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Blackbody Radiation & Planck’s Hypothesis

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## Blackbody Radiation & Planck’s Hypothesis

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**Blackbody Radiation & Planck’s Hypothesis**• A blackbody is any object that absorbs all light incident upon it • Shiny & reflective objects are poor blackbodies • Recall: good absorbers and also good emitters • Ideally we imagine a box with a small hole that very little light (EM radiation) can reflect back out**Consider heating blackbodies to various temperatures and**recording intensity of radiation at differing frequencies • At both low and high freq. there is very little radiation • The rad. Peaks at an intermediate freq. • This distribution holds true regardless of the material • Note: As temp. increases – area under curve increases • This represents total energy • As temp. increases – peak moves to higher frequency**The temperature therefore indicates its emitted color and**vice versa • We can determine star temperature (surface) by analyzing its color • Red stars are fairly cool, like the bolt shown • But White, or Blue-White stars are very hot • Our sun is intermediate**Planck’s Quantum Hypothesis**• Attempts to explain blackbody radiation using classical physics failed miserably • At low temps. Prediction & exp match well • At high temps. Classical prediction explodes to infinity • Very different from experimental result • Referred to as the Ultraviolet Catastrophe**German physicists Max Planck diligently tried to solve this**issue • He “stumbled” upon a mathematical formula that matched the experiment • He then needed to derive the physical formula • The only way was to assume energy (in the form of EM radiation) way quantized • Little “packets” of energy** E αf**• Inserting a constant, h E = nhf • Where n = number of packets and h = planck’s constant h = 6.63 x 10-34 J • s • One of our fundamental constants of nature • This tells us that energy can only change in quantum jumps, a very tiny amount not experienced everyday**Planck was not satisfied and believed (along with other**physicists) that it was a purely mathematical solution, not a “real” physical one • It does explain the exp. quite well: • The > f, the > quantum of energy needed • As frequency increased, the amount of energy needed for small jumps increased as well • The object only has a certain amount of energy to supply • Therefore: radiation drops to zero at high frequency**Photons & the Photoelectric Effect**• Planck believed that the atoms of a blackbody vibrated with discrete frequencies (like standing waves) • But, at the time light was considered a wave therefore no connection • Einstein took the idea of quanta of energy and applied it to light – called photons**Each photon has energy based on its frequency** E = n h f • A beam of light can be thought of as a beam of particles • More intense = more particles • Since each photon have small amounts of energy, there must be tremendous numbers of them**Einstein applied this model to the photoelectric effect**issue • Light hitting the surface of metals can cause electrons to be ejected • The effect could not be explained using the wave theory of light • We can determine the number of e ejected by connecting the apparatus to a simple circuit**The minimum amount of energy needed to eject e = work**function, W0 • Metal dependent • Usually a few eV • If an e is given energy by light that exceeds W0, the additional amount goes into kinetic energy of e Kmax = E – W0 • Classical physics predicts • light of any frequency should eject e as long as intensity is high enough • The K of e should increase with intensity**These do not agree with experiment:**• There is a minimum frequency required – the cutoff frequency, f0 • If f < f0 no e regardless of the intensity • The Kmax of e depends only on the frequency • Increasing intensity about f0 only increases the number of e • Both of these are explained using the photon model of light • Changing intensity only changes the number of photons • E is ejected only if the photon has sufficient energy (at least equal to the work function) • The is the cutoff frequency, f0**If f > f0, the e leaves metal with some K**• If f < f0, no e are ejected regardless of intensity • Since energy is that of a photon Kmax = hf – W0 • Therefore, Kmax depends linearly on frequency • A plot of Kmax for Na & Au shows different cutoff frequencies, but the same slope, h**Photons & the Photoelectric Effect**• Quantization of light – Albert Einstein (1905) • Based on properties of EM waves • Emitted radiation should be quantized • Quantum (packet of light) – photon • Each photon has energy E = h f • Little bundles of light energy • Connection between wave & particle nature of light**Einstein used this to explain the photoelectric effect**• Certain metallic materials are photosensitive • Light striking material emits electrons (e) • The radiant energy supplies the work necessary to free the e – photoelectrons**When photocell is illuminated with monochromatic light,**characteristic curves are obtained • Photocurrent until a saturation current is reached • All emitted e reach anode • voltage has no effect on current**Classically: > the intensity, the > energy of e**• K of e can be tested by reversing voltage • Only e with enough K (eV) make it to the negative plate & contribute to the current • As voltage , then is made negative, current • At some voltage V0, the stopping potential, no current will flow**The max K (Kmax) is related to stopping potential**• eV = the work needed to stop e Kmax = eV0 • When f of light is varied, the Kmax is found to depend linearly on f • No photoemission is observed below cutoff frequency, f0**Emission begins the instant (~10-9 s) even with low**intensity light • Classically, time is required to “build up” energy • Since light can be considered a “bundle of energy”, E = hf • The e absorb whole photon or nothing • Since e are bound by attractive forces, work must be done • Conservation of energy hf = K+ φ where φ = amount of work (energy) needed to free e • Part of energy of photon “frees” e & the rest is carried away as K**Least tightly bound will have maximum K**• Energy needed = work function, φ0 hf = Kmax + φ0 • Other e require more energy & the K is less • Increasing light intensity, increases # of photons thus increasing # of e • Does not change energy of individual photons • Photon energy depends on frequency • Below a certain freq. no e are dislodged • When Kmax = 0 the minimum cutoff frequency, f0 • Hf0 = Kmax + φ0 = 0 + φ0 f0 = φ0 / h**Photon has enough energy to free e, but no extra to give it**K • Sometimes called threshold frequency • Light below this (no matter how many) will not dislodge e