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NUCLEAR PHYSICS Building Atoms, Molecules and solids U(r) +e r a r n=3 n=2 y6 n=1 y5 +e yeven +e r y4 y3 y2 y1 Last week of Course Today -- Atoms, Molecules, Solids States with many electrons – filled according to the Pauli exclusion principle Next time – Consequences of quantum mechanics Metals, insulators, semiconductors, superconductors, lasers, . . Overview Nuclear Spin and Magnetic Resonance Imaging (MRI) Atomic Configurations States in atoms with many electrons – filled according to the Pauli exclusion principle Molecular Wavefunctions: origins of covalent bonds Example: H + H H2 Electron energy bands in Solids States in atoms with many electrons – filled according to the Pauli exclusion principle Electron Spin and Nuclear Spin Last time: Electrons have ‘spin’, so we need FOUR quantum numbers to specify the full electronic state of a hydrogen atom 1 1 n, l, ml, ms (where ms = - /2 and + /2) Actually, the proton in the H atom ALSO has a spin, which is described by an additional quantum number, mp The energy difference between the two proton spin states in a magnetic field is 660 times smaller than for electron spin states! But… There are many more unpaired proton spins than unpaired electron spins in ordinary matter. Our bodies have many unpaired protons in H2O. Detect them …... In order to image tissue of various types, Magnetic Resonance Imaging detects the small difference in the numbers of “up” and “down” hydrogen proton spins generated when the object studied is placed in a magnetic field. Nobel Prize (2003): www.beckman.uiuc.edu/research/mri.html Lauterbur (UIUC) Nuclear Spin and MRI: Example Magnetic resonance imaging (MRI) depends on the absorption of electromagnetic radiation by the nuclear spin of the hydrogen atoms in our bodies. The nucleus is a proton with spin ½, so in a magnetic field B there are two energy states. The energy difference between these states is DE=2mpB, with mp = 1.41 x 10-26 J /Tesla. B=0 B0 DE = 2mpB B Question 1: The person to be scanned by an MRI machine is placed in a strong (1 Tesla) magnetic field. What is the energy difference between spin-up and spin-down proton states in this field? Question 2: What photon frequency, f, will be absorbed by this energy difference? Nuclear Spin and MRI: Example Magnetic resonance imaging (MRI) depends on the absorption of electromagnetic radiation by the nuclear spin of the hydrogen atoms in our bodies. The nucleus is a proton with spin ½, so in a magnetic field B there are two energy states. The energy difference between these states is DE=2mpB, with mp = 1.41 x 10-26 J /Tesla. B=0 B0 DE = 2mpB B Question 1: The person to be scanned by an MRI machine is placed in a strong (1 Tesla) magnetic field. What is the energy difference between spin-up and spin-down proton states in this field? Solution: DE = 2mpB = 2 x (1.41 x 10-26 J/T) x 1 T = 2.82 x 10-26 J x 1 eV/ 1.6 x 10-19 J = 1.76 x 10-7 eV Nuclear Spin and MRI: Example Magnetic resonance imaging (MRI) depends on the absorption of electromagnetic radiation by the nuclear spin of the hydrogen atoms in our bodies. The nucleus is a proton with spin ½, so in a magnetic field B there are two energy states. The energy difference between these states is DE=2mpB, with mp = 1.41 x 10-26 J /Tesla. B=0 B0 DE = 2mpB B Question 2: What photon frequency, f, will be absorbed by this energy difference? Solution: Energy conservation: so: Ephoton = hf = DE f = DE / h = 2.82 x 10-26 J / 6.626 x 10-34 J-sec = 42 MHz Radio waves Act 1 We just saw that radio-wave photons with energy 1.7 x 10-7 eV can cause a nuclear spin to flip. What is the angular momentum of each photon? a. 0 b. ħ/2 c. ħ Act 1 Solution We just saw that radio-wave photons with energy 1.7 x 10-7 eV can cause a nuclear spin to flip. What is the angular momentum of each photon? a. 0 b. ħ/2 c. ħ The nuclear spin has changed from to (or vice versa). That is, its z-component has changed by ħ. Conservation of angular momentum requires that the photon have carried (at least) this much in. Pauli Exclusion Principle Let’s start building more complicated atoms to study the Periodic Table. For atoms with many electrons (e.g., carbon: 6, iron: 26, etc.) what energies do the electrons have? “Pauli Exclusion Principle” (1925) No two electrons can be in the same quantum state. For example, in a given atom they cannot have the same set of quantum numbers n, l, ml, ms. This means that each atomic orbital (n,l,ml) can hold 2 electrons: ms = ±½. Important consequence: Electrons do not pile up in the lowest energy state. It’s more like filling a bucket with water. They are distributed among the higher energy levels according to the Exclusion Principle. Particles that obey this principle are called “fermions”. Protons and neutrons are also fermions, but photons are not. Filling Atomic Orbitals According to the Pauli Principle n s p d f g l=0 1 2 3 4 4 Energy En 13.6 eV 2 Z 2 n Lecture 11 In a multi-electron atom, the H-atom energy level diagram is distorted by Coulomb repulsion between electrons. Nevertheless, the H-atom diagram is useful (with some caveats) for figuring out the order in which orbitals are filled. 