Exercise: A Heteronuclear Diatomic Molecule

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The applet uses the divided square well potential to model electron states in a diatomic molecule. U = 100 eV is the height of the dividing barrier and 2w = 0.50 Å is its width. The barrier is centered at x = b and 2a = 4.50 Å is the combined width of wells plus barrier. The case b = 0 describes potential wells of equal width (= 2.00 Å); this is the homonuclear molecule, i.e., one formed from identical atoms. Heteronuclear molecules can be modeled by taking b nonzero. In this exercise we will find the bonding and anti-bonding orbitals for the electron that derive from the 'atomic' ground states of two dissimilar atoms, modeled as square wells of the same height but unequal widths, 1.75 Å and 2.25 Å (requires b = 0.25 Å – see 1st Technical Note below). All parameter values appear on the Math tab of the applet, along with the electron mass = 511 keV/c2. The list to the right of the graph includes placeholders for two stationary states of the electron in this environment.

Instructions for use

  1. Show one of the stationary state waveforms by right-clicking the first placeholder in the list, clicking the visibility icon beside the "Real" label in the Colors | Visibilities field, then choosing the OK button. The displayed wavefunction has a noticeable discontinuity. Since quantum wavefunctions must be everywhere continuous, the energy of this state cannot be one of the allowed energies for the electron.
  2. Use a combination of manual and automated search techniques to quickly find the energy of the bonding orbital. Switch to the Math tab, where the energy of this state is recorded as E0 = 0. Right click anywhere in the value field for E0, and select "Edit Parameter..." from the popup menu to activate the Energy Editor. Next, return to the Graphics: [x] tab and re-position the editor dialog so as to afford an unobstructed view of the waveform. Move the slider rightward to increase the energy of this state (recorded in the energy field, labeled E) until the discontinuity is noticeably reduced, or even 'reversed'. Next, activate the automated search function by typing "0.1" in the auto-search range field (labeled E±) and pressing the Enter key. The editor responds by inspecting energies in the range E ± 0.1, looking for a waveforms with fractional wave mismatch no larger than the [default] value shown in the tolerance field, labeled δψ. If the search succeeds, the new energy is displayed in the energy field and the tolerance field is updated with the actual wave discontinuity (see 2nd Technical Note below); A failed search is marked by the requested tolerance being displayed in red, to indicate that the desired wave discontinuity was not, in fact, realized. In the event of a failed search, simply widen the search range and repeat the process. The bonding orbital has the least energy and is nodeless; if you have found another, continue searching at lower energies. Finish by selecting OK to end the edit session with the current settings.

    When an automated search finishes (successfully or not), the editor reverts to manual mode; as a consequence, after editing any search parameters, you must return to the auto-search range field and press Enter to initiate another search using the new values.

    Even with an automated search, it is good practice to follow-up a successful search result by resetting the match point (click on ), as this will generate an improved waveform and a more accurate value for the wave discontinuity at the match point.

  3. Repeat the above procedure for the second placeholder in the list, using it to find the anti-bonding orbital. This state has one node and is the next lowest in energy.

    If an automated search returns the value from the previous step, narrow the auto-search range entry so as to exclude that value from the search.

    Notice that the symmetry evident in the stationary states for the homonuclear case is absent here; the bonding orbital is concentrated in the wider well, the anti-bonding orbital in the narrower well. Compared to the homonuclear case, the energy 'splitting' also is far greater in the present case. In fact, this is not a simple 'splitting' at all, since each well in isolation has a different ground state energy.

Technical Notes