Chapter 3 Crystal Binding Chapter 3 Crystal Binding Introduction What forces hold solids together? electrostatic forces between electrons, ions, nuclei Pauli principle There are attractive forces and[.]
Chapter 3: Crystal Binding Introduction What forces hold solids together? - electrostatic forces between electrons, ions, nuclei - Pauli principle There are attractive forces and repulsive forces Cohesive energy = energy that must be added to the crystal in order to separate it into neutral, free atoms, far apart There is a correlation between cohesive energy and - melting temperature - bulk modulii NOT with ionization energy (energy needed to excite an electron from the lattice) Ionization Energy • • • • • • This is the energy needed to form a positive ion: Eg Rb + Br → Rb+ + Br- - I + E I – Ionization Energy E – Electron Affinity (energy gained in forming a negative ion) Neglected: Coulomb energy of attraction between Rb+ and BrExercise: Calculate how much energy it costs to create RbBr from a pair at infinite distance FClBrILi+ Na+ K+ Rb+ Cs+ E/I (eV) 3.6 3.7 3.5 3.2 5.4 5.1 4.3 4.2 3.9 Cohesive Energies Transition metals – strongly bound Alkali metals – Intermediate values Inert gases – weakly bound The cohesive energy is related to the melting points (See tables in Chapter of Kittel) It is also related to the bulk modulus: ∆p B=− ∆V / VO (this is how pressure changes when you change the volume) Types of Binding Inert gases –van der Waals forces Ionic crystals – Na+Cl- (electrostatic charge) Covalent crystals – diamond (electronic overlap – chemical bonding) Metals (sea of electrons) Hydrogen bond (special case of bonding) Crystals of Inert Gases • • • • • Inert gases form the simplest crystals The electron distribution is made of filled shells They are insulators, and have low melting temperatures (why?) The atoms have very high ionization energies (why?) Since the atoms have filled electronic shells, they pack as close together as possible (nearly ideal c/a ratios), in the FCC stucture (except for He3 and He4) Properties of Ideal Gases The special case of He3 and He4 • • • He3 and He4 not solidify, even at zero temperature (!) This is entirely a quantum effect Remember, as atoms cool down to form solids, they can still vibrate within the lattice (there is what is called zero-point motion) This effect is strongest for low-mass atoms Frequency of oscillation f = 2π k m At low temperatures, the oscillations of He3 and He4 atoms are still significant In fact, the average fluctuation from it’s equilibrium position is 30-40% of the nearestneighbour distance This is enough to prevent solid formation Spring constant (V= ½ kx2) Mass of oscillator (From Harmonic Oscillator) A lattice of oscillating atoms More on zero-point motion • Another way to look at this is from the point of view of the Heisenburg Uncertainty Principle: Uncertainty in position ∆x∆p ≈ h Planck’s constant/2π Uncertainty of momentum • So, to calculate the spread in energy of a particle (ie How much it is fluctuating): (∆p ) h ∆E ≈ = 2m 2m (∆x) • • This is large if m is small, so this effect is largest for He It is even larger for electrons (where, in this case, it actually gives rise to the Pauli Principle – no two electrons can have the same quantum state) Homework: Calculate, in eV, how large of an effect this is for Neon What holds an inert gas crystal together? • What holds the crystal together? • The electronic distribution cannot be significantly distorted from free atoms – the cohesive energy of an atom in a crystal is less than % or less of the ionization of an atomic electron • Therefore, not much energy is available to distort the free atom charge distributions • But, we will find that small distortions of the electron clouds, which cause instantaneous and temporary attractive forces (due to electrostatic charge), will be enough to hold these crystals together Small, induced dipoles hold these solids together • These are called van der Waals forces Van der Waals Interaction • Consider two atoms, which are surrounded by electronic clouds of charge At any given time, these clouds can distort and give rise to temporary dipoles of charge + - + - • These charges can interact via dipolar forces It is these forces which hold atoms together in inert gas solids (which is pretty amazing, if you think about it) Van der Waals Interaction • For now, we are going to look at two atoms, and show that they can give rise to this interaction We will put identical atoms at a distance R from each other, where R is much bigger than the electron clouds about each atom • These charge clouds can be distorted a bit (so when there is more charge in one part, it has a negative sign, and less, positive) Since these distortions are only temporary, we will describe them as harmonic oscillators (what is oscillating is the charge) Van der Waals Interaction • Each oscillator will have a charge +/- e, with separations x1 and x2 For now, we will just let them oscillate about the x-axis Let p1 and p2 be the momenta The force constant is C (spring constant, k) The Hamiltonian describing the total energy of this system is: • Each oscillator will have a frequency defined by C = mωO2 (this is what we wrote earlier for the Harmonic Oscillator) Our total energy will be H = H0 + H1 (harmonic oscillator + dipole contribution from coloumb energy) • Van der Waals Interaction • Let H1 be the coulomb interaction energy of the two oscillators This is just due to the induced charge • This reduces to: (on next slide) Van der Waals Interaction • For |x1|, |x2| < R, we then can expand this to give: e2 e2 e2 e2 H1 = + − − x x x − R R(1 + ) R (1 + ) R(1 − x2 ) R R R • Using the binomial expansion: (1+ x)-1 = 1-x+x2+… (x is small) e2 e2 x1 − x2 x1 − x2 e2 x1 x1 e2 − x2 − x2 + − − + − − + H1 = + (1 − ) ( ) ( ) R R R R R R R R R R e (− x1 + x2 + x1 − x2 ) e 2 2 + − − − H1 = x x x x [( ) 2 ] R R e2 2 2 H1 = x1 − x1 x2 + x2 − x1 − x2 R [ ] 2 Van der Waals Interaction • With the problem of the Harmonic Oscillator, it is convenient to change coordinates So, instead of using x1 and x2, we are going to use xs and xa These stand for symmetric and antisymmetric modes of motion (why?) This is also called diagonalizing the Hamiltonian to find it’s normal modes of vibration • We can also put the momenta in the same coordinate system, calling p1 and p2 now ps and pa Van der Waals Interaction • • Now our total Hamiltonian is H = H0 + H1 (harmonic oscillator part + dipole part) So, putting it all together, with the new coordinates, we have: Spring constant • • What is the frequency of these oscillators? Look for 1/2Cx2 (or 1/2kx2) terms, then use C = mωo2 Expanding the root • What we have done here is expand the root: 2e / ω = ωO [1 ± ( )] CR 2 2e 2e ω = ωO [1 ± ( ) − ( ) + ] CR CR Because : n(n − 1) x + (1 ± x) = ± nx + n