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## Chapter 5

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**Chapter 5**The Wavelike Properties of Particles**CHAPTER 5**Wave Properties of Matter and Quantum Mechanics • De Broglie Waves • Electron Scattering • Wave Motion • Waves or Particles? • Uncertainty Principle • Probability, Wave Functions, and the Copenhagen Interpretation • Particle in a Box Louis de Broglie (1892-1987) I thus arrived at the overall concept which guided my studies: for both matter and radiations, light in particular, it is necessary to introduce the corpuscle concept and the wave concept at the same time. - Louis de Broglie, 1929**The Wavelike Properties of Particles**• The de Broglie Hypothesis • Measurements of Particles Wavelengths • Wave Packets • The Probabilistic Interpretation of the Wave Function • The Uncertainty Principle • Some Consequences of Uncertainty Principle • Wave-Particle Duality**De Broglie Waves**If a light-wave could also act like a particle, why shouldn’t matter-particles also act like waves? • In his thesis in 1923, Prince Louis V. de Broglie suggested that mass particles should have wave properties similar to electromagnetic radiation. • The energy can be written as: hf = pc = plf • Thus the wavelength of a matter wave is called the De Broglie wavelength: Louis V. de Broglie(1892-1987)**The de Broglie Hypothesis**Since light seems to have both wave and particle properties, it is natural to ask whether matter (electrons, protons) might also have both wave and particle characteristics. For the wavelength of electron, de Broglie chose: λ = h/p f = E/h whereEis the total energy, pis the momentum, and λis called the de Broglie wavelength of the particle.**The de Broglie Hypothesis**For photons these same equations results directly from Einstein’s quantization of radiationE = hfand equation for an energy of a photon with zero rest energyE = pc: Using relativistic mechanics de Broglie demonstrated, that this equation can also be applied to particles with mass and used them to physical interpretation of Bohr’s hydrogen-like atom.**The de Broglie Wavelength**Using de Broglie relation let’s find the wavelength of a10-6gparticle moving with a speed10-6m/s: Since the wavelength found in this example is so small, much smaller than any possible apertures, diffraction or interference of such waves can not be observed.**The de Broglie Wavelength**The situation are different for low energy electrons and other microscopic particles. Consider a particle with kinetic energyK.**The de Broglie Wavelength**The situation are different for low energy electrons and other microscopic particles. Consider a particle with kinetic energyK. Its momentum is found from Its wavelength is then**The de Broglie Wavelength**If we multiply the numerator and denominator bycwe obtain: Where mc2=0.511MeVfor electrons, andKin electron-volts.**The de Broglie Wavelength**We obtained the electron wavelength: Similarly, for proton (mc2 = 938 MeVfor protons)**The de Broglie Wavelength**For the molecules of a stationary gas at the absolute temperatureT, the square average speed of the moleculev2is determined by Maxwell’s Law Then the momentum of the molecule is: Knowing that the mass ofHeatom, for instance, is6.7x10-24g, (kB=1.38x10-23J/K) we obtain for He wavelength:**The de Broglie Wavelength**Similarly, for the molecule of hydrogen . and for the thermal neutrons This calculations shows, that for the acceleratedelectrons, foratoms of helium, hydrogen moleculesunder the room temperature, forthermal neutronsand other “slow” light particles de Broglie wavelength is on the same order as for softX-rays.**Show that the wavelength of a nonrelativistic neutron is**where Knis the kinetic energy of the neutron in electron-volts. (b) What is the wavelength of a 1.00-keV neutron?**Show that the wavelength of a nonrelativistic neutron is**where Knis the kinetic energy of the neutron in electron-volts. (b) What is the wavelength of a 1.00-keV neutron? Kinetic energy, K, in this equation is in Joules (a)**Show that the wavelength of a nonrelativistic neutron is**where Knis the kinetic energy of the neutron in electron-volts. (b) What is the wavelength of a 1.00-keV neutron? (b)**The nucleus of an atom is on the order of 10–14 m in**diameter. For an electron to be confined to a nucleus, its de Broglie wavelength would have to be on this order of magnitude or smaller. (a) What would be the kinetic energy of an electron confined to this region? (b) Given that typical binding energies of electrons in atoms are measured to be on the order of a few eV, would you expect to find an electron in a nucleus?**The nucleus of an atom is on the order of 10–14 m in**diameter. For an electron to be confined to a nucleus, its de Broglie wavelength would have to be on this order of magnitude or smaller. (a) What would be the kinetic energy of an electron confined to this region? (b) Given that typical binding energies of electrons in atoms are measured to be on the order of a few eV, would you expect to find an electron in a nucleus? (a)**The nucleus of an atom is on the order of 10–14 m in**diameter. For an electron to be confined to a nucleus, its de Broglie wavelength would have to be on this order of magnitude or smaller. (a) What would be the kinetic energy of an electron confined to this region? (b) Given that typical binding energies of electrons in atoms are measured to be on the order of a few eV, would you expect to find an electron in a nucleus? , (b) With its**The calculations show, that for the acceleratedelectrons,**foratoms of helium, hydrogen moleculesunder the room temperature, forthermal neutronsand other “slow” light particles de Broglie wavelength is on the same order as for softX-rays. So, we can expect, that diffraction can be observed for this particles**Electron Interference and Diffraction**The electron wave interference was discovered in 1927 by C.J. Davisson and L.H.Germer as they were studying electron scattering from a nickel target at the Bell Telephone Laboratories. After heating the target to remove an oxide coating that had accumulate after accidental break in the vacuum system, they found that the scattered electron intensity is a function of the scattered angle and show maxima and minima. Their target had crystallized during the heating, and by accident they had observed electron diffraction. Then Davisson and Germer prepared a target from a single crystal of nickel and investigated this phenomenon.