[Big-Daddy]

http://icho2014.hus.edu.vn/document/01-24-2014%20IChO46-Preparatory.pdf
I got the results:

3.1.1. ΔU = h²(N+1)/(8m_EL²) where N is the number of delocalized pi electrons in one molecule of the species.
3.1.2.
[tex]\lambda = \frac{8m_EL^2c}{h \cdot (N+1)}[/tex]
3.2.1. L(BD) = 7.0 Å, L(HT) = 9.8 Å, L(OT) = 12.6 Å
3.2.2. λ(BD) = 95.04 nm, λ(HT) = 166.7 nm, λ(OT) = 249.3 nm
3.3. ?
3.4.1. L(BD) = 13.0 Å, L(HT) = 14.2 Å, L(OT) = 19.3 Å
3.4.2. Since the last method comes from experimental data, of course it must be the most accurate wrt. experimental data, by definition?

My questions: Do you get similar results for 3.2.1., 3.2.2, 3.4.1?
Regarding 3.1.1: am I right to think ΔU=ΔH=ΔE here, and in general ΔE=ΔU (this of course is just ΔU for one molecule - to get for one mole we'd multiply by Avogadro's constant)?
Regarding 3.1.2: I assumed only one photon could be absorbed to cause transition HOMO electron  LUMO electron. I wonder if this is safe.
Regarding 3.3: What is the question asking? How is it telling us to determine L?

I thought it might be something to do with just seeing a C-C-C as a triangle and determining the base length and then multiplying across all such triangles from one phenyl to the other, but why would this be a good estimate of L?
Also, 3.4.2. seems a bit silly the way I read it now - you can't give experimental data, get a derived value of L from it then compare it to theoretical values of L and ask which, theoretical or experimental, fits closer to the experimental data ...

[Rutherford]

Got the same except 3.2.2 where I got 324, 453 and 583nm (I think that the last molecule has to absorb in the visible region, and all are close to the experimental ones except maybe the second) and 3.4.2 where I got 7.1, 8.6 and 12.6Å.

No Avogadro constant.
One photon excites one particle in a box.
The triangle usage seems as a bad estimation (use the law of cosine), as I got the box lengths to be 6.1, 8.4 and 10.8Å.

[Big-Daddy]

Got the same except 3.2.2 where I got 324, 453 and 583nm (I think that the last molecule has to absorb in the visible region, and all are close to the experimental ones except maybe the second) and 3.4.2 where I got 7.1, 8.6 and 12.6Å.

I see the difference. I have assumed that the phenyl delocalized electrons will join the conjugated system, you have assumed they will not. Are you sure the delocalization is restricted only to within the linear chain between the phenyl groups?

One photon excites one particle in a box.

Is that a fact or just intuition? - Do you have a source?

The triangle usage seems as a bad estimation (use the law of cosine), as I got the box lengths to be 6.1, 8.4 and 10.8Å.

So they are asking for the triangle method ... very odd way, but I guess because this is a planar system it might not be that bad an estimation.

[Rutherford]

Why did you assume that, when it is stated in the problem that the electrons are delocalised in the space between the two phenyl groups?

One photon of a certain wavelength whose energy corresponds to HOMO-LUMO transition of one electron will excite one electron.

How bad the estimation is, you can see by checking the obtained values with the experimental ones.

[Big-Daddy]

Why did you assume that, when it is stated in the problem that the electrons are delocalised in the space between the two phenyl groups?

I guess it is because it seems to me on looking at the structures that the delocalization should cross the entire molecule. Is there a good reason why the delocalized system of the phenyl groups will not join that of the alkenes between them? Maybe I haven't studied that field enough.

[Rutherford]

The delocalization includes the phenyl rings, but it would be more complex to apply the particle in a box model. I think that the only reason is simplicity.

[Big-Daddy]

The delocalization includes the phenyl rings, but it would be more complex to apply the particle in a box model. I think that the only reason is simplicity.

How can this be a decent approximation? The number of delocalized electrons is 2k+1 in the chain between the two phenyl groups and 2k+13 including them, so the highest orbital is n(HOMO)=k+1/2 in the first case and n(HOMO)=k+13/2 in the second. When there are 2 double bonds between the phenyl groups, k=2, and n(HOMO)² = 6.25 for the first case but n(HOMO)² = 72.25 for the second case. The difference in the principle quantum number for the HOMO is more than an order of magnitude (!), if you neglect the phenyl electrons being part of the delocalized system.

[Rutherford]

I didn't say that it is okay, but they said so .