Document Sample

“It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper, and it came back to hit you!” --E. Rutherford (on the „discovery‟ of the nucleus) Overview of the Rest of the Course Up to now: General properties and equations of quantum mechanics Time-independent Schrodinger’s Equation (SEQ) and eigenstates. Time dependent SEQ, superposition of eigenstates, time dependence. Collapse of the wave function, Schrodinger’s cat Tunneling This week: 3 dimensions, angular momentum, electron spin, H atom Exclusion principle, periodic table of atoms Next week: Molecules and solids, consequences of Q. M. Metals, insulators, semiconductors, superconductors, lasers, . . Final Exam: Monday, March 8 Homework 6: Due next Saturday (March 6), 8 am Lecture 11: 3D Potentials and the Hydrogen Atom 1 r / ao ( x, y, z ) ( x) ( y) ( z ) (r ) e ao 3 z 11 1 r = a0 L P(r)x) 0.5 L g( x 0 L 0 0 0 0 r2 4a0 4 0 x 4 2 h 2 2 2 13.6 eV En nx n y nz En x n y nz 2 2 8mL n Today 3-Dimensional Potential Well: Product Wavefunctions Concept of degeneracy Schrödinger’s Equation for the Hydrogen Atom: Semi-quantitative picture from uncertainty principle Ground state solution* Spherically-symmetric excited states (“s-states”)* *contains details beyond what we expect you to learn here Quantum Particles in 3D Potentials A real (2D) “quantum dot” So far, we have considered quantum particles bound in one-dimensional potentials. This situation can be applicable to certain physical systems but it lacks some of the features of many real 3D quantum systems, such as atoms and artificial structures. http://pages.unibas.ch/phys-meso/Pictures/pictures.html One consequence of confining a quantum particle in two or three dimensions is “degeneracy” -- the existence of several quantum states at the same energy level. To illustrate this important point in a simple system, we extend our favorite potential -- the infinite square well -- to three dimensions. Particle in a 3D Box (1) The extension of the Schrödinger Equation (SEQ) to 3D is straightforward in Cartesian (x,y,z) coordinates: 2 2 2 2 2 2 U ( x, y , z ) E where ( x, y , z ) 2m x y z 2 1 Kinetic energy term 2m p 2 x py pz 2 2 Let’s solve this SEQ for the particle in a 3D box: z outside box, x or y or z < 0 U(x,y,z) = 0 inside box L outside box, x or y or z > L x y L This simple U(x,y,z) can be “separated”: L U(x,y,z) = U(x) + U(y) + U(z) U = if any of the three terms = . Particle in a 3D Box (2) Whenever U(x,y,z) can be written as the sum of functions of the individual coordinates, we can write some wave functions as products of functions of the individual coordinates: (discussion is in the supplementary slides) ( x, y , z ) f ( x )g ( y )h( z ) 2D wave functions: n ny For the 3D square well, each function is sin x x sin y L L simply the solution to the 1D square well (nx,ny) = (1,1) problem: L (nx,ny) = (1,2) L 2 y n y 2 h n fn ( x ) N sin x x Enx x 0 L 2m 2L 0 x x Similarly for y and z. L L (nx,ny) = (2,1) (nx,ny) = (2,2) Each function contributes to the energy. L L The total energy is the sum: y y Ex + Ey + Ez = E 0 0 x x http://www.falstad.com/qm2dbox/ L L Particle in a 3D Box (3) The energy eigenstates and energy values in a 3D cubical box are: nx ny nz z N sin x sin y sin z L L L 2 h En x ny nz 8mL 2 nx ny nz 2 2 2 L x where nx,ny, and nz can each have values 1,2,3,…. y L L This problem illustrates 2 important points. Three quantum numbers (nx,ny,nz) are needed to identify the state of this three-dimensional system. That is true for every 3D system. More than one state can have the same energy: “Degeneracy”. Degeneracy reflects an underlying symmetry in the problem. 