3 2 Example: Na (Z = 11) 1s2 2s2 2p6 3s1 1 Z = atomic number = # protons l label #orbitals (2l+1) 0 s 1 1 p 3 2 d 5 3 f 7 Act 2: Pauli Exclusion Principle 1. Which of the following states (n,l,ml,ms) is/are NOT allowed? a. b. c. d. e. (2, 1, 1, -1/2) (4, 0, 0, 1/2) (3, 2, 3, -1/2) (5, 2, 2, 1/2) (4, 4, 2, -1/2) 2. Which of the following atomic electron configurations violates the Pauli Exclusion Principle? a. b. c. d. e. 1s2, 2s2, 2p6, 3d10 1s2, 2s2, 2p6, 3d4 1s2, 2s2, 2p8, 3d8 1s1, 2s2, 2p6, 3d5 1s2, 2s2, 2p3, 3d11 Act 2: Solution 1. Which of the following states (n,l,ml,ms) is/are NOT allowed? a. b. c. d. e. (2, 1, 1, -1/2) (4, 0, 0, 1/2) (3, 2, 3, -1/2) (5, 2, 2, 1/2) (4, 4, 2, -1/2) ml > l l=n 2. Which of the following atomic electron configurations violates the Pauli Exclusion Principle? a. b. c. d. e. 1s2, 2s2, 2p6, 3d10 1s2, 2s2, 2p6, 3d4 1s2, 2s2, 2p8, 3d8 1s1, 2s2, 2p6, 3d5 1s2, 2s2, 2p3, 3d11 Act 2: Solution 1. Which of the following states (n,l,ml,ms) is/are NOT allowed? a. b. c. d. e. (2, 1, 1, -1/2) (4, 0, 0, 1/2) (3, 2, 3, -1/2) (5, 2, 2, 1/2) (4, 4, 2, -1/2) ml > l l=n 2. Which of the following atomic electron configurations violates the Pauli Exclusion Principle? a. b. c. d. e. 1s2, 2s2, 2p6, 3d10 1s2, 2s2, 2p6, 3d4 1s2, 2s2, 2p8, 3d8 1s1, 2s2, 2p6, 3d5 1s2, 2s2, 2p3, 3d11 Only 6 p-states. Only 10 d-states. Filling Procedure for Atomic Orbitals Due to electron-electron interactions, the hydrogen levels fail to give us the correct filling order as we go higher in the periodic table. The actual filling order is given in the table below. Electrons are added by proceeding along the arrows shown. Home exercise: Bromine is an element with Z = 35. Find its electronic configuration (e.g., 1s2 2s2 2p6 …). Note: The chemical properties of an atom are determined by the electrons in the highest n orbitals, because they are on the surface of the atom. Act 3: Pauli Exclusion Principle – Part 2 The Pauli exclusion principle applies to all fermions in all situations (not just to electrons in atoms). Consider electrons in a 2-dimensional infinite square well potential. 1. How many electrons can be in the first excited energy level? a. 1 b. 2 c. 3 d. 4 e. 5 Hint: Remember the (nx,ny) quantum numbers. 2. If there are 4 electrons in the well, what is the energy of the most energetic one (ignoring e-e interactions, and assuming the total energy is as low as possible)? a. (h2/8mL2) x 2 b. (h2/8mL2) x 5 c. (h2/8mL2) x 10 Act 3: Solution The Pauli exclusion principle applies to all fermions in all situations (not just to electrons in atoms). Consider electrons in a 2-dimensional infinite square well potential. 1. How many electrons can be in the first excited energy level? a. 1 b. 2 c. 3 d. 4 e. 5 The first excited energy level has (nx,ny) = (1,2) or (2,1). That is, it is degenerate. Each of these can hold two electrons (spin up and down). 2. If there are 4 electrons in the well, what is the energy of the most energetic one (ignoring e-e interactions, and assuming the total energy is as low as possible)? a. (h2/8mL2) x 2 b. (h2/8mL2) x 5 c. (h2/8mL2) x 10 Act 3: Solution The Pauli exclusion principle applies to all fermions in all situations (not just to electrons in atoms). Consider electrons in a 2-dimensional infinite square well potential. 1. How many electrons can be in the first excited energy level? a. 1 b. 2 c. 3 d. 4 e. 5 The first excited energy level has (nx,ny) = (1,2) or (2,1). That is, it is degenerate. Each of these can hold two electrons (spin up and down). 2. If there are 4 electrons in the well, what is the energy of the most energetic one (ignoring e-e interactions, and assuming the total energy is as low as possible)? a. (h2/8mL2) x 2 b. (h2/8mL2) x 5 c. (h2/8mL2) x 10 Two electrons are in the (1,1) state, and two are in the (2,1) or (1,2) state. So, Emax = (12+22)(h2/8mL2). Bonding Between Atoms -- How can two neutral objects stick together? H + H H2 +e Let’s represent the atom in space by its Coulomb potential centered on the proton (+e): r n=3 n=2 n=1 +e The potential energy of the two protons in an H2 molecule look something like this: +e r The energy levels for this potential are more complicated, so we consider a simpler potential that we already know a lot about. Particle in a Finite Square Well Potential This has all of the qualitative features of molecular bonding, but is easier to analyze.. The ‘atomic’ potential: Bound states The ‘molecular’ potential: Consider what happens when two “atoms” approach one other. There is one electron, which can be in either well. ‘Molecular’ Wave