1. Making Light How do we make light?

2. 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 review Heat and Light first. Later in this part we will consider Atoms and Light.

3. Heat and Light The way we see most things is by shining light on them, and then looking at the light reflected from the object. The way we see stars is not this way. We see the light that comes solely from the object itself rather than light reflected from some other source. This type of radiation is called “blackbody radiation” since there is no reflected light involved, and things that don’t reflect light normally look black.

4. Heat and Light A good absorber is also a good emitter, and a poor absorber is a poor emitter. We use the symbol  to indicate the blackness (  =1) or the whiteness (  =0) of an object.

5. Heat and Light 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 (whiteness) of the object,   Temperature must be in Kelvin, where size of one Kelvin is same as size of one degree Celsius, but T=0K is absolute zero, and T=273K = 0 o C (freezing).]

6. Heat and Light: Experimental Results At 310 Kelvin (=37 o C = 98.6 o F), only get IR Intensity wavelength UV IRblue yellow red

7. Experimental Results At much higher temperatures, get visible look at blue/red ratio to get temperature Intensity wavelength UV IRblue yellow red

8. Experimental Results P total =  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 wavelength UV IRblue yellow red

9. 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=297K, and pick a surface area, say 1.5m 2, that is at a temperature of 93 o F=34 o C=307K: P emitted =  AT 4 = (5.67x10 -8 W/m 2 K 4 )*(.97)*(1.5m 2 )*(307K) 4 = 733 Watts emitted!

10. Example continued But this is not the whole story: besides emitting radiation, we receive radiation from the outside: P absorbed =  AT 4 = (5.67x10 -8 W/m 2 K 4 )*(.97)*(1.5m 2 )*(297K) 4 = 642 Watts absorbed! Hence, the net power emitted by the body via radiation is: P net = 733 Watts - 642 Watts = 91 Watts. The peak of this radiation is at:  peak = b/T = 2.9x10 -3 m*K / 307K = 9.5  m which is in the infrared (as expected).

11. Heat and Light: Wave Theory wave theory: UV catastrophe intensity wavelength experiment

12. Heat and Light: Planck’s idea Planck found that he could match the curve and DERIVE the empirical relations: –P =  AT 4 where  = 5.67 x 10 -8 m 2 *K 4 –  max = b/T where b = 2.9 x 10 -3 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.

13. How to Make Light The wave theory combined with the equipartition of energy theory failed to explain blackbody radiation (heat and light). 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.

14. 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.

15. 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!

16. Photons and Colors Electron volts are useful size units of energy 1 eV = 1.6 x 10 -19 Coul * 1V = 1.6 x 10 -19 J. radio photon: hf = 6.63 x 10 -34 J*s * 1 x 10 6 /s = 6.63 x 10 -28 J = 4 x 10 -15 eV red photon: f = c/  3 x 10 8 m/s / 7 x 10 -7 m = 4.3 x 10 14 Hz, red photon energy = 1.78 eV blue:  = 400 nm; photon energy = 3.11 eV.

17. Temperature of the Sun When we look at the visible spectra of the sun, we see that it’s intensity peaks at about 500 nm (green light). From the equation: = b/T (where b = 2.9 x 10 -3 m*K) we get: T = b/ = (2.9 x 10 -3 m*K) / 500 x 10 -9 m  6,000 K.

18. Power output of the sun From the relation: P =  AT 4 where  = 5.67 x 10 -8 m 2 *K 4 and the size of the sun (radius = 700,000 km = 7 x 10 8 m; A = area = 4  r 2 = 6 x 10 18 m 2 ), we get: P = (1) * (5.67 x 10 -8 m 2 *K 4 ) * (6x10 18 m 2 ) * (6000K) 4  4 x 10 26 Watts.

19. Intensity of sunlight at the earth’s orbit At the earth’s distance from the sun (93 million miles = 1.5 x 10 11 m), the intensity of sunlight we receive is about I = P/A = (4 x 10 26 Watts)/(4*  *[1.5x10 11 m] 2 )  1,600 W/m 2. However, due to reflection off the atmosphere and due to day/night and slanting angle of sunlight during the day, the average intensity striking the earth is only about 250 W/m 2.

20. Atoms and Light When we excite atoms with energy, we find that the atoms emit light. However, they do not emit light in a continuous spectrum (all colors) like hot objects do. Rather, they emit only certain colors of light, which we call a discrete spectrum, or an emission spectrum. Each element emits its own individual spectrum. Several examples will be demonstrated in class.

21. 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.

22. Absorption Spectra In addition to the continuous spectrum (all the colors) of hot objects, and the discrete or absorption spectra of hot gases of atoms, we find a third type of spectrum: an absorption spectrum. This occurs when a hot object shines light through a relatively cold cloud of gas, and the cloud absorbs just those colors that the gas emits when it is hot. An absorption spectrum is just a continuous spectrum with certain lines MISSING.

23. Spectra We find that our atmosphere and the atmospheres of stars (including the sun) give us absorption spectra. We also find emission spectra from the hotter areas of the atmospheres of stars (including the sun).

24. Doppler Effect We are all familiar with the change in pitch of a train horn as the train approaches and then passes us at a railroad crossing. Does a similar change in color (light’s “pitch”) occur when a light source approaches or recedes? We don’t notice this for light – but that is because the speed of light is so much faster than that of sound that the effect is so much smaller. However, that effect does exist for light. In fact, we can use it in radar to detect speeds – in traffic speed traps.

25. Doppler Effect for light If a source is approaching us, the pitch of sound is higher which means the frequency of the sound is higher. For light, this effect also shifts the frequency and hence the color of the light. For a spectral line in the middle of the visible spectrum (yellow), the shift is toward the blue. Thus, this shifting of frequency to a higher value is called a “Doppler blue shift”. If the source is receding, the yellow spectral line will be shifted towards the red, and so this shifting of frequency to a lower value is called a “Doppler red shift”.

26. Red shift and blue shift Regular spectrum for hydrogen: Blue-shifted spectrum for hydrogen: (first line is in UV) Red-shifted spectrum for hydrogen: (third line is in IR) Notice that the colors change, but the spacing remains the same.

27. Doppler Effect for light Note that a line in the blue will be shifted towards the ultraviolet if the source is moving towards us, but this is still called a “blue shift”. A line in the red will be shifted towards the infrared if the source is moving away from us, and again this is still called a “red shift”.

28. Doppler Shifts Because we know the emission spectra of many elements when they are at rest in the lab, we can look at emission spectra from objects outside the lab and measure how far the spectral lines have been shifted. We can even see the difference in lines coming from one side of an object compared to the opposite side of the object if that object is spinning!

29. Doppler shifts due to rotations Blue shifted light Red shifted light Spinning object