3 equivalent directions, because it’s a cube, not a rectangle. Act 1 Consider a particle in a 2D well, with Lx = Ly = L. 1. Compare the energies of the (2,2), (1,3), and (3,1) states? a. E(2,2) > E(1,3) = E(3,1) b. E(2,2) = E(1,3) = E(3,1) c. E(2,2) < E(1,3) = E(3,1) 2. If we squeeze the box in the x-direction (i.e., Lx < Ly) compare E(1,3) with E(3,1). a. E(1,3) < E(3,1) b. E(1,3) = E(3,1) c. E(1,3) > E(3,1) Act 1: Solution Consider a particle in a 2D well, with Lx = Ly = L. 1. Compare the energies of the (2,2), (1,3), and (3,1) states? a. E(2,2) > E(1,3) = E(3,1) b. E(2,2) = E(1,3) = E(3,1) E(1,3) = E(3,1) = E0 (12 + 32) = 10 E0 c. E(2,2) < E(1,3) = E(3,1) E(2,2) = E0 (22 + 22) = 8 E0 2 h E0 2 8mL 2. If we squeeze the box in the x-direction (i.e., Lx < Ly) compare E(1,3) with E(3,1). a. E(1,3) < E(3,1) b. E(1,3) = E(3,1) c. E(1,3) > E(3,1) Act 1: Solution Consider a particle in a 2D well, with Lx = Ly = L. 1. Compare the energies of the (2,2), (1,3), and (3,1) states? a. E(2,2) > E(1,3) = E(3,1) b. E(2,2) = E(1,3) = E(3,1) E(1,3) = E(3,1) = E0 (12 + 32) = 10 E0 c. E(2,2) < E(1,3) = E(3,1) E(2,2) = E0 (22 + 22) = 8 E0 2 h E0 2 8mL 2. If we squeeze the box in the x-direction (i.e., Lx < Ly) compare E(1,3) with E(3,1). a. E(1,3) < E(3,1) Because Lx < Ly, for a given n, the x b. E(1,3) = E(3,1) motion contributes more to E than the y motion. This effect is larger for larger nx. c. E(1,3) > E(3,1) Therefore, E(3,1) > E(1,3). Example: Lx = ½ , Ly = 1. E(1,3) 124 + 321 = 13 We say, “The degeneracy E(3,1) 324 + 121 = 37 has been lifted.” Energy Level Exercise (1) z Now back to the 3D cubic box: Show energies and label (nx,ny,nz) for the first 11 states of the particle in the 3D box, and write the L degeneracy D for each allowed energy. x Use Eo= h2/8mL2. y L L E (nx,nynz) n 2 h 2 2 2 En n y nz x n y nz 2 x 8 mL nx,ny,nz = 1,2,3,... 6Eo (2,1,1) (1,2,1) (1,1,2) D=3 The first four are given. 3Eo (1,1,1) D=1 Exercise (1) Solution z Now back to the 3D cubic box: Show energies and label (nx,ny,nz) for the first 11 states of the particle in the 3D box, and write the L degeneracy D for each allowed energy. x Use Eo= h2/8mL2. y L L E (nx,nynz) 12Eo (2,2,2) D=1 n 2 11Eo (3,1,1) (1,3,1) (1,1,3) D=3 h 2 2 2 En n y nz x n y nz 2 x 8 mL 9Eo (2,2,1) (2,1,2) (1,2,2) D=3 nx,ny,nz = 1,2,3,... 6Eo (2,1,1) (1,2,1) (1,1,2) D=3 3Eo (1,1,1) D=1 Act 2 For a cubical box, we just saw that the 5th energy level is at 12 E0, with a degeneracy of 1 and quantum numbers (2,2,2). 1. What is the energy of the next energy level? a. 13E0 b. 14E0 c. 15E0 2. What is the degeneracy of this energy level? a. 2 b. 4 c. 6 Act 2: Solution For a cubical box, we just saw that the 5th energy level is at 12 E0, with a degeneracy of 1 and quantum numbers (2,2,2). 1. What is the energy of the next energy level? a. 13E0 b. 14E0 c. 15E0 E1,2,3 = E0 (12 + 22 + 32) = 14 E0 2. What is the degeneracy of this energy level? a. 2 b. 4 c. 6 Act 2: Solution For a cubical box, we just saw that the 5th energy level is at 12 E0, with a degeneracy of 1 and quantum numbers (2,2,2). 