work function Φ Inside metal Now you change the frequency to violet light without changing the # of photons per second. Electrons over large range of energy have equal chance of absorbing photons. photoelectric_en.jar Potential energy of electrons Electrons over large range of energy have about equal chance of absorbing photons. Say you shine blue light on Ephot a metal plate the metal emits n electrons per sec. Ephot Electron potential energy Q1: Ephot work function Φ metal What happens to the number of electrons/sec. coming out of the metal? a. fewer electrons / sec. b. same # of electrons/sec but electrons are faster c. more electrons/sec. d. not enough information e. no change because light is a wave absorb blue light and have enough energy to leave absorb blue light, but don’t come out so the more energy the light has, the more electrons that come out, until so much energy that every electron comes out. (violet and ultraviolet would not be very different in this case) Typical energies Photon Energies: Each photon has: Energy = Planks constant * Frequency (Energy in Joules) Application of the PE effect: The photo multiplier tube (PMT) (Energy in eV) E=hf=(6.626*10-34 J-s)*(f in Hz) E=hc/λ = (1.99*10-25 J-m)/(λ m) Red Photon: 650 nm c. more electrons come out with violet E=hf= (4.14*10-15 eV-s)*(f in Hz) E= hc/λ = (1240 eV-nm)/(λ nm) Ephoton = 1240 eV-nm = 1.91 eV 650 nm Electron Work functions of metals (in eV): Aluminum 4.08 eV Cesium 2.1 Lead 4.14 Potassium 2.3 Beryllium 5.0 eV Cobalt 5.0 Magnesium 3.68 Platinum 6.35 Cadmium 4.07 eV Copper 4.7 Mercury 4.5 Selenium 5.11 Calcium Carbon 2.9 4.81 Gold Iron 5.1 4.5 Nickel Niobium 5.01 4.3 Silver Sodium 4.73 2.28 Uranium 3.6 Zinc 4.3 Q2: Photomultiplier tubes: An application of photoelectric effect. Most sensitive way to detect visible light, see single photons Q3: V Metal plate big voltage current electron multiplier: gives pulse of current for each photoelectron 1 2 3 4 5 Time (millisec) What would be the best choice of these materials to detect visible light? a. Platinum Φ = 6.35 eV b. Magnesium = 3.68 eV c. Nickel = 5.01 eV d. lead = 4.14 eV e. Sodium = 2.28 eV A photon at 300 nm kicks out an electron with a maximum initial kinetic energy KEλ=300nm. If the wavelength is halved to 150 nm, and the photon hits an electron in the same metal, what would be the maximum possible initial kinetic energy KEλ=150nm of the released electron? a. less than ½KEλ=300nm. b. ½KEλ=300nm c. = KEλ=300nm d. 2 x KEλ=300nm e. more than 2 x KE λ=300nm (remember hill/kicker analogy, draw pictures to reason out answer, don’t just pick answer without careful reasoning) Q4: KE = photon energy-energy to get out = hf – Φ KE300 If λ is ½ then, f twice as big, Ephot150 = 2 hf300 V Old KE300 = hf300 - energy to get out Shine in light of 300 nm. The most energetic electrons come out with initial kinetic energy KEλ=300nm. A voltage difference of 1.8 V is required to stop these electrons. What is the work function Φ for this metal? (e.g. the minimum amount of energy needed to kick electron out of metal?) New KEnew= 2hf300 - energy to get out Energy so KEnew is more than twice as big. hf150 Remember: The energy of a photon is hf (= hc/λ). In units of eV you can write: E= hc/λ = (1240 eV-nm)/(λnm) a. 1.2 eV b. 1.8 eV c. 4.1 eV d. 5.9 eV e. none of the above hf300 KE300 (Your answer should be within ~0.1eV) KE300 This is the end of our discussion about the PE effect V CQ: Shine in light of 300 nm, most energetic electrons come out with kinetic energy, KE300. A voltage diff of 1.8 V is required to stop these electrons. What is the work function Φ for this plate? (e.g. the minimum amount of energy needed to kick e out of metal?) Energy is conserved so: a. 1.2 eV b. 2.9 eV Ephot= energy need to exit (Φ) + electron’s left over energy c. 6.4 eV d. 11.3 eV so Φ=E – electron’s energy phot e. none When electron stops, all of initial KE has been of the converted to electrostatic potential energy: above electron energy = q*∆V = e x 1.8V = 1.8 eV, and Ephot = 1240 eV nm/300 nm = 4.1 eV. Next: More about photons and the particle-wave duality All these fifty years of conscious brooding have brought me no nearer to the answer to the question, 'What are light quanta?' Nowadays every Tom, Dick and Harry thinks he knows it, but he is mistaken! (Albert Einstein, 1954) So Φ = 4.1eV - 1.8 eV = 2.3 eV Electromagnetic Waves and Photons are describing the same thing! Common confusion: Recap: Intensity ∝ Emax2 but does not depend on frequency! Need to have a model where these views/perspectives all fit together and are consistent with each other! X vs. X Interference was definitive test that light is a wave! “Why do higher frequency gamma rays carry more energy (hf) than lower frequency radio waves, but frequency has nothing to do with intensity?” Do not confuse the energy carried by a beam of light with the energy of a single quantum particle of light! 1. Light, Radio waves, X-rays are all electromagnetic waves! 2. The electromagnetic wave is made up from lots of photons. 3. Photons could be thought of as mini E/M wave segments (each has a specific energy hf and a wavelength c/f ) Electromagnetic Wave E Travels Straight Made up from photons lots of photons This picture can help explain wave view and particle view. “Particle” = little chunk of the electromagnetic wave. Energy of photon (hf) is in its oscillating E and B fields. (Sometimes it also helps to think of a photon as a tiny particle/energy packet). When is it important to think about particle aspect of light? Only if your detector is good enough to see individual photons! If you think of photons as particles you probably automatically think of them as perfectly localized - like a tiny billiard ball at a coordinate (x, y, z). Watch out! This is not a good analogy for photons!! • Billiard balls never produce a diffraction pattern • Billiard balls have no measurable wavelength/frequency • Billiard balls cannot go through two slits at the same time (photons can; electrons too! Will show later) Answers to clicker questions Q1: C Q2: E Q3: E Q4: E Examples where important to think about particle behavior: Photoelectric effect: Individual electrons popping out of metal Lasers: Electrons in atoms transitioning between energy levels Molecular bonds: Chemical bonds breaking with light Quantum optics: Shot noise in a beam of light. Examples where we don’t need to think about particle behavior (detection system isn’t good enough and wave behavior is easier): Heating: Energy absorbed in microwave or by black asphalt. Optics: Light bending through lenses or passing through slits Laser beam: Treat it just like a beam of lightT (Understanding the inner workings of a laser requires photon picture.)