**The Davisson-Germer experiment.**Low energy electrons scattered at angleΦfrom a nickel crystal are detected in an ionization chamber. The kinetic energy of electrons could be varied by changing the accelerating voltage on the electron gun.**Scattered intensity vs detector angle for 54-ev electrons.**Polar plot of the data. The intensity at each angle is indicated by the distance of the point from the origin. Scattered angle Φis plotted clockwise started at the vertical axis.**The same data plotted on a Cartesian graph. The intensity**scale are the same on the both graphs. In each plot there is maximum intensity atΦ=50º, as predicted for Bragg scattering of waves having wavelengthλ = h/p.**Scattering of electron by crystal. Electron waves are**strongly scattered if the Bragg conditionnλ = D SinΦis met.**In 1925, Davisson and Germer experimentally observed that**electrons were diffracted (much like x-rays) in nickel crystals. George P. Thomson (1892–1975), son of J. J. Thomson, reported seeing electron diffraction in transmission experiments on celluloid, gold, aluminum, and platinum. A randomly oriented polycrystalline sample of SnO2 produces rings.**Test of the de Broglie formulaλ = h/p. The wavelength is**computed from a plot of the diffraction data for electrons plotted against V0-1/2, whereV0is the accelerating voltage. The straight line is1.226V0-1/2nm as predicted fromλ = h/(2mE)-1/2**Test of the de Broglie formulaλ = h/p. The wavelength is**computed from a plot of the diffraction data plotted against V0-1/2, whereV0is the accelerating voltage. The straight line is1.226V0-1/2nm as predicted fromλ = h/(2mE)-1/2**A series of a polar graphs of Davisson and Germer’s data**at electron accelerating potential from36 Vto 68 V. Note the development of the peak atΦ = 50ºto a maximum whenV0 = 54 V.**Variation of the scattered electron intensity with**wavelength for constantΦ. The incident beam in this case was10º from the normal, the resulting diffraction causing the measured peaks to be slightly shifted from the positions computed fromnλ = D Sin Φ.**Schematic arrangement used for producing a diffraction**pattern from a polycrystalline aluminum target.**Diffraction pattern produced byx-raysof wavelength0.071**nmand an aluminum foil target.**Diffraction pattern produced by600-eVelectrons and an**aluminum foil target ( de Broigle wavelength of about0.05 nm: )**Diffraction pattern produced by600-eVelectrons and an**aluminum foil target ( de Broigle wavelength of about0.05 nm: ) .**Diffraction pattern produced by0.0568-eVneutrons (de Broglie**wavelength of0.120 nm) and a target of polycrystalline copper. Note the similarity in the pattern produced byx-rays,electrons, and neutrons.**Diffraction pattern produced by0.0568-eVneutrons (de Broglie**wavelength of0.120 nm) and a target of polycrystalline copper. Note the similarity in the pattern produced byx-rays,electrons, and neutrons.**In the Davisson–Germer experiment, 54.0-eV electrons were**diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed at φ = 50.0°, what was the lattice spacing a between the vertical rows of atoms in the figure? (It is not the same as the spacing between the horizontal rows of atoms.)**In the Davisson–Germer experiment, 54.0-eV electrons were**diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed at φ = 50.0°, what was the lattice spacing a between the vertical rows of atoms in the figure? (It is not the same as the spacing between the horizontal rows of atoms.)**In the Davisson–Germer experiment, 54.0-eV electrons were**diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed at φ = 50.0°, what was the lattice spacing a between the vertical rows of atoms in the figure? (It is not the same as the spacing between the horizontal rows of atoms.)**A photon has an energy equal to the kinetic energy of a**particle moving with a speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having a speed of 0.00100c? (c) What value does the ratio of the two wavelengths approach at high particle speeds?(d) At low particle speeds?**A photon has an energy equal to the kinetic energy of a**particle moving with a speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having a speed of 0.00100c? (c) What value does the ratio of the two wavelengths approach at high particle speeds? (d) At low particle speeds? For a particle: For a photon:**A photon has an energy equal to the kinetic energy of a**particle moving with a speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having a speed of 0.00100c? (c) What value does the ratio of the two wavelengths approach at high particle speeds?(d) At low particle speeds? (a) (b)**A photon has an energy equal to the kinetic energy of a**particle moving with a speed of 0.900c. (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having a speed of 0.00100c? (c) What value does the ratio of the two wavelengths approach at high particle speed? (d) At low particle speed? As (c) becomes nearly equal to γ. and , (d)**What is “waving”? For matter it is the probability of**finding the particle that waves. Classical waves are the solution of the classical wave equation Harmonic waves of amplitudey0, frequency fand periodT: where the angular frequencyωand the wave numberkare defined by and the wave or phase velocityvpis given by**If the film were to be observed at various stages, such**as after being struck by28 electronsthe pattern of individually exposed grains will be similar to shown here.**After exposure by about1000 electronsthe pattern will be**similar to this.**And again for exposure of about10,000 electronswe will**obtained a pattern like this.**Two source interference pattern. If the sources are coherent**and in phase, the waves from the sources interfere constructively at points for which the path difference dsinθis an integer number of wavelength.**Grows of two-slits interference pattern. The photo is an**actual two-slit electron interference pattern in which the film was exposed to millions of electrons. The pattern is identical to that usually obtained with photons.