1. What is the energy of the next energy level? a. 13E0 b. 14E0 c. 15E0 E1,2,3 = E0 (12 + 22 + 32) = 14 E0 2. What is the degeneracy of this energy level? a. 2 b. 4 c. 6 Any ordering of the three numbers will give the same energy. Because they are all different (distinguishable), the answer is 3! = 6. Exercise (2) z Now consider a non-cubic box: The box is stretched along the y-direction. L1 What will happen to the energy levels? x L2 > L1 y E L1 11Eo 9Eo 6Eo 3Eo Exercise (2) Solution z Now consider a non-cubic box: The box is stretched along the y-direction. L1 What will happen to the energy levels? x 2 2 h h En x ny nz 8mL 2 n x 2 nz 2 8mL n 2 y 2 y L2 > L1 1 2 E L1 11Eo The others are left for you. 9Eo 1: The symmetry is “broken” for y, so the 3-fold degeneracy is lowered. A 2-fold” degeneracy remains, 6Eo (2,1,1) (1,1,2) D=2 because x and z are still symmetric. (1,2,1) D=1 2: There is an overall lowering of 3Eo energies due to decreased (1,1,1) D=1 confinement along y. Another 3D System: The Atom -electrons confined in Coulomb field of a nucleus Early hints of the quantum nature of atoms: Discrete Emission and Absorption spectra Atomic hydrogen When excited in an electrical discharge, atoms emitted radiation only at discrete wavelengths Different emission spectra for different atoms (nm) Geiger-Marsden (Rutherford) Experiment (1911): Au Measured angular dependence of a particles (He ions) scattered from gold foil. Mostly scattering at small angles supported the v “plum pudding” model. But… Occasional scatterings at large angles Something massive in there! Conclusion: Most of atomic mass is concentrated in a small region of the atom a nucleus! Rutherford Experiment Hydrogen atom qualitatively Why doesn’t the electron collapse into the nucleus, where its potential energy is lowest? We must balance two effects: As the electron moves closer to the nucleus, e 2 U its potential energy decreases: r 2 However, as it becomes more and more p KE confined, its kinetic energy increases: r 2m r 2 e 2 2 Therefore, the total energy is: E KE PE 2 2m r r 2 This function has a minimum at: r a 0 0 .0 5 3 n m m e 2 At this radius, E m e 1 3 .6 e V 2 4 The “Bohr radius” of the H atom. 2 2 This is the ground state energy of the hydrogen atom. Heisenberg’s uncertainty principle prevents the atom’s collapse. Act 3 Consider an electron around a nucleus that has two protons, like an ionized Helium atom. 1. Compare the “effective Bohr radius” a0,He with the usual Bohr radius for hydrogen, a0: a. a0,He > a0 Hint: b. a0,He = a0 Look at the expression for a0 on the previous slide. What changes c. a0,He < a0 when we change H to He? 2. What is the ratio of ground state energies E0,He/E0,H? a. E0,He/E0,H = 1 b. E0,He/E0,H = 2 c. E0,He/E0,H = 4 Act 3 - Solution Consider an electron around a nucleus that has two protons, like an ionized Helium atom. 1. Compare the “effective Bohr radius” a0,He with the usual Bohr radius for hydrogen, a0: Look at how a0 depends on the charge: a. a0,He > a0 2 2 a0 a0 a0, He m e m (2e)e 2 b. a0,He = a0 2 c. a0,He < a0 This should make sense: more charge stronger attraction electron “sits” closer to the nucleus 2. What is the ratio of ground state energies E0,He/E0,H? a. E0,He/E0,H = 1 b. E0,He/E0,H = 2 c. E0,He/E0,H = 4 Act 3 - Solution Consider an electron around a nucleus that has two protons, (an ionized Helium atom). 1. Compare the “effective Bohr radius” a0,He with the usual Bohr radius for hydrogen, a0: Look at how a0 depends on the charge: a. a0,He > a0 2 2 a0 a0 a0, He m e m (2e)e 2 b. a0,He = a0 2 c. a0,He < a0 This should make sense: more charge stronger attraction electron “sits” closer to the nucleus 2. What is the ratio of ground state energies E0,He/E0,H? a. E0,He/E0,H = 1 Clearly the electron will be more tightly b. E0,He/E0,H = 2 bound, so |E0,He| > |E0,H| . How much c. E0,He/E0,H = 4 more tightly? Look at E0: e (2e)e 2 E0, H E0, He ao 4 E0, H 2ao 2 2 In general, for a “hydrogenic” atom with Z protons: E0, Z Z 2 E0, H Potential Energy for the Hydrogen Atom To solve this quantum mechanics problem, we U(r) must specify the potential energy of the electron. We assume that the Coulomb force binds the 0 r electron to the nucleus. e2 This problem does not separate in Cartesian U (r ) coordinates, because we cannot write r U(x,y,z) = Ux(x)+Uy(y)+Uz(z). 