Chem 110AHomework 9: Due Friday December 12 2014Chemistry 110AProfessor McCurdy1. (Problem 9-4 in McQuarrie and Simon) Determine the normalized wave function for? – = c (1sA – 1sB )which is one of the eigenfunctions found in the simple variational calculation on H2 performed in your text.This problem is asking you to find the normalization constant c in terms of the overlap*integral S = 1sA 1sB = ? ? 1sA (r) ?1sB (r) d 3r given in Eq. 9.11.2. (Problem 9-8 in McQuarrie and Simon)In the linear variational calculation on the H2 molecule the matrix elements that appearareH AA = 1sA H 1sAH BB = 1sB H 1sBwhere H is the Hamiltonian H AB = 1sA H 1sBShow that H AA = H BB = -1 / 2 Jand that H AB = -S / 2 Kin the simple molecular-orbital treatment of H2 . H is the Born-OppenheimerHamiltonian for H2 and the quantities J and K are defined in Eqs.(9.23) and (9.24) in thetext respectively.Atomic units are being used in these expressions in the text. Remember to use the factthat the each atomic orbital in this basis is an eigenfunction of part of theHamiltonian.3. (Problem 9-10 in McQuarrie and Simon)Show that for the antibonding orbital of H2 the eigenvalue of the simple LCAOJ -K1 J -Ktreatment is E = E1s =- 1- S2 1- SYou can do this one of two ways: Either: Use the result of problem 1 above for thenormalized antibonding MO and take the expectation value of the Hamiltonian makinguse of the results in problem 2. Or use the results of problem 2 above together with thedefinition of the overlap integral to set up the 2×2 secular determinant and solve the1 Chem 110Aresulting quadratic equation for E. In the latter case you get two solutions one for thebonding orbital and one for the antibonding one.4. (Problem 9-12 in McQuarrie and Simon)”Use simple molecular-orbital theory to explain why the dissociation energy of N2 isgreater than that of N2 but that the dissociation energy of O2 is greater than that of O2.”This is a freshman chemistry question about bond order and requires no math beyondaddition and subtraction.5. (Problem 9-23 in McQuarrie and Simon) A common light source used inphotoelectron spectroscopy is a helium discharge which generates light at 58.4 nm. Aphotoelectron spectrometer measures the kinetic energy of the electrons ejected when themolecule absorbs this light. What is the largest electron binding energy than be measuredusing this radiation source? Explain how a measurement of the kinetic energy of thephotoejected electrons can be used to determine the energy of the occupied molecularorbitals of a molecule.This problem simply asks you to quantify the basic idea behind photoelectronspectroscopy as shown in Figures 9.16 and 9.17 of the text as well as in figures inchapter 10 and later chapters.-Below is a figure of a photoelectron spectrum taken of the iron dimer anion Fe2 . Itshows both the kinetic energy of the ejected electron in the processFe2 hv ! Fe2 e .and a scale with the corresponding binding energy. In this case the binding energies ofthe orbitals involved are so small that visible light can be used in the experiment. Fromthis figure find the energy of the photon source used in this experiment. Express youranswer in eV and find the wavelength of the light – what color was it?2