Chapter 41

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## Chapter 41

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**Chapter 41**Quantum Mechanics**de Broglie’s Idea**• Particles had wave characteristics just like waves had particle characteristics (e.g. E&M wave photon) • The de Broglie wavelength of a particle of mass, m, is • The frequency of matter waves is**Uncertainty Principle and wave character**• Mathematically, • For a wave with λ one has Δx~λ and by deBroglie Δp~h/λ • Thus ΔxΔp~h**Quantum mechanical dual nature (Light)**• All things have a dual particle/wave nature. Which property is important/noticeable depends on the experiment • The light two slit experiment reveals wave nature of light • The Compton experiment, photoelectric effect reveals the particle nature of light**Quantum mechanical dual nature (particles)**• De Broglie’s hypothesis suggested a wave character for particles. • This wave nature revealed itself in the Davisson-Germer experiment. • Also in the particle two-slit experiment**The particle two slit experiment II.**• Which slit does a particle go through? • The answer is both! Probability wave**The Wave Function ψ**• What are these matter waves of? • They are probability waves described by a complex valued function of x and t i.e. Ψ(x,t) • The probability to find the particle in some dx is given**Properties of the Wave Function**• Normalization • Finite probability • Expectation value (average)**Particle in 1D Hard Box I**• Classically the particle has equal probability to be anywhere in the box • But if xi and pi are known then one knows exactly where the ball is at any later time**Particle in a 1D Hard Wall Box II**• Quantum mechanically L=nλ/2 or λ=2L/n (n=1,2,3…) • The wave function (guess) ψ(x)=Asin(kx) with k=2π/λ=nπ /L • Giving ψ(x)=Asin(n πx/L). A=√(2/L) by normalization**Particle in a 1D Hard Wall Box III**• The energy spectrum is found as follows • De Broglie p=h/λ=hn/2L (n=1, 2, 3, …) • Energy En = p2 /2m = h2 n2 / (8 m L2 ) • n=0 not allowed. Lowest energy is n=1 the zero point energy. This is different from classical mechanics where E=0 is possible. This is connected with the Heisenberg uncertainty principle.**The Schrödinger Equation for ψ I**• In 1926 Schrödinger proposed a wave equation that describes the manner in which de Broglie matter waves change in space and time • Schrödinger’s wave equation is a key element in quantum mechanics • Schrödinger’s wave equation is generally solved for the wave function, Ψ**The Schrödinger Equation for ψ II**• The 1D Schrödinger Equation for a particle with energy E, moving in a potential U(x), having mass, m, is • The full (complex) wave function is**Particle in 1D box (again)**• In terms of Schrödinger’s Equation the 1D box problem becomes • Where • Solutions • Boundary Conditions give**The Schrödinger Cat Problem**• It’s OK for electrons and other small things but what about marcoscopic things like a cat?**Emission Spectra**• A gas at low pressure has a voltage applied to it • A gas emits light characteristic of the gas • When the emitted light is analyzed with a spectrometer, a series of discrete bright lines is observed • Each line has a different wavelength and color • This series of lines is called an emission spectrum**Emission Spectrum of Hydrogen – Equation**• The wavelengths of hydrogen’s spectral lines can be found from • RH is the Rydberg constant • RH = 1.0973732 x 107 m-1 • n is an integer, n = 1, 2, 3, … • The spectral lines correspond to different values of n**Spectral Lines of Hydrogen**• The Balmer Series has lines whose wavelengths are given by the preceding equation • Examples of spectral lines • n = 3, λ = 656.3 nm • n = 4, λ = 486.1 nm**The Bohr Theory of Hydrogen**• In 1913 Bohr provided an explanation of atomic spectra that includes some features of the currently accepted theory • His model includes both classical and non-classical ideas • His model included an attempt to explain why the atom was stable**Bohr’s Assumptions for Hydrogen**• The electron moves in circular orbits around the proton under the influence of the Coulomb force of attraction • The Coulomb force produces the centripetal acceleration**Bohr’s Assumptions, cont**• Only certain electron orbits are stable • These are the orbits in which the