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4/9/2017 Infrared radiation AP PHYSICS 2 UNIT 7 Quantum Physics, atomic, and nuclear physics Hot star CHAPTER 26 Quantum Optics Hot coal Hot objects radiate energy Black body radiation A black body absorbs all incident light and converts the light's energy into thermal energy (no light is reflected). The black body then radiates electromagnetic waves based solely on its temperature. SPECTRAL CURVE Hotter objects higher intensity (power) Cooler objects (higher wavelength) Model of a black body Hot objects radiate energy. Infrared – Electromagnetic spectrum What characterizes a black body? If we raise the temperature of the box and again measure the spectrum of the EM radiation being emitted from the hole, we find that: The total power output from the hole is now greater. The spectral curve rises at all wavelengths. The peak of the power per small wavelength interval shifts to a shorter wavelength. 1 4/9/2017 Studies of black body radiation Joseph Stefan The total radiation output (power per unit area of the emitting object) increases dramatically as the temperature of the black body increases. Stefan's law: 𝑃 = 𝜀∙𝐴 Maximum Wavelength Wilhem Wien In 1893, the German physicist Wilhelm Wien (1864–1928) quantified a second aspect of black body radiation. He defined a relationship for the maximum wavelength at which a black body emits radiation of maximum power per wavelength: 𝑃 = 𝜎 ∙ 𝑇4 ∙ 𝐴 𝜎 = 5.67𝑥10−8 𝑊 𝑚2 ∙ 𝐾 4 WHITEBOARD What is the max for: Human T = 310 K Molten iron T = 1810 K The range of frequencies and wavelengths of electromagnetic waves is called the electromagnetic spectrum. 𝜆𝑀𝐴𝑋 = 9354 𝑛𝑚 𝜆𝑀𝐴𝑋 = 1602 𝑛𝑚 WHITEBOARD The maximum power per wavelength of light from the Sun is at a wavelength of about 510 nm, which corresponds to yellow light. What is the surface temperature of the Sun? 𝜆𝑀𝐴𝑋 = 𝑇= 2.90 ∗ 10−3 𝑇 2.90 ∗ 10−3 𝜆𝑀𝐴𝑋 The electromagnetic spectrum 𝑇= The color of stars Very hot stars (around 10,000 K) look blue. Even hotter stars are generally invisible to our eyes. 2.90 ∗ 10−3 510 ∗ 10−9 𝑇 = 5686 𝐾 2 4/9/2017 The ultraviolet catastrophe No models built on the physics of the late 19th century made predictions that were consistent with the drop in intensity at high frequencies. This problem became known as the ultraviolet catastrophe. Energy Quantization In classical physics, standing wave frequencies are quantized. The energy of these standing waves can take on any positive value, because the energy of the wave depends on the amplitude of the wave and not on its frequency. In quantum physics, the energy of the oscillator is quantized, rather than its frequency. Planck's hypothesis Planck hypothesized that charged particles could radiate energy only in discrete portions called quanta. He proposed that “a” microscopic oscillating charged particle had some kind of fundamental portion of energy, which was proportional to the frequency of its oscillation. According to Planck, the particle could emit amounts of energy equal only to multiples of this fundamental portion. Planck's hypothesis Using his energy quantization model, Planck was able to derive a mathematical function describing the black body spectral curve: 𝐸0 = ℎ𝑓 The proportionality constant h became known as Planck's constant. Photon Mathematical model that describes the black body spectral curve E0 = hf f = Frequency of the electromagnetic radiation h = Planck’s constant −𝟑𝟒 𝒉 = 𝟔. 𝟔𝟑 ∗ 𝟏𝟎 𝑱∙𝒔 The Photoelectric Effect 3 4/9/2017 PHOTOELECTRIC EFFECT The Experimental Setup The photoelectric effect was first observed in 1887 by Heinrich Hertz Hertz helped establish the photoelectric effect (which was later explained by Albert Einstein) when he noticed that a charged object loses its charge more readily when illuminated by ultraviolet light. In 1887, he made observations of the photoelectric effect and of the production and reception of electromagnetic (EM) waves, published in the journal Annalen der Physik. His receiver consisted of a coil with a spark gap, whereby a spark would be seen upon detection of EM waves. He placed the apparatus in a darkened box to see