functions and Energies “Atomic’ wave functions: yA ‘Molecular’ Wavefunctions: L = 1 nm 1.505 eV nm 2 E 0.4 eV 2 (2L) 2 ‘atomic’ states 2 ‘molecular’ states yeven yodd When the wells are far apart, the ‘atomic’ functions don’t overlap. The single electron can be in either well with E = 0.4 eV. ‘Molecular’ Wave Functions and Energies Wells far apart: d Degenerate states: yeven 1.505 eV nm2 E 0.4 eV (2L)2 yodd Wells closer together: yeven yodd L = 1 nm 1.505 eV nm2 E 0.4 eV (2L)2 d ‘Atomic’ states are beginning to overlap and distort. yeven and yodd are not the same. The degeneracy is broken: Eeven < Eodd (why?) yeven: no nodes, yodd: one node. Act 4: Symmetric vs. Antisymmetric states d yeven yodd What will happen to the energy of yeven as the two wells come together (i.e., as d is reduced)? [Hint: think of the limit as d 0] a. Eeven decreases. b. Eeven stays the same. c. Eeven increases. Act 4: Solution d yeven yodd What will happen to the energy of yeven as the two wells come together (i.e., as d is reduced)? [Hint: think of the limit as d 0] a. Eeven decreases. b. Eeven stays the same. c. Eeven increases. As the two wells come together, the barrier disappears, and the wave function spreads out over a single double-width well. Therefore the energy goes down (by a factor of 4). Energy as a Function of Well Separation When the wells just touch (d = 0, becoming one well) we know the energies: 2L = 2 nm yeven 1.505 eV nm2 E1 0.1 eV (4L)2 (n = 1 state) yodd 1.505 eV nm2 2 E2 2 0.4 eV (4L)2 (n = 2 state) As the wells are brought together, the even state always has lower kinetic energy (smaller curvature). The odd state stays at about the same energy. odd 0.4 eV f 0.1 eV Splitting between even and odd states: even d DE = 0.4 – 0.1 eV = 0.3 eV Molecular Wave functions and Energies with the Coulomb Potential To understand real molecular bonding, we must deal with two issues: The atomic potential is not a square well. There is more than one electron in the well. +e r Atomic ground state (1s): yA n=1 Molecular states: +e +e +e r +e yodd yeven Bonding state Antibonding state r Energy as a Function of Atom Separation The even and odd states behave similarly to the square well, but there is also repulsion between the nuclei that prevents them from coming too close. Schematic picture for the total energy of the nuclei and one electron: Binding energy Let’s consider what happens when there is more than one electron: Anti-bonding state Bonding state Equilibrium bond length 2 electrons (two neutral H atoms): Both electrons occupy the bonding state (with different ms). This is H2. 4 electrons (two neutral He atoms). Two electrons must be in the anti-bonding state. The repulsive force cancels the bonding, and the atoms don’t stick. The He2 molecule does not exist. d Electron states in a crystal (1) +e Again start with a simple atomic state: r yA n=1 Bring N atoms together forming a 1-d crystal (a periodic lattice). N atomic states N crystal states. What do these crystal states look like? Like molecular bonding, the total wave function is (approximately) just a superposition of 1-atom orbitals. Electron states in a crystal (2) The lowest energy combination is just the sum of the atomic states. This is a generalization of the 2-atom bonding state. No nodes! The highest energy state is the one where every adjacent pair of atoms has a minus sign: N-1 nodes! There are N states, with energies lying between these extremes. Kruse Energy bands Electron Wavefunctions and Energy Band Highest energy wavefunction Energy y6 y5 y4 y3 y2 Closely spaced energy levels form a “band” – a continuum of energies between the max and min energies y1 Lowest energy wavefunction FYI: These states are called “Bloch states” after Felix Bloch who derived the mathematical form in 1929. They can be written as: y n (x) u(x)eik x n where u is an atomic-like function and the exponential is a convenient way to represent both sin and cos functions Energy Bands and Band Gaps In a crystal the number of atoms is very large and the states approach a continuum of energies between the lowest and highest a “band” of energies. A band has exactly enough states to hold 2 electron per atom (spin up and down). Each 1-atom state leads to an energy band. So a crystal has multiple energy bands. Band gaps (regions of disallowed energies) lie between the bands. 2s band Band gap 1s band Summary Atomic configurations States in atoms with many electrons – filled according to the Pauli exclusion principle Molecular wave functions: origins of covalent bonds Example: H + H H2 Electron energy bands in crystals Bands and band gaps are properties of waves in periodic systems. There is a continuous range of energies for “allowed” states of an electron in a crystal. A Band Gap is a range of energies where there are no allowed states Bands are filled according to the Pauli exclusion principle