Making Light How do we make light?

Making Light How do we make light? –Heat and Light: Incandescent Lighting (3-5% efficient) –Atoms and Light: Fluorescent Lighting (20-40% efficient) We’ll consider Heat and Light first. Later in this part we will consider Atoms and Light.

Blackbody Radiation: What is a blackbody? A BLACK object absorbs all the light incident on it. A WHITE object reflects all the light incident on it, usually in a diffuse way rather than in a specular (mirror-like) way.

Blackbody Radiation: The light from a blackbody then is light that comes solely from the object itself rather than being reflected from some other source. A good way of making a blackbody is to force reflected light to make lots of reflections: inside a bottle with a small opening.

Blackbody Radiation: If very hot objects glow (such as the filaments of light bulbs and electric burners), do all warm objects glow? Do we glow? (Are we warm? Are you HOT?)

Blackbody Radiation: What are the parameters associated with the making of light from warm objects?

Blackbody Radiation: What are the parameters associated with the making of light from warm objects? – Temperature of the object. – Surface area of the object. – Color of the object ? (If black objects absorb better than white objects, will black objects emit better than white objects?)

Blackbody Radiation: Consider the following way of making your stove hot and your freezer cold:

Blackbody Radiation: Put a white object in an insulated and evacuated box with a black object. The black object will absorb the radiation from the white object and become hot, while the white object will reflect the radiation from the black object and become cool. Put the white object in the freezer, and the black object in the stove.

Blackbody Radiation: Does this violate Conservation of Energy?

Blackbody Radiation: Does this violate Conservation of Energy? NO Does this violate the Second Law of Thermodynamics (entropy tends to increase) ?

Blackbody Radiation: Does this violate Conservation of Energy? NO Does this violate the Second Law of Thermodynamics (entropy tends to increase) ? YES This means that a good absorber is also a good emitter, and a poor absorber is a poor emitter. Use the symbol  to indicate the blackness (  =1) or the whiteness (  =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 of the object, 

Blackbody Radiation: Is the  for us close to 0 or 1? (i.e., are we white or black?) We emit light in the IR, not the visible. So what is our  for the IR?

Blackbody Radiation: So what is our  for the IR? Have you ever been near a fire on a cold night? Have you noticed that your front can get hot at the same time your back can get cold? Can your hand block this heat from the fire? Is this due to convection or radiation?

Blackbody radiation: For humans in the IR, we are all fairly good absorbers (black). An estimated value for  for us then is about.97.

Blackbody Radiation: Experimental Results At 310 Kelvin, only get IR Intensit y (log scale) wavelength UV IRblue yellow red

Blackbody Radiation: Experimental Results At much higher temperatures, get visible look at blue/red ratio to get temperature Intensit y (log scale) wavelength UV IRblue yellow red

Blackbody Radiation: Experimental Results P total =  AT 4 where  = 5.67 x W/m 2 *K 4  peak = b/T where b = 2.9 x m*K Intensit y (log scale) wavelength UV IRblue yellow red

Blackbody Radiation: Example Given that you eat 2000 Calories/day, your power output is around 100 Watts. Given that your body surface temperature is about 90 o F, and Given that your surface area is about 1.5 m 2,

Blackbody Radiation: Example Given P total = 100 Watts Given that T body = 90 o F Given that A = 1.5 m 2 WHAT IS THE POWER EMITTED VIA RADIATION?

Blackbody Radiation: Example P emitted =  AT 4 –  =.97 –  = 5.67 x W/m 2 *K 4 – T = (90-32)*5/9 (in K) = 305 K – A = 1.5 m 2 P emitted = 714 Watts (compared to 100 Watts generated!)

Blackbody Radiation: Example need to consider power absorbed at room T P absorbed =  AT 4 –  =.97 –  = 5.67 x W/m 2 *K 4 – T = (90-72)*5/9 (in K) = 295 K – A = 1.5 m 2 P absorbed = 625 Watts (compared to 714 Watts emitted!)

Blackbody Radiation: Example Total power lost by radiation = 714 W W = 89 Watts (Power generated is 100 Watts.) Power also lost by convection (with air) and by evaporation.

Blackbody Radiation: Example At colder temperatures, our emitted power stays about the same while our absorbed power gets much lower. This means that we will get cold unless –we generate more power, or –our skin gets colder, or –we reflect the IR back into our bodies. Use metal foil for insulation!

Blackbody Radiation: Wave Theory 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. From thermodynamics, we have the equipartition of energy: Each mode on average has an energy proportional to the Temperature of the object.

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 work 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 wavelength experiment

Blackbody Radiation: Planck’s idea Need to turn the curve down when gets small (or frequency gets large). Keep standing wave idea and number of modes. Look at equipartition theory and how the energy per mode got to be kT (where k is Boltzmann’s constant: k = 1.38 x J/K.

Blackbody Radiation: Planck’s idea E avg =  E i /  1 =  P(E)*E /  P(E) where P(E) is the probability of having energy, E. From probability theory (see page 5 of Study Guide for Part 3), we have the Boltzmann probability distribution function: P(E) = Ae -E/kT. If we assume that energy is continuous, then the summation can become an integral:

BOLTZMANN DISTRIBUTION Probability of one atom having n units of energy is based on equal likelihood of any possible state. Following is a listing of all possible states for two cases.

BOLTZMANN DISTRIBUTION CASE I: four atoms having three units of energy : ABCD ABCD ABCD (3000) 4 (2100) 12 (1110)

BOLTZMANN DISTRIBUTION Case I: Prob of atom A having n of 3 units: P(3) = 1/20 =.05 P(2) = 3/20 =.15 P(1) = 6/20 =.30 P(0) =10/20 =.50

BOLTZMANN DISTRIBUTION CASE II: four atoms having five units of energy: Prob of atom A having: 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

Plot of P(E) vs E

Plot of E*P(E) vs E

Blackbody Radiation: Planck’s idea E avg =  LIM  E->0 [  P(E) /  P(E)] = = = Area under the curve / 1 = kT.

Blackbody Radiation: Planck’s idea Planck recalled that the SUM only became the INTEGRAL if you let  E go to zero. Planck’s idea was NOT to let  E go to zero. 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, E avg gets smaller: E*P(E) = A*E*e -E/kT. 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. The mathematical details of the actual summation are considered in PHYS 447 (Modern Physics).

Blackbody Radiation: Planck’s idea To get the curve to fall at small wavelengths (big frequencies) Planck tried the simplest relation:  E = (constant) * f since we need to decrease the average energy per mode more as the wavelengths get smaller - and the frequency gets bigger.

Blackbody Radiation: Planck’s idea Planck found that he could match the curve and DERIVE both empirical relations: – P =  AT 4 where  = 5.67 x m 2 *K 4 –  max = b/T where b = 2.9 x m*K with the simplest relation:  E = (constant) * f if the constant = 6.63 x 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, 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 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 eV = 1.6 x Coul * 1V = 1.6 x J. radio photon: hf = 6.63 x J*s * 1 x 10 6 /s = 6.63 x J = 4 x eV red photon: f = c/  3 x 10 8 m/s / 7 x m = 4.3 x Hz, red photon energy = 1.78 eV blue:  = 400 nm; photon energy = 3.11 eV.

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.