the spark better. He observed that the maximum spark length was reduced when in the box. A glass panel placed between the source of EM waves and the receiver absorbed ultraviolet radiation (UV) that assisted the electrons in jumping across the gap. When removed, the spark length would increase. He observed no decrease in spark length when he substituted quartz for glass, as quartz does not absorb UV radiation. Hertz concluded his months of investigation and reported the results obtained. He did not further pursue investigation of this effect, nor did he make any attempt at explaining how the observed phenomenon was brought about Photoelectric current produced immediately The intensity and wavelength/frequency of the incident light can be changed Vacuum tube for a free path from emitter to collector plate The voltage of the battery can be adjusted to change the electric field between the emitting and collecting plates. The amount of EPE that the electrons gain (KE that they lose) if they cross the full gap is 𝑼𝑬 = 𝒒 ∙ ∆𝑽 Wave Model of Light – Light is a wave (of crossed electric and magnetic fields.) Absorbing one quantum of light of sufficient energy could instantly eject an electron from the metal. 𝑼𝑬 = 𝒒 ∙ ∆𝑽 𝑼𝑬 = ∆𝑲𝑬 Energy does not need to accumulate in the metal before we see an effect. The wave model of light states that the energy of a beam of light depends on its intensity (amplitude of the E and B fields) What actually happens Predictions made by the Wave Model of Light (Caution: These are the predictions made by a disproved model of light) “If light is a wave, its energy depends on its amplitude (intensity).” Low-Intensity Light High-Intensity Light “Therefore the electrons ejected by high-intensity light should have more kinetic energy than the electrons ejected by low-intensity light.” Prediction Low-Intensity Light High-Intensity Light The intensity of the incident light does not affect the energy of the ejected electrons! However, high-intensity light will eject more electrons per second from the metal. 4 4/9/2017 Conclusion Evolution of ideas concerning the nature of light The wave model of light cannot be correct! The energy of light does not depend on its intensity. Another model of light is needed to explain the way that light interacts with matter. This discovery, made in the year 1900, sparked the birth of quantum physics. Photoelectric Effect: Whiteboard I Ultraviolet light is incident upon a sodium surface, ejecting electrons into the surroundings. The intensity of the light is then steadily increased. a) Sketch a graph the maximum kinetic energy of the ejected electrons as a function of the intensity of the light (KE vs. Intensity) b) Sketch a graph the current through the circuit as a function of the intensity of the light. (I vs. Intensity) Kmax The intensity of light does not affect the energy that it transfers to each electron. The electrons gain Kinetic energy. Where is the energy coming from? (energy must be conserved) Intensity I ENERGY The intensity of light affects the number of photons that are incident on the metal. Intensity 5 4/9/2017 It wasn’t Amplitude . . . . It is the frequency ! Photon A photon is a discrete portion of electromagnetic radiation that has energy Ephoton = hf f = Frequency of the electromagnetic radiation h = Planck’s constant 𝒉 = 𝟔. 𝟔𝟑 ∗ 𝟏𝟎−𝟑𝟒 𝑱 ∙ 𝒔 An extremely tiny number! Higher frequency photons have more energy WHITEBOARD Photoelectric Effect: Whiteboard II An electron at rest absorbs a photon of wavelength 450 nm. Estimate the energy of a quantum of: e Radio waves (frequency of 1 x 106 Hz) E = 6.63x10-28 J Infrared radiation (3 x 1013 Hz) E = 1.99x10-20 J Visible light (5 x 1014 Hz) UV radiation (1 x 1015 Hz). E = 3.32x10-19 J E = 6.63x10-19 J An electron at rest absorbs a photon of wavelength 450 nm. What is its recoil velocity? me = 9.1 × 10-31 kg qe = 1.6 × 10-19 C c = 3 × 108 m/s h = 6.63 × 10-34 J*s Hint: You will need to use another relationship that relates the frequency and wavelength of light! An electron at rest absorbs a photon of wavelength 450 nm. v v Energy Conservation! 