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I'm briefly studying the model of a particle in a linear box when we have a chain of insaturated carbons... I have a chain with 3 bonds and 4 atoms of carbon.

Applying the expression

$\lambda = \frac{8cmL^2}{ n^2 h}$ (instead of n I'm calculating the variation of n)

So make c and m the light speed and the mass of the electron, respectively and I make L the length of the box ((3+1) x distance Carbon-carbon (1.41 x 10^-10 m) and I suppose the transition with less energy is $3^2-2^2$.

I get to 210 nm and the correct answer should be 207 nm... Am I making any incorrect approach? Thanks!

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    $\begingroup$ Ok, first of all you need to be way clearer in what you are saying. It took me a while to figure out what you were even trying to calculate, because you didn't say what $\lambda$ represented. For a while I thought you were calculating the de Broglie wavelength of the electron in the orbital. It's easy to use maths to write "the variation in $n$". All you need to do is: $$\lambda = \frac{8cmL^2}{(n_2^2-n_1^2)h}$$ where $n_2$ is the upper level and $n_1$ is the lower level. Then say $n_1 = 2$, $n_2 = 3$. Anyway, my answer is 209.76 nm. $\endgroup$ – orthocresol Apr 23 '16 at 0:00
  • $\begingroup$ What is a linear box? How carbons come into picture? Where does the 207nm as an answer come from? $\endgroup$ – Greg Apr 23 '16 at 3:37
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    $\begingroup$ 210 instead of 207? I'd say that's an excellent agreement for such a crude approximation. $\endgroup$ – Ivan Neretin Apr 23 '16 at 6:18

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