From Hayward's Quantum Mechanics for Chemists [1, p. 36]

2.3. Calculate the wavelength of light that will be absorbed when a it electron in hexa-1,3,5-triene, $\ce{CH2=CH—CH=CH—CH=CH2},$ is promoted from the highest occupied level to the lowest unoccupied level. The average $\ce{C—C}$ bond length in hexatriene can be taken to be $\pu{144 pm}.$ Compare your answer with the experimentally observed wavelength of $\pu{258 nm}.$


2.3. Wavelength of light is $\pu{352 nm}$ (for box length of $\pu{864 nm})$


Since $L = 5\times(\pu{144E-12 m}),$

$$E = \frac{7h^2}{8mL^2} = \pu{8.14E-19 J}.\tag{1}$$

After substituting the values I get the answer

$$\lambda = \frac{hc}{E} = \pu{244 nm}.\tag{2}$$

The textbook answer is $\pu{352 nm}.$ What confuses me is that they state that the length of the box is $\pu{864 nm}$ — but there are five bonds joining the six carbons together. So, would not we multiply the average bond length by $5$ and not $6?$

Is it acceptable to take the length of the box as the average bond length multiplied by the number of atoms?


  1. Hayward, D. O. Quantum Mechanics for Chemists; Tutorial chemistry texts; Royal Society of Chemistry: Cambridge, UK, 2002. ISBN 978-0-85404-607-2.
  • $\begingroup$ Where is 7 coming from? The value of $\lambda$ is too high. $\endgroup$
    – AChem
    Dec 30, 2022 at 5:46
  • 1
    $\begingroup$ Please check the units, m, mass of electron in kg, c in m/s etc. then the longest transition wavelength is found as the difference from the top filled energy level to the next one. The answer given is correct for the box length of 6 , perhaps someone else knows why 6 is used. $\endgroup$
    – porphyrin
    Dec 30, 2022 at 8:43
  • $\begingroup$ My apologies- I have edited this question and the proposed alternative answer using a box length of 5 should now be correct. $\endgroup$ Dec 30, 2022 at 10:06
  • $\begingroup$ I remember reading that the pi electron cloud extends half a bond distance further than the last atom on the two ends of the molecule, which would add one more bond in total. $\endgroup$
    – Siva
    Dec 30, 2022 at 10:41
  • $\begingroup$ @porphyrin A length of 6 is typically used if you compare the particle on a ring model for benzene with the corresponding linear particle in a box model, see for example doi.org/10.1021/acs.jchemed.2c00523 $\endgroup$
    – Loong
    Jan 1 at 20:26

1 Answer 1


For this type of question, we not only consider the length of the bonds between the carbon atoms, but we also have to add on the atomic radii at both ends of the structure. I assume the atomic radii of carbon atoms taken in the text is $\pu{72 nm}$, which means doing $\pu{144 nm}\cdot5 + \pu{72 nm}\cdot2 = \pu{864 nm}$.

We can then solve it like you did doing $E = (4^2-3^2)\frac{h^2}{8mL^2}$ and the applying $\lambda = \frac{hc}{E}$. Ultimately, however, the addition of the two carbon radii to the length of the box would explain the textbook result (see page 6 of the link below).

Source: https://nanohub.org/courses/OED/01a/asset/5958#:~:text=Using%20the%20simple%20model%20shown,estimated%20to%20be%200.867%20nm.

  • 1
    $\begingroup$ But what is the reason for adding the atomic radii, it makes no sense and is just a fix, i.e. not rational or scientific, so as to get closer to the expected answer. $\endgroup$
    – porphyrin
    Jan 19 at 8:19
  • $\begingroup$ @porphyrin I mean, it is simply a model, but for it, I think that's what you have to include. From ChemLibreText: "A diagram of the particle-in-a-box potential energy superimposed on a somewhat more realistic potential. The bond length is given by β, the overshoot by δ, and the length of the box by L = bβ + 2δ, where b is the number of bonds." $\endgroup$
    – M.L
    Jan 19 at 17:32
  • $\begingroup$ yes of course its a only model but my point was that it the extension bit is not really based on anything other than to try to fit a number by fiddling the calculation by adding an arbitrary parameter, $\delta$ etc. The real point is that even though PIB is so simple it gets quite close in energy terms to something that be measured. That should be enough. Actually it is surprising it gets so close. $\endgroup$
    – porphyrin
    Jan 20 at 14:07

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