𝑬𝒑𝒉𝒐𝒕𝒐𝒏 = 𝑲𝑬𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 𝒉𝒇 = 𝒄 = 𝝀𝒇 𝒎 ∙ 𝒗𝟐 𝟐 𝒇= 𝒄 𝝀 𝒉𝒄 𝒎 ∙ 𝒗𝟐 = 𝝀 𝟐 𝒗= 𝟐𝒉𝒄 𝝀𝒎 vfinal = 985,610 m/s 𝒇= 𝒄 𝝀 fphoton = 6.7 × 1014 Hz Light has extremely high frequencies! 6 4/9/2017 Work function (Φ) Work function (Φ) The work function of a metal is the minimum amount of energy that is needed to eject electrons from the surface of the metal (remove a free electron from a metal). Electrons that are deeper below the surface will require more energy than Φ to be ejected. The electrons on the surface will require the least amount of energy to be ejected, and will have the greatest kinetic energy when they leave the metal. The work function is the bare minimum needed to eject a surface electron. Work function Φ The work function has units of energy but is measured in electron volts (eV) because it is typically very small. The work function has units of energy but is measured in electron volts because it is typically very small. Work function Φ The greater the work function of a metal, the more tightly the free electrons are bound to the metal and the more energy must be added to separate them. Electron Volt: Energy to move one electron through a potential difference V = 1 V. 𝑼𝑬 = 𝒒 ∙ ∆𝑽 𝒒=𝒆 ∆𝑽 = 𝟏 𝑽 𝑼𝑬 = 𝟏. 𝟔 ∗ 𝟏𝟎−𝟏𝟗 𝑪 ∙ 𝟏 𝑽 𝟏𝒆𝑽 = 𝟏. 𝟔 ∗ 𝟏𝟎−𝟏𝟗 𝑱 When an electron in the metal absorbs a photon, Stopping potential Leftover kinetic energy = Photon energy – Energy needed to break it free KE = hf – (Energy binding electron to the metal) If the electron is a surface electron, then the photon only had to spend Φ to break it free. The electron will be ejected with the maximum possible kinetic energy! KEmax = hf - Φ This relationship can be used to calculate the KE of the most energetic ejected electrons. v0 By applying a voltage across the plates, we can determine exactly how much kinetic energy those surface electrons were ejected with! UE = qΔV ΔV The voltage that is able to just barely stop the most energetic electrons is called the stopping potential. Energy conservation as they cross the gap gives: qeΔVstop = KEmax 7 4/9/2017 The net result: Photoelectric Effect: Whiteboard III qeΔVstop = hf - Φ When violet light of wavelength 400 nm is incident upon a Calcium surface, a voltage of -0.21 Volts is required to completely stop the photoelectric current. In order for the current to drop to zero, you must stop the most energetic electrons (surface electrons). What is the work function for Calcium? (in Joules and eV) If you stop those, then none of the electrons will be able to complete the circuit. qe = -1.6 × 10-19 C c = 3 × 108 m/s h = 6.63 × 10-34 J*s 𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑 = 𝒉𝒇 − 𝝓 Photoelectric Effect: Whiteboard IV 𝝓 = 𝒉𝒇 −𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑 What would be the stopping potential if ultraviolet light of wavelength 250 nm were incident on the calcium surface? 𝝓=𝒉 𝒄 − 𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑 𝝀 𝑼𝒔𝒆 𝒒 = −𝒆 𝝓 = 𝟒. 𝟔𝟒 ∗ 𝟏𝟎−𝟏𝟗 𝑱 𝝓 = 𝟐. 𝟗 𝒆𝑽 qe = -1.6 × 10-19 C c = 3 × 108 m/s 𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑 = 𝒉𝒇 − 𝝓 ∆𝑽𝒔𝒕𝒐𝒑 = 𝒉𝒇 − 𝝓 𝒒 𝒉𝒄 𝝓 ∆𝑽𝒔𝒕𝒐𝒑 = + 𝝀𝒒 𝒒 ∆𝑽𝒔𝒕𝒐𝒑 = −𝟐. 𝟎𝟕 𝑽 h = 6.63 × 10-34 J*s 𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑 = 𝒉𝒇 − 𝝓 in Joules -0.21 V for violet light -2.07 V for ultraviolet light When divided by q, it will be the same magnitude as eV (-q then +) in V A greater voltage is needed to stop the electrons now, since UV light has more energy (higher frequency)! A greater voltage is needed to stop the electrons, since UV light has more energy (higher frequency)! Constant for Calcium (2.9 eV) 𝒄 𝒇= 𝝀 7.50x1014 Hz for violet light 1.20x1015 Hz for ultraviolet light 8 4/9/2017 Photoelectric Effect: Whiteboard V Ultraviolet light is incident upon a sodium surface, ejecting electrons into the surroundings. The frequency of the light is then steadily increased. Which of the following happens as a result? a) Electrons are ejected at the same rate, and have the same maximum kinetic energy. b) Electrons are ejected at the same rate, and have a greater maximum kinetic energy. c) Electrons are ejected at a greater rate, and have the same maximum kinetic