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Chapter 38 Photons, Electrons & Atoms Need for Quantum Physics Problems remained from classical mechanics that relativity didn’t explain Blackbody Radiation Photoelectric Effect Emission of electrons by an illuminated metal X Rays The electromagnetic radiation emitted by a heated object How & why are they produced? Why would f changes upon collisions? Spectral Lines Emission of sharp spectral lines by gas atoms in an electric discharge tube Development of Quantum Physics 1900 to 1930 Development of ideas of quantum mechanics Also called wave mechanics Highly successful in explaining the behavior of atoms, molecules, and nuclei Involved a large number of physicists Planck introduced basic ideas Mathematical developments and interpretations involved such people as Einstein, Bohr, Schrödinger, de Broglie, Heisenberg, Born and Dirac Blackbody Radiation An object at any temperature emits electromagnetic radiation Sometimes called thermal radiation Stefan-Boltzmann Law describes the total power intensity radiated I = σT4 The spectrum of the radiation depends on the temperature and properties of the object Blackbody Radiation Graph Experimental data for distribution of energy in blackbody radiation As the temperature increases, the total amount of energy increases Shown by the area under the curve As the temperature increases, the peak of the distribution shifts to shorter wavelengths Wien’s Displacement Law The wavelength of the peak of the blackbody distribution was found to follow Wein’s Displacement Law λmax T = 0.2898 x 10-2 m • K λmax is the wavelength at which the curve peaks T is the absolute temperature of the object emitting the radiation The Ultraviolet Catastrophe Classical theory did not match the experimental data At long wavelengths, the match is good At short wavelengths, classical theory predicted infinite energy At short wavelengths, experiment showed no energy This contradiction is called the ultraviolet catastrophe Rayleigh’s cavity model • EM wave in cavity (light) • Normal Modes (resonants) • Equipartition Principle E~kT • Power Radiation Intensity I = 2πckT/λ4 • Didn’t explain experimental results for small λ Planck’s Resolution Planck hypothesized that the blackbody radiation was produced by electromagnetic resonators Resonators were submicroscopic charged oscillators (electrons in a box??) The resonators could only have discrete energies En = n h ƒ n is called the quantum number ƒ is the frequency of vibration h is Planck’s constant, 6.626 x 10-34 J s Key point is quantized energy states Max Planck 1858 – 1947 Introduced a “quantum of action,” h Awarded Nobel Prize in 1918 for discovering the quantized nature of energy Planck Radiation Law I= 2πhc2/[λ5(ehc/λkT – 1)] I for large λ, I ~ 1/λ4 λ for small λ, I ~ e-hc/λkT Electromagnetic radiation is quantized ! Photoelectric Effect When light is incident on certain metallic surfaces, electrons are emitted from the surface This is called the photoelectric effect The emitted electrons are called photoelectrons The effect was first discovered by Hertz The successful explanation of the effect was given by Einstein in 1905 Received Nobel Prize in 1921 for paper on electromagnetic radiation, of which the photoelectric effect was a part Photoelectric Effect Schematic When light strikes E (Cathode), photoelectrons are emitted Electrons collected at C (small Anode) and passing through the ammeter create a current in the circuit C is maintained at a positive potential by the power supply Photoelectric Current/Voltage Graph The current increases with intensity, but reaches a saturation level for large ∆V’s No current flows for voltages less than or equal to –∆Vs, the stopping potential The stopping potential is independent of the radiation intensity More About Photoelectric Effect The stopping potential is independent of the radiation intensity The maximum kinetic energy of the photoelectrons is related to the stopping potential: KEmax = e∆Vo Features Not Explained by Classical Physics/Wave Theory No electrons are emitted if the incident light frequency is below some cutoff frequency that is characteristic of the material being illuminated (Poynting S ~ |Efield|2) The maximum kinetic energy of the photoelectrons is independent of the light intensity More Features Not Explained The maximum kinetic energy of the photoelectrons increases linearly with increasing light frequency Electrons are emitted from the surface almost instantaneously, even at low intensities (E ~ P∆t) Einstein’s Explanation A tiny packet of light energy, called a photon, would be emitted when a quantized oscillator jumped from one energy level to the next lower one Extended Planck’s idea of quantization to electromagnetic radiation The photon’s energy would be E = hƒ Each photon can give all its energy to an electron in the metal The maximum kinetic energy of the