every second from one watt of yellow light? Power = Energy / time = Energy per photon * number of photons / time = hf * n/sec; f =c, so f = c/ Power = (hc/ ) * n/sec P = 1 Watt = (6. 63 x 10 -34 J-s * 3 x 108 m/s / 5. 5 x 10 -7 m) * n/sec; n/sec = 2. 8 x 1018 photons per second.

Making and Absorbing Light The photon theory with E = hf was useful in explaining the blackbody radiation. • Is it useful in explaining other experiments? • We’ll consider next the photoelectric effect.

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

Photoelectric Effect The more light, the more energy is deposited on the plate and the more electrons are able to be ejected. The more electrons that reach the plate, the more current will flow. The amount of current will be limited by the amount of electrons emitted and collected by the plate. light Electron flow Normal current flow ejected electron A - + V We could use this as a light detector: the more light, the higher the current.

Photoelectric Effect The more light, the more electrons will collect on the plate; if they are resisted by a reverse voltage source, the electrons will collect until the voltage on the plate is higher than the voltage of the battery, and then a reverse current will flow. But the more electrons on light the plate, the fewer electrons will be able to make it to the plate. ejected electron Only the few very A high energy electrons + will be able to make it, and so the current will V decrease as the reverse voltage is increased.

Photoelectric Effect The following graphs illustrate what the wave theory predicts will happen: current light intensity current Can’t have light of zero frequency Voltage current frequency of light

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

Photoelectric Effect In addition, we see a connect between V-stop and f above fcutoff: V-stop No current below fcutoff so no need for Vstop fcutoff frequency

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 fcutoff.

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, called the work function), then the following equation would apply: hf = W + KE The 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 for the electron (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 highest hill, we could arrive at the extra energy of the electron. Hill sizes in electrical terms are in VOLTS: KE = PE = q. Vstop.

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 voltage - height of the electric “hill”; special case: Vstop = 0, when f = fcutoff , so hfcutoff = W.

Photoelectric Effect - Example Most metals have a work function on the order of several electron volts. For example, Copper has a work function of 4. 5 e. V. Therefore, the cut-off frequency for light ejecting electrons from copper is: hfcutoff = 4. 5 e. V, 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.

Computer Homework The computer homework program on Photons (Vol 5, #5) deals with both Blackbody Radiation and the Photoelectric Effect.

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

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 From theory of relativity, E = mc 2, and since photons have E = hf, they have mass and so 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 back from an electron so = 180 o) 2 h/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 ( ≈ 10 -7 m) but NOT for x-ray and -ray light ( ≈ 10 -10 m and smaller).