1 9 10 Nm / C 9 2 2 However, we can separate the potential in 4 0 spherical coordinates (r,,f), because: U(r,,f) = Ur(r) + U() + Uf(f) Therefore, we will be able to write: e r , , f R r f 2 0 0 r Question: How many quantum numbers will be needed to describe the hydrogen wave function? Product Wave Function in Spherical Coordinates We saw that because U depends only on the radius, the problem is separable. The hydrogen SEQ can be solved (but not by us). We will show you the solutions and discuss their physical significance. r z We can write: y nlm r , , f R nl r Y lm , f x f There are three quantum numbers: n “principal” (n 1) l “orbital” (0 l < n) This is what we called m “magnetic” (-l m +l) f before. The Ylm are called “spherical harmonics.” Today, we will only consider l = 0 and m = 0. These are called “s-states”. This simplifies the problem, because Y00(,f) is a constant and the wave function has no angular dependence: n 00 (r , ,f ) Rn 0 (r ) Note: Some of this nomenclature dates back to the 19th century, and has no physical significance. Radial Eigenstates of Hydrogen Here are graphs of the s-state wave functions, Rno(r) , for the electron in the Coulomb potential of the proton. The zeros in the subscripts are a reminder that these are states with l = 0 (zero angular momentum!). 1 1 1 1 3 2 R10 R20 f( x) 0.5 0.5 d4( x) R30 h( x) 0 0 0 00 .2 0 0 .5 00 2 4 00 5 10a0 10 00 5 r 10 15a0 15 rx 4a0 r 0 x 15 0 4 0 x 10 r r / 3 a0 2 r / a0 r r / 2a0 2r R1,0 (r ) e R2,0 (r ) 1 e R3,0 (r ) 3 2 e 2a0 a0 3a0 13.6 eV 2 a0 0.053 nm En me e 2 2 n Probability Density of Electrons 2 = Probability density = Probability per unit volume Rn02 for s-states. The density of dots plotted below is proportional to Rn02. 1s state 2s state 1 1 1 1 R10 0.5 R20 f( x) 0.5 h( x) 0 0 00 .2 0 00 0 10a0 4a0 rx 0 5 10 rx 2 4 0 4 0 10 Radial Probability Densities for s-states Summary of wave functions and radial probability densities for some s-states. 1 1 1 1 P10 The radial probability density has an extra R10 factor of r2 because there is more volume f( x) 0.5 g( x) 0.5 at large r. That is, P(r) r2|(r)|2. 0 00 0 4a0 0 00 0 This means that: 00 r 2 4 0 2 r 4 4a0 1 1 x 4 0 x 4 The most likely r is not 0 !!! R20 .5 0.4 P20 Even though that’s where h( x) 0.5 |(r)|2 is largest. h2( x) 0.2 00 .2 00 0 5 10a0 10 00 0 0 10a0 r x 10 0 0 r 5 x 10 10 3 P30 2 R30 ( x) 0 .5 0 http://www.falstad.com/qmatom/ 00 0 5 r x 10 15a0 15 15 0 r 20a0 radial wave functions radial probability densities, P(r) Example: Optical Transitions in Hydrogen An electron, initially excited to the n = 3 energy E level of the hydrogen atom, falls to the n = 2 level, 0 emitting a photon in the process. What are the n=3 energy and wavelength of the photon emitted? n=2 n=1 -15 eV Solution An electron, initially excited to the n = 3 energy E level of the hydrogen atom, falls to the n = 2 level, 0 emitting a photon in the process. What are the n=3 energy and wavelength of the photon emitted? n=2 13.6 eV En 2 n n=1 1 1 Eni n f 13.6 2 2 eV -15 eV n nf i 1 1 E photon E32 13.6 eV 1.9eV Atomic hydrogen 9 4 hc 1240eV nm 656nm E photon 1.9eV (nm) Question: Which transition is this? You will measure several transitions in Lab. = 486 nm) Next week: Laboratory 4 Cross Focusing Light Collimating Eyepiece hairs lens Grating shield lens Slit Light source Deflected beam Undeflected beam 13.6eV En 2 n Example Problem 1 Consider the three lowest energy states of the hydrogen atom. What wavelengths of light will be emitted when the electron jumps from one state to another? Solution: E21 = 10.2 eV E = -13.6 eV/n2, so E1 = -13.6 eV, E2 = -3.4 eV, E31 = 12.1 eV and E3 = -1.5 eV. There are three jumps to E32 = 1.9 eV consider, 2-to-1, 3-to-1, and 3-to-2. The photon carries away the energy that the electron loses. = h/p = hc/E hc = 1240 eV·nm 21 = 122 nm Two wavelengths are in the ultraviolet. 31 = 102 nm 32 = 653 nm The 3-to-2 transition gives a visible (red) photon. Example Problem 2 What is the normalization constant N? 1 1 R10 r / ao 100 (r , ,f ) NR10 (r ) Ne f( x) 0.5 0 2 0 The probability density is = N2 exp(-2r/ao). 00 0 r 2 x 4a0 4 4 In 3D, this means “probability per unit volume”. Require that the total probability = 1: 2 dV = 1 dV = r2 sin dr d df With spherical symmetry, the angular integrals give 4, so we are left with: 2 r / a0 1 4 N r dr 1 2 2 2 e N (integral table) a 3 0 o 1 r / ao Ground-state wave function 100 (r ) e ao 3 for hydrogen Example Problem 3 1 Estimate the probability of finding the 1 electron within a small sphere of radius r R10 (r) rs = 0.2 ao at the origin. f( x) 0.5 rs If it says “estimate”, don’t “integrate”. 0 00 00 2 4a0 4 The wave function is nearly constant near r = 0: 0 x 4 r / ao ( r ) Ne 1 0 / ao 1 (0) e ao 3 ao 3 Simply multiply 2 by the volume V = (4/3)rs3. 3 4 rs 2 Probability (0) V 0.0107 3 ao Example Problem 4 At what radius are you most likely to 1 1 r / ao find the electron? ( r ) Ne Looks like a no-brainer. r = 0, of course! f( x) 0.5 Well, that’s not the answer. r 0 0 You must find the probability P(r)r that the 00 rmax 2 4a0 4 electron is in a shell of thickness r at radius r. 0 x 4 The volume, V, of the shell increases with rmax radius: 2 Radial Probability P ( r )r ( r ) V 1 1 P(r)r Cr e r 2 -2 r / ao 4r22 Set dP/dr = 0 to find rmax = a0 g( x) 0.5 V = 4r2 r r 0 0 00 r2 4a0 4 r 0 x 4 Supplement: Separation of Variables In the 3D box, the SEQ is: 2 2 2 2 NOTE: 2 2 U ( x ) U ( y ) U ( z ) E 2m x y z Partial derivatives. 2 Let’s see if separation of variables works. Substitute this expression for into the SEQ: ( x, y , z ) f ( x )g ( y )h( z ) 2 2 d f 2 d g d h 2 NOTE: gh 2 fh 2 fg 2 U ( x ) U ( y ) U (z ) fgh Efgh Total derivatives. 2m dx dy dz Divide by fgh: 2 1 d 2f 1 d 2 g 1 d 2 h 2 U ( x ) U ( y ) U ( z ) E 2m f x 2 2 g dy h dz Supplement: Separation of Variables Regroup: 2 2 1d f 2 2 1d g 2 2 1d h U ( x ) U ( y ) U ( z ) E 2m f x 2 2 2 2m g dy 2m h dz A function of x A function of y A function of z We have three functions, each depending on a different variable, that must sum to a constant. Therefore, each function must be a constant: 2 2 1d f U( x ) Ex 2m f x 2 2 2 1d g 2 U ( y ) Ey Each function, f(x), g(y), and h(z) 2m g dy satisfies its own 1D SEQ. 2 2 1d h 2 U ( z ) Ez 2m h dz E x E y Ez E

DOCUMENT INFO

Shared By:

Categories:

Tags:
Lecture Notes, PowerPoint Presentation, PHYSICS DEPARTMENT, Unruh's interferometer, Modern Physics, Cambridge University Press, Final Exam, Search physics, Publishing Services, Hour Exams

Stats:

views: | 421 |

posted: | 5/23/2010 |

language: | English |

pages: | 38 |

OTHER DOCS BY decree

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.