Modern Physics. Unit 1: Classical Models and the Birth of Modern Physics Lecture 1.4: Blackbody Radiation and Photoelectric Effect

Transcription

1 Modern Physics Unit 1: Classical Models and the Birth of Modern Physics Lecture 1.4: Blackbody Radiation and Photoelectric Effect Ron Reifenberger Professor of Physics Purdue University 1

2 I. Blackbody Radiation (1859/1879/1884/1895) Combining Thermodynamics with E & M Hot objects emit radiation. What is the nature of this radiation? Kirchoff s challenge (1859): measure energy vs. frequency for heat radiation

3 Stefan s Law (1879) T oven A The plan: Measure P (power emitted), infer I or S P AVERAGE then infer u tot. AT 4 P = eσ AT Boltzmann s Derivation (1884) 4 P = power radiated by hot object (in Watts) e = emissivity of object; 0<e<1 σ= Stefan constant = x 10-8 W/(m K 4 ) A = area of emitting (hot) object T = temperature of hot object (in K) 3

4 Wien and Lummer (1895) punch a hole in the side of a completely closed oven and spectrally measure what comes out. This was good idea! Power meter 1. Assume the energy flux of radiation is uniform over a hole of area A. P = c t I or S V AVERAGE [units: (J/s)/m ] Hot oven Area A. The power passing through the hole is related to the light intensity I emitted by oven: P = I da 3. In a time interval t, the amount of energy U (in J) passing through the area A is: I U = I A t = I A = A = utot V c c I since = utot ( Lecture L1.03) c 4. Recall from Lecture L1.03 for EM plane wave: u tot = 1 [( ε o E o ) ] [units: (J/m 3 ] 4

5 Summary: Blackbody Radiation - light emission from objects heated to a temperature T Blackbody Thermal Radiation (from experiment): Continuous light emission with no well defined emission lines Light spectrum (to first approximation) does not depend on material that is heated, only on absolute temperature T Energy density of radiation field (u tot ) in equilibrium with temperature of heated object thermodynamics 5

6 Spectral Distribution of Blackbody Radiation PHET, University Colorado at Boulder, see: 6

7 What s going on (Rayleigh, 1900)? a) Charges in the hot walls emit light at all frequencies b) Equilibrium is established; only certain light modes persist L y λ 1 λ λ 3 λ 4 λ 5 λ 6 λ 7 T E at surface is finite; damped mode E at surface is zero; undamped mode L x c) The shorter the wavelength, the more modes can fit inside the cavity c) Assume each persistent mode has same average energy = k B T 3 ( kb = J / K) d) Finally, count ALL possible modes and sum them up to obtain the spectral energy density = u tot (f) (the energy stored in the radiation field inside the cavity vs. frequency (or wavelength) 7

8 Classical physics unable to explain shape of measured light spectrum Experiment Emitted Intensity [ (W/m )/µm] UV visible 6000 K 3000 K IR Classical prediction - radiation field inside cavity is comprised of standing EM waves. Increases without bound as wavelength decreases. Rayleigh-Jeans (1905) 1000 nm 000 nm Wavelength,λ 8

9 Planck s Remarkable Hypothesis (1900) Planck assumes that a blackbody is made of atomic oscillators (presumably electrons) that emit or radiate light which is quantized in energy (photons), ie E = hf; c=fλ h= 6.66 x Js (fit to data) f= frequency of radiation Furthermore, a cavity may contain only a quantized number 0, 1,, 3,... of photons at a specific frequency f (or wavelength λ). Planck finds expression for the energy density u(f) (J/m 3 ) between frequency f and f+df (empirical): 3.0 8π f hf u( f ) df = df 3 hf/ kbt c e k = J / K B It follows that the number of photons at a frequency f is given by 1 n( f ) = hf/ kbt e 1 < n > 1.0 IR T=500 K T=000 K T=4000 K photon energy hf (in ev) visible More about u(f) later in the course 9