liberated photoelectron is KEmax = hƒ – Φ Φ is called the work function of the metal Light Intensity Therefore, a more intense beam of light will contain more photons, but the energy of each photon does not change. Explanation of Classical “Problems” The effect is not observed below a certain cutoff frequency since the photon energy must be greater than or equal to the work function Without this, electrons are not emitted, regardless of the intensity of the light The maximum KE depends only on the frequency and the work function, not on the intensity More Explanations The maximum KE increases with increasing frequency The effect is instantaneous since there is a one-to-one interaction between the photon and the electron Verification of Einstein’s Theory Experimental observations of a linear relationship between KE and frequency confirm Einstein’s theory The x-intercept is the cutoff frequency Cutoff Wavelength The cutoff wavelength is related to the work function hc λc = φ Wavelengths greater than λC incident on a material with a work function φ don’t result in the emission of photoelectrons Photocells Photocells are an application of the photoelectric effect When light of sufficiently high frequency falls on the cell, a current is produced Examples Solar cells, IR remote controls …etc X-Rays Discovered and named by Roentgen in 1895 Later identified as electromagnetic radiation with short wavelengths Wavelengths less than for ultraviolet Wavelengths are typically about 0.1 nm X-rays have the ability to penetrate most materials with relative ease Production of X-rays, 1 X-rays are produced when high-speed electrons are suddenly slowed down Can be caused by the electron striking a metal target Heat generated by current in the filament causes electrons to be emitted These freed electrons are accelerated toward a dense metal target The target is held at a higher potential than the filament X-ray Spectrum The x-ray spectrum has two distinct components Continuous broad spectrum Depends on voltage applied to the tube Sometimes called bremsstrahlung The sharp, intense lines depend on the nature of the target material Production of X-rays, 2 An electron passes near a target nucleus The electron is deflected from its path by its attraction to the nucleus This produces an acceleration It will emit electromagnetic radiation when it is accelerated Wavelengths Produced If the electron loses all of its energy in the collision, the initial energy of the electron is completely transformed into a photon The wavelength can be found from e∆V = h ƒmax = hc λmin Wavelengths Produced, cont Not all radiation produced is at this wavelength Many electrons undergo more than one collision before being stopped This results in the continuous spectrum produced Diffraction of X-rays by Crystals For diffraction to occur, the spacing between the lines must be approximately equal to the wavelength of the radiation to be measured The regular array of atoms in a crystal can act as a three-dimensional grating for diffracting X-rays Schematic for X-ray Diffraction A beam of X-rays with a continuous range of wavelengths is incident on the crystal The diffracted radiation is very intense in certain directions These directions correspond to constructive interference from waves reflected from the layers of the crystal The diffraction pattern is detected by photographic film Photo of X-ray Diffraction Pattern The array of spots is called a Laue pattern The crystal structure is determined by analyzing the positions and intensities of the various spots Bragg’s Law The beam reflected from the lower surface travels farther than the one reflected from the upper surface If the path difference equals some integral multiple of the wavelength, constructive interference occurs Bragg’s Law gives the conditions for constructive interference 2 d sin θ = m λ, m = 1, 2, 3… Arthur Holly Compton 1892 – 1962 Discovered the Compton effect Worked with cosmic rays Director of the lab at U of Chicago Shared Nobel Prize in 1927 The Compton Effect Compton directed a beam of x-rays toward a block of graphite He found that the scattered x-rays had a slightly longer wavelength that the incident x-rays This means they also had less energy The amount of energy reduction depended on the angle at which the xrays were scattered The change in wavelength is called the Compton shift Compton Scattering Compton assumed the photons acted like other particles in collisions Energy and momentum were conserved The shift in wavelength is h ∆λ = λ − λo = (1 − cos θ ) mec Compton Scattering, final The quantity h/mec is called the Compton wavelength Compton wavelength = 0.002 43 nm Very small compared to visible light The Compton shift depends on the scattering angle and not on the wavelength Experiments confirm the results of Compton scattering and strongly support the photon concept Photons and Electromagnetic Waves Light has a