energy. d) Electrons are ejected at a greater rate, and have a greater maximum kinetic energy. Threshold or cutoff frequency Ultraviolet light is incident upon a sodium surface, ejecting electrons into the surroundings. The frequency of the light is then steadily increased. Electrons are ejected at a greater rate, and have a greater maximum kinetic energy. Light of higher frequency has more energy per photon, and is able to eject electrons that are deeper within the metal. This results in more ejected electrons, and a greater current through the circuit! Also, electrons from the surface of the metal that absorb those photons are ejected with greater KE. f = fthreshold f > fthreshold f >> fthreshold The light has just enough energy to just barely eject the surface electrons from the metal. The light has enough energy to eject electrons from deeper within the metal, and the ones from the surface are ejected with some spare KE. The light has enough energy to eject electrons from very deep within the metal, surface electrons are ejected with lots of spare KE. More electrons, and some leftover energy for KE More electrons, and lots of leftover energy for KE 𝑲𝑬𝒎𝒂𝒙 = 𝒉𝒇 − 𝝓 When light with the threshold frequency is incident on the metal, electrons at the very surface are just barely able to absorb enough energy to escape. But they have no leftover energy once they do! 𝟎 = 𝒉𝒇 − 𝝓 𝒇𝒄𝒖𝒕𝒐𝒇𝒇 𝝓 = 𝒉 Very few electrons, and no leftover energy for KE Cutoff frequency HAPPY SPRING BREAK With low-frequency light, no electrons at all will be ejected from the metal! This is because each photon does not have enough energy to overcome the work function of the metal. KEmax fcutoff f 9 4/9/2017 No photocurrent below cutoff frequency Cutoff frequencies for selected metals We can express the cutoff frequency in terms of the work function of the metal and Planck's constant: if the energy of one quantum is less than the work function of the metal, and no photocurrent is produced. An Amazing Discovery! The wave model of light predicted that if you shine really bright red light on a metal, the combined energy of all of that light should be able to give at least some of the electrons enough energy to be ejected. How does light interact with matter? When light interacts with matter, it’s either all or nothing. Light transfers its energy to matter in individual packets (quanta) of amount hf. However, light doesn’t work that way! Light can only transfer its energy in distinct amounts of quantity hf Since hf for red light is not enough to overcome the work function of the metal, the electrons will never be able to absorb any of the red light’s energy! What is “Quantum”? A quantum is the smallest possible amount of something that can exist in the physical world. If some property is quantized, it means that it is made up of indivisible “building blocks”, or quanta. Electric charge is quantized, because there cannot exist an amount of electric charge that is smaller than 1.6 × 10-19 C. Mass is quantized, because there cannot exist an amount of mass that is smaller than 9.1 × 10-31 kg (mass of an electron). And now we know that energy is quantized, because it travels in indivisible “packets” (photons) with energies hf. These packets cannot be combined or split, and will either be absorbed completely or not at all. Therefore, increasing the intensity of the light will do nothing if none of the individual hf packets have enough energy to eject an electron! The Photon Model of Light Light is made up of individual packets of energy, called photons. Photons are neither waves nor particles, but rather have properties of each in different scenarios. When a photon travels through empty space, it behaves like a wave. It it spreads out in all directions, obeys the superposition principle, and it can create an interference pattern. When a photon interacts with matter, it behaves like a particle. It completely transmits all its energy in an amount hf. 10 4/9/2017 Photons The Wave-Particle Duality Physicists started to think of light as being composed of particle-like photons (the quanta of light). This marks the beginning of a