Particle in a Box
Question: Can a simple function like y(x) = Sin[n*Pi*x/a] describe an electron in a molecule?
Answer: yes.
The Basic Model System
1) Consider a box with no lid with infinitely high walls. This box represents the "molecular frame" for which the electron can exist.
2) When a particle/electron is placed inside this box, it has a particular "energy." Considering that "energy" is a measurable quantity in systems described by classical mechanics, Q.M. postulate 2 tells us that there exists a corresponding operator that can be used to predict the energy or as we say this observable.
3) As with all "operator algebra" problems, there are 3 parts:
- - the operator
- - the (eigen)function
- - the (eigen)value = observable.
The operator in this case, like many, is the Hamiltonian operator, the eigenfunction is given above, and we need to solve for the eigenvalue or energy.
Initial Tasks
0) Is the wavefunction y(x) = Sin[n*Pi*x/a] normalized? No...normalize it...answer = y(x) = Sqrt[a/2]*Sin[n*Pi*x/a]
1) Convince yourself that the function y(x) = Sqrt[a/2]*Sin[n*Pi*x/a] is an eigenfunction of the Hamiltonian operator.
- You ask, how do i do that?...just operate on the function with the Hamiltonian operator and if you return a "value" and the function, then the function is an eigenfunction for the Hamiltonian operator. You do need to know that the potential energy, V(x), inside of the box and it is equal to zero...so the Hamiltonian operator is a bit more simple in this model system. Please "do this" work in your lab notebook.
2) Notice that although this function is an eigenfunction of the Hamiltonian operator, there may be other solutions...any ideas?
- Please plot this function in Mathematica..
- Note: Boundary conditions y(0) = 0 and y(a) = 0
- See figure 4.2/4.3 Reproduce these in Mathematica
- why are these offset as they are?
3) Use the "Boltzmann" distribution to calculate the number of particles in the n=1 verses the n=2 state...essentially reproduce graph in example problem 2.1.
The Chemical System
Pi electrons in conjugated molecules can be treated as "particles in a box." See section 5.3.
- 1,4-diphenyl-1,3-butadiene (4 pi-electrons)
- 1,6-diphenyl-1,3,5-hexatriene (6 pi-electrons)
- 1,8-diphenyl-1,3,5,7-octatetraene (8 pi-electrons)
1) What is an estimate of the "box length" (ie molecular frame).
2) What is the ΔE between the HOMO and LUMO
3) Use the "Boltzmann" distribution to calculate the number of particles in the n(HOMO) verses the n(LUMO) state...essentially reproduce graph in example problem 2.1.
Data Collection
1) collect UV-Vis data for a conjugated dye.
2) record structure in lab notebook.
3) can you predict the box length based on lambda max?
Results
3,3′Diethylthiacarbocyanine iodide_AK
3,3'-Diethylthiacarbocyanine iodide_MA
3,3'-Diethylthiacarbocyanine iodide_CF
1,1%27-diethyl-2,2%27-cyanine_iodide