10 Planck s idea had an impact similar to that of Newton s in the early 1700 s Why is this quantization hypothesis so surprising? E (or B) position Recall from lecture L1.03: Energy transported by an EM plane wave: 1 EB S = E B = = ε oce µ o µ o 1 S = cε oeo AVERAGE Light is a wave (only way to explain interference and diffraction effects discovered back in the early 1800 s). It is well established that the energy transported by a wave is continuous and can be varied by adjusting the wave s amplitude. Nowhere does f appear? These well accepted ideas were completely contradicted by Planck s quantum hypothesis which assigned the energy in an EM wave to the frequency of the wave. 10

11 II. The Photoelectric Effect: Ejection of electrons from a material due to illumination by light photons in, electrons out What you can measure experimentally: 1. Intensity of Light. Frequency (color) of Light 3. Composition of Target 5. Electron Current (Number of Electrons) 4. Energy of Ejected Electrons 11

12 Photoelectric Effect Check out the photoelectric simulation at 1

13 What was expected For a given wavelength of light, as the intensity increases, the maximum energy of ejected electrons should also increase. For a given wavelength of light, at low enough intensities, there should be a time delay in electron emission. Electrons should be emitted for all wavelengths of light. None of these expectations were met by the experimental results! 13

14 Modern Picture of Photoemission (requires quantization of light energy) (hf) Kinetic Energy=K 0 (hf) Energy Photon energy Electron energy Work function, W (or φ) E = hf ( ) max K = hf W METAL VACUUUM h=planck s constant = 6.66x10-34 m kg/s 14

15 Note: Sometimes, the spectral radiance is used in lackbody calculations: 3 hf B( f) = hf kt B c e Other times, the spectral energy density is used instead: 3 8π hf u( f) = hf 3 kt B c e 1 1 It is evident that the two are related by 4π u( f) = B( f) c Appendix A: How do you get a T 4 dependence for power radiated by a Blackbody? Calculate u tot( T), the total energy density inside a cavity, by integrating Planck's u(f) over f: 3 3 hf kt B 8πhf 8πhf e utot = u( f ) df = df = df hf hf hf kt B 0 3 kt B kt B c e 1 c e e hf kt let x = ; df = B dx kt h B 4 3 x 4 3 x 8πh kt B xe 8πh kt xe tot = 3 = x x 3 x B 0 ( ) c h e e 0 ( 1 e ) u dx dx c h 4 kt 3 x x x B 8π h = x e 1 e 3 ( e )... dx c h let x = v ; dx = vdv; x dx = v dv 4 v 7 v 4v v π h kt B utot = e e e 3 c h.. dv αv n n v 3 now e v dv = so for example e dv = n = = 4 α u tot ( n )!. α 7 α v ; 7, α π h kt B = c h

16 4 8π h kt B = c h π you can show that =, so h kt B π 3 π 8 π = c h 8 = ch ( kt) 5 4 π kb / ( ) u 3 tot σ define σ = W K m = T 15ch c The energy that leaves the cavity per unit time thru a hole of area Ais P Acutot Ac T A T 4 4 c σ = = = σ 4 4 Where does the factor of ¼ come from? B 4 Stefan-Boltzmann Law Imagine an energy density u tot (T ) transported through a hollow cylinder of cross-section area A at speed c. What radiation intensity per unit time would emerge from the cylinder? 16

17 Well, you might expect ½ Ac u tot (T) to emerge from one end and ½ Ac u tot (T) to emerge from the other end. So if you are just looking at one end of the cylinder, you might expect to see just ½ Ac u tot (T). That gives a factor of ½. So why the factor of ¼? Well, inside a 3-dimnesional cavity, the radiation is not aimed directly at the hole of area A. It s hitting the hole from all inside directions, at different angles θ to the vertical. This implies the radiation effectively sees a smaller area, by a factor cos θ. Also, only a fraction of the radiation at an angle θ leaving the hole will strike a detector positioned at some distance in front of the hole. This fraction is also proportional to cos θ. This discussion implies we need to calculate the average value of cos θ between 0 and π/. π π 1 cos θdθ ( θ+ sin θcos θ) cos θ = = = π π dθ 0 So ultimately we have two factors of ½ which gives the factor of ¼ in the equation above. 17