Blackbody Radiation & Planck’s Hypothesis - PowerPoint PPT Presentation

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Blackbody Radiation & Planck’s Hypothesis

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  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6.  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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. 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

  21. 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