atom does not emit energy in the form of electromagnetic radiation • Therefore, the energy of the atom remains constant and classical mechanics can be used to describe the electron’s motion • Radiation is emitted by the atom when the electron “jumps” from a more energetic initial state to a lower state • The “jump” cannot be treated classically**Bohr’s Assumptions, final**• The electron’s “jump,” continued • The frequency emitted in the “jump” is related to the change in the atom’s energy • It is generally not the same as the frequency of the electron’s orbital motion • The size of the allowed electron orbits is determined by a condition imposed on the electron’s orbital angular momentum**Mathematics of Bohr’s Assumptions and Results**• Electron’s orbital angular momentum • me v r = n ħ where n = 1, 2, 3, … • The total energy of the atom • The energy can also be expressed as**Bohr Radius**• The radii of the Bohr orbits are quantized • This shows that the electron can only exist in certain allowed orbits determined by the integer n • When n = 1, the orbit has the smallest radius, called the Bohr radius, ao • ao = 0.0529 nm**Radii and Energy of Orbits**• A general expression for the radius of any orbit in a hydrogen atom is • rn = n2 ao • The energy of any orbit is • En = - 13.6 eV/ n2**Specific Energy Levels**• The lowest energy state is called the ground state • This corresponds to n = 1 • Energy is –13.6 eV • The next energy level has an energy of –3.40 eV • The energies can be compiled in an energy level diagram • The ionization energy is the energy needed to completely remove the electron from the atom • The ionization energy for hydrogen is 13.6 eV**Energy Level Diagram**• The value of RH from Bohr’s analysis is in excellent agreement with the experimental value • A more generalized equation can be used to find the wavelengths of any spectral lines**Generalized Equation**• For the Balmer series, nf = 2 • For the Lyman series, nf = 1 • Whenever an transition occurs between a state, ni to another state, nf (where ni > nf), a photon is emitted • The photon has a frequency f = (Ei – Ef)/h and wavelength λ**Successes of the Bohr Theory**• Explained several features of the hydrogen spectrum • Accounts for Balmer and other series • Predicts a value for RH that agrees with the experimental value • Gives an expression for the radius of the atom • Predicts energy levels of hydrogen • Gives a model of what the atom looks like and how it behaves • Can be extended to “hydrogen-like” atoms • Those with one electron • Ze2 needs to be substituted for e2 in equations • Z is the atomic number of the element**A non-relativistic electron and a non-relativistic proton**are moving and have the same de Broglie wavelength. Which of the following are also the same for the two particles: (a) speed, (b) kinetic energy, (c) momentum, (d) frequency? QUICK QUIZ 27.4**(c). Two particles with the same de Broglie wavelength will**have the same momentum p = mv. If the electron and proton have the same momentum, they cannot have the same speed because of the difference in their masses. For the same reason, remembering that KE = p2/2m, they cannot have the same kinetic energy. Because the kinetic energy is the only type of energy an isolated particle can have, and we have argued that the particles have different energies, Equation 27.15 tells us that the particles do not have the same frequency. QUICK QUIZ 27.4 ANSWER**We have seen two wavelengths assigned to the electron, the**Compton wavelength and the de Broglie wavelength. Which is an actual physical wavelength associated with the electron: (a) the Compton wavelength, (b) the de Broglie wavelength, (c) both wavelengths, (d) neither wavelength? QUICK QUIZ 27.5**(b). The Compton wavelength, λC = h/mec, is a combination**of constants and has no relation to the motion of the electron. The de Broglie wavelength, λ= h/mev, is associated with the motion of the electron through its momentum. QUICK QUIZ 27.5 ANSWER**The Electron Microscope**• The electron microscope depends on the wave characteristics of electrons • Microscopes can only resolve details that are slightly smaller than the wavelength of the radiation used to illuminate the object • The electrons can be accelerated to high energies and have small wavelengths