Nov Chem Problem Set 10: solution to Problem 2 a) Using the given functions S( R), J '( R), K '( R ), a plot of E( R) 1/ 2 (which reach

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6 Nov Chem. 4 Problem Set : solution to Problem 2 a) Using the given functions S( ), J '( ), K '( ), a plot of E( ) / 2 (which reaches the asymptotic value of as ) vs. is shown in Fig.. delep( ) Fig.. delep()=e() + ½ vs.. [The dissociation limit is also indicated.] As can be gleaned from Fig., this curve has a minimum at 2.5 a. u. or 2.5*.53=.33Å, at which the energy of the ground state relative to the dissociation limit is ca..65.a.u. or.65*27.2=.77ev. b) i) Using the dependent functions provided in the statement of the problem, a plot of E( 2, ) is shown in Fig. 2. The function clearly shows a unique minimum, corresponding to the value.24, and that E(2;.24).588 a.u. ii) epeating the exercise in part i) for each of the requested values of, we obtain the plot of E( ) / 2 shown in Fig. 3. As expected, the value of E obtained at a given value of is always lower for the optimized version of the calculation (part b) than for the simpler version corresponding to (hydrogen s orbitals).

7 iii) The minimum of the optimized E( ) curve occurs at 2 a. u. or 2*.53=.6Å, and corresponds to a dissociation energy of.88a.u. or.88*27.2=2.394ev...2 Ebar( Z 2) Z Fig. 2. Ebar(Z,2)= E( ; ) vs. for =2. iv) According to the Variational Principle, the exact solution of the Schrödinger Eq. (which corresponds to experimental reality!) gives an E( ) curve which is everywhere lower in energy than the optimized result plotted in Fig. 3. Thus, the experimentally measured dissociation energy for H 2 should be larger than 2.394eV. (In fact, the measured value of this dissociation energy is 2.78eV.) N.B.: A printout of the.xmcd worksheet used to generate the numerical results presented above is attached below. 2

8 .5.4 optimized Z Z= (hydrogen) dissociation lim Egs()+/ Fig. 3. Egs=ground state energy vs.. 3

9 Study of electronic structure of H2+. Focus on g.s. energy (and eigenfunction): Begin with the simplist version (using : linear combo of standard [Z=] hydrogen s orbitals: S( ) e 2 3 Jp( ) e 2 Kp( ) S( ) e ( ) delep( ) Jp( ) Kp( ) S( ) Note: E_gs = -/2 + delep() delep( ) Note: binding curve has delep_min = -.65, at _min=2.5 (approximately). 2.5*.53=.325 Angst.,.65*27.2=.77eV.

10 Now, define analogs of S,Jp,Kp which are valid for hydrogenic s orbitals with effective nuclear charge Z: Sz( Z ) e Z Z ( Z) 2 3 Jpz( Z ) ( Z ) Z Z e 2Z Kpz( Z ) Sz( Z ) ( Z 2) Z( Z) e Z Option to check Sz, Jpz, Kpz vs the standard S,J,K (i.e., with Z=): 3 Kpz( ) Kp( ) Ebar(Z,)= expectation of H in LCAO of s oritals with effective nuclear charge Z at separation (in a.u.): Ebar( Z ) Z 2 2 Jpz( Z ) Kpz( Z ) Sz( Z ) Z.5.5 2

11 Note: Fix, scan Z=effective nuclear charge..2 Ebar( Z2) Z Note: For =2., find minimum at Z=.24, with value for Ebar=-.588 To determine the ground state potential energy curve by applying the variational principle at each nuclear separation, it is simplest to simply plot Ebar(Z,) vs Z for fixed (above), read off the minimum value of Ebar ** (this is the variational estimate of the g.s. energy), make a table (below), and plot it. ** Note: right-click on graph; choose Trace (<-- click on curve, Trace will determine numerical values). ESULTS: au Egs.465 au.25 Egs.54 au.5 2 Egs au.75 3 Egs au 2 Egs au 2.25 Egs

12 au Egs au 2.75 Egs au 3 Egs j Note: plot here E_gs () + /2, so that the dissociation limit ( --> infinity), corresponds to E_gs =.5.4 optimized Z Z= (hydrogen) dissociation lim Egs()+/2 Egs j 2 delep( ) au j NB: Minimum in E_gs ("variationally optimized) is approximately at =2. a.u. = 2.*.53=.6 Angst. The corresponding dissocation energy is E_dis= =.88 a.u. =.88*27.2=2.394eV. Note: Experimentally measured dissociation E for H2+ =2.78 ev.