dual nature. It exhibits both wave and particle characteristics The photoelectric effect and Compton scattering offer evidence for the particle nature of light Applies to all electromagnetic radiation Different frequencies allow one or the other characteristic to be more easily observed When light and matter interact, light behaves as if it were composed of particles Interference and diffraction offer evidence of the wave nature of light Rutherford Scattering Rutherford did a very clever experiment with thin Au foil and alpha particles. Scattering was nearly absent or very dramatic, leading him to conclude that the atom was mostly empty space around a dense (+) center. (Muffin - raisin model) Bohr Atom (in Rutherford’s Lab) Classical physics predicts that a moving electron would emit radiation and decay. (not observed!) Bohr postulated stable orbits at quantized energy levels that do not decay. hf=hω=Ei – Ef. (single photon-electron) Ln= mvnrn = nh. rn = n2ao. ao=Bohr radius = 0.53Å Reduced mass Hydrogen Atom Hydrogen has one set of lines in the visible spectrum (Balmer), three sets of lines in the IR (Paschen, Braclett and Pfund series) and one in the UV (Lyman series). Hydrogen Spectrum 1/λ = R(1/nf2 – 1/n2) R = Rydberg Constant = 1.097 x 107 m-1 nf = 1 (Lyman Series) - UV nf = 2 (Balmer Series) – visible nf = 3, 4, 5 (Paschen, Brackett, & Pfund Series) – IR Hydrogen-like atoms • Solutions for H apply to He+ because everything hinges on the system being a “two-body problem.” Li++ works also, but soon things become overly artificial. Maxwell-Boltzmann Statistics Population of the orbital state ~ e-E /kT =#of electrons in the energy state Ei =Probability to find an electron at the energy state Ei Probability of an electron goes thru a transition from Ei to Ej is proportional to e-(E - E )/kT k is the Boltzmann’s constant = 1.38 x 10-23 J/K i i j Example (#38.21) A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on" to a particular lead nucleus and stops 6.50×10−14 m away from the center of the nucleus. (This point is well outside the nucleus). Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64×10−27 kg. (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. U(r) = k q1q2/r =(9x109)(2)(82)(1.6x10-19)2/(6.5x10-14) = 5.82x10-13 J 1 eV = 1.6x10-19 J, so U(r) = 3.63 MeV (b) What initial kinetic energy did the alpha particle have? Conservation of energy: initial KE = final PE = 3.63 MeV=5.82x10-13 J (c) What was the initial speed of the alpha particle? KE = ½ mv2 = 5.82x10-13 J for the alpha particle m = 6.64×10−27 kg, so v= 1.32 x 107 m/s = 0.044c γ = [1 – (v/c)2]-1/2 = 1.00097 non-relativistic is ok Lasers To achieve laser action, three conditions must be met The system must be in a state of population inversion The excited state of the system must be a metastable state More atoms in an excited state than the ground state Its lifetime must be long compared to the normal lifetime of an excited state The emitted photons must be confined in the system long enough to allow them to stimulate further emission from other excited atoms This is achieved by using reflecting mirrors Laser Beam – He Ne Example The energy level diagram for Ne in a He-Ne laser The mixture of helium and neon is confined to a glass tube sealed at the ends by mirrors A high voltage applied causes electrons to sweep through the tube, producing excited states When the electron falls to E2 from E*3 in Ne, a 632.8 nm photon is emitted Production of a Laser Beam (#38.30) Example – PRK Photorefractive keratectomy (PRK) is a laserbased surgery process that corrects near- and farsightedness by removing part of the lens of the eye to change its curvature and hence focal length. This procedure can remove layers 0.25 µm thick in pulses lasting 12 ns with a laser beam of wavelength 193 nm. Low-intensity beams can be used because each individual photon has enough energy to break the covalent bonds of the tissue. (a) In what part of the electromagnetic spectrum does this light lie? Visible light has wavelengths from about 400 nm to about 700 nm. 193 nm is shorter than 400 nm so it hc is in Ultra Violet (UV) −18 E = = 1.03 × 10 (b) What is the energy of a single photon? λ (c) If a 1.50 mW beam is used, how many photons are delivered to the lens in each pulse? E NE P = tot = t t so J = 6.44 eV Pt (1.50 × 10 −3 W)(12.0 × 10−9 s) N= = = 1.75 × 107 photons −18 E 1.03 × 10 J Example – recoil An atom with mass m emits a photon of wavelength λ. (a) What is the recoil speed of the atom? Assume a non-relativistic velocity and conserve momentum zero mass E = pc (#38.52) h ⇒ mv = λ or h v= . mλ (b) What is the kinetic energy K of the recoiling atom? 2 1 1 h h2 2 K = mv = m = 2 2 mλ 2mλ 2 (c) Find the ratio K/E, where E is the energy of the emitted photon. K h2 λ h Recoil becomes an important concern for small m and = ⋅ = . E 2mλ 2 hc 2mcλ small λ since this ratio becomes large in those limits.