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Unformatted text preview: CHEM 120A Problem Set 9 Solutions David Hoffman , Doran Bennett , Tara Yacovitch 4/26/2010 1. (a) We’re asked to verify that  Φ k i =  ψ k i ψ k i is also and eigen solution of our Hamiltonian. Let’s see we know that  ψ k i is degenerate with  ψ k i which means they have the same energy eigenvalue associated with them (you can prove this to yourself by finding the action of H on both eigenstates keeping in mind that cosine is an even function). If they both have the same energy eigenvalues then we have, H  Φ k i = H  ψ k i  H  ψ k i = E  ψ k i  E  ψ k i = E  Φ k i (1a) The second part of this problem involves evaluating Φ k (0) which I’m confident you are capable of doing. (b) We want to show that the boundary condition is satisfied for certain values of k . Before proceeding note that, Φ k ( x ) = sin( kx ) (1b) Continuing, Φ k ( b + Mb ) = sin( kx ) = sin( πs ) = 0 (1c) The final equality holds because s is an integer. (c) Here we have a long, polyunsaturated alkyl chain and we want to estimate the energies of the ions relative to the neutral species. We know that we have n possible k values and that there are likewise n different wave functions (you might say “orbitals”) corresponding to these k values. Each orbital can be occupied by two electrons, so the energy of the neutral system is, E = 2 n/ 2 X s =1 α + 2 β cos πs ( n + 1) (1d) Where n is the number of π electrons for n even and, E = α + 2 β cos π (( n 1) / 2 + 1) n + 1 + 2 ( n 1) / 2 X s =1 α + 2 β cos πs n + 1 (1e) 1 For j odd. For l = 0 that means that j = n . Assuming that we have an even number of electrons and setting the zero of energy to be equal to α the difference in energy between our neutral molecule and the 1 ion is, Δ E = 2 β cos π ( n/ 2 + 1) n + 1 = 2 β cos π 2 1 + 1 n + 1 (1f) While the difference in energy between our neutral molecule and the +1 ion is,...
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 Spring '10
 CHandler
 Electron, Light, Eigenvalue, eigenvector and eigenspace, Band gap

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