new era in physics, where the distinction between waves and particles vanishes. To explain interference and diffraction phenomena, the photons also had to have wave-like properties. A photon behaves like a wave when traveling through empty space, but collapses and transfers all of its energy when it interacts with a proton or electron. This is called the dual particle-wave property of photons. The Photon Model is the currently accepted and exhaustively-tested model of light. The Wave-Particle Duality A photon behaves like a wave when traveling through empty space, but collapses and transfers all of its energy when it interacts (collides) with a proton or electron. Energy vv==cc momentum m = 0 kg RELATIVISTIC ENERGY & MOMENTUM (vphoton = c) Relativistic Energy 𝑬= 𝟏− 𝒎𝒄𝟐 𝒗 𝟏− 𝒄 𝒗 𝒄 𝟐 = 𝟐 𝒎𝒄𝟐 𝑬 Relativistic Momentum 𝒑= 𝟏− 𝒎𝒗 𝒗 𝟏− 𝒄 𝒗 𝒄 𝟐 = 𝟐 𝒎𝒗 𝒑 RELATIVISTIC ENERGY & MOMENTUM (vphoton = c) 𝒎𝒄𝟐 𝒎𝒗 = 𝑬 𝒑 𝒄𝟐 𝒗 = 𝑬 𝒑 𝒄𝟐 𝒄 = 𝑬 𝒑 𝒄 𝟏 = 𝑬 𝒑 𝐩= 𝐩= 𝑬 𝒄 𝒉𝒇 𝝀𝒇 𝐩= 𝒉 𝝀 11 4/9/2017 Photon momentum Whiteboard Photons participate in collisions with electrons inside metals (the photoelectric effect). This suggests that photons momentum. must have 𝒉 𝐩= 𝝀 What is the momentum of one of these photons? In 1922, Arthur Compton performed a testing experiment to determine whether this expression was reasonable. 𝑵= 𝑬𝒕𝒐𝒕𝒂𝒍 𝑬𝟏 𝟏 ∗ 𝟏𝟎−𝟑 𝑱 𝟕. 𝟗𝟓𝟔 ∗ 𝟏𝟎−𝟏𝟓 𝑱 𝐩= 𝑵 = 𝟏. 𝟐𝟔 ∗ 𝟏𝟎𝟏𝟏 𝒑𝒉𝒐𝒕𝒐𝒏𝒔 𝐩= 𝑬 = 𝒉𝒇 𝑬= 𝒉𝒄 𝝀 𝑬 = 𝟕. 𝟗𝟓𝟔 ∗ 𝟏𝟎−𝟏𝟓 𝑱 Wave-like and particle-like properties of photons Whiteboard 𝑵= Determine the number of photons (N) absorbed during a single chest X-ray if the body absorbs a total of 1x10-3 J of 0.025 nm wavelength X-ray photons. 𝒉 𝝀 𝟔. 𝟔𝟑 ∗ 𝟏𝟎−𝟑𝟒 𝟎. 𝟎𝟐𝟓 ∗ 𝟏𝟎−𝟗 𝒑 = 𝟐. 𝟔𝟓𝟐 ∗ 𝟏𝟎−𝟐𝟑 DOUBLE SLIT EXPERIMENT 𝒌𝒈 ∙ 𝒎 𝒔 DOUBLE SLIT EXPERIMENT (PARTICLE MODEL) Only two bright bands should appear (images of the slits themselves). As intensity decreases, individual flashes are expected. 12 4/9/2017 DOUBLE SLIT EXPERIMENT (WAVE MODEL) Many alternating bright and dark bands appear. As intensity decreases, bright bands disappear. Cathode ray tubes Physicists discovered that although a physical gap was present between the cathode and the anode in a cathode ray tube, when the cathode was heated, a current appeared in the circuit and the tube would glow. Conceptual Exercise 26.8 Imagine that you can see the beam of electrons shooting from the cathode to the anode of a cathode ray tube. A. Draw the direction of the electric force exerted on the electron. B. Draw the direction of the magnetic force exerted on the electron. C. What would happen if you placed the tube in a region with both an electric field and a magnetic field? DOUBLE SLIT EXPERIMENT (DUAL WAVE-PARTICLE (PHOTON) MODEL) Many alternating bright and dark bands appear. As intensity decreases, individual flashes are expected. The accidental discovery of X-rays The first X-ray image of a human was made by Roentgen of his wife Anna Bertha's hand on December 22, 1895. Explanation of X-rays Not deflected by either electric or magnetic fields Do not cause the viewing screen to become electrically charged Cause photographic paper to darken Produce a diffraction pattern of dark and bright bands on the screen after passing through a single narrow slit Can be polarized Can go through many materials that are opaque to visible light Can ionize gases 13 4/9/2017 Producing X-ray photons WHITEBOARD In a cathode ray tube used to produce X-rays for imaging in a hospital, the potential difference between the cathode and the anode is 90.0 kV (90,000 V). If the X-rays are generated by electrons in the tube that stop abruptly when they collide with the anode and emit photons, what are the energy and the wavelength of each photon? Conservation of energy. KE becomes radiating energy (x-rays) The Compton effect and X-ray interference Ephoton = KE Ephoton = qV Ephoton = 1.44x10-14 J Ephoton = 90,000 eV Ephoton = hf Ephoton = hc/ =hc/Ephoton =0.014 nm Compton effect If a photon of wavelength i scatters off a charged particle of mass m and moves at an angle relative to its initial direction of motion, the scattered photon will then have a longer wavelength f given by the following equation: 14