Length of a 1D box in hexa-1,3,5-triene

### Problem


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 $\pu{258 nm}.$
> 
> 
> 


### Answer



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


### Question


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?


### Reference


1. Hayward, D. O. *Quantum Mechanics for Chemists*; Tutorial chemistry texts; Royal Society of Chemistry: Cambridge, UK, **2002**. ISBN 978-0-85404-607-2.




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](https://nanohub.org/courses/OED/01a/asset/5958#:%7E:text=Using%20the%20simple%20model%20shown,estimated%20to%20be%200.867%20nm).


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