# Excitation energy of carotene using the particle in a box model

I'm practicing for an exam and I came across the following question:

The linear, conjugated π-electron system of a carotene molecule comprises 11 atoms and the distance between two atoms is 1.4 Å. Calculate the excitation energy and the wavelength of the radiation for an electron in the described system using the particle in a box model.

My (probably wrong) approach of solving this problem is as follows:

Assume the carotene molecule can be modeled as a particle in a 1 dimensional box of length 15.4 Å. An electron is excited when it goes from the ground state up to the next energy level, so the excitation energy is $\Delta E=E_2 - E_1$ with $E_n=\frac{n^2h^2}{8mL}$.

So calculating the excitation energy with $h=6.626*10^{-34} J*s$, $m=9.109*10^{-31} kg$ and $L=1.54*10^{-9}m$ gives:

$\Delta E =\frac{2^2h^2}{8mL} - \frac{1^2h^2}{8mL} = 3*\frac{h^2}{8mL} = 3*\frac{(6.626*10^{-34})^{2}}{8*9.109*10^{-31}*1.54*10^{-9}}=1.174*10^{-28} J$.

Using the formula $\lambda = \frac{hc}{\Delta E}$ to calculate the wavelength gives:

$\lambda = \frac{hc}{\Delta E} = \frac{6.626*10^{-34} * 2.998*10^{8}}{1.174*10^{-28}} = 1692.1m$.

That's 1692 meters. Quite obviously wrong... Now I don't have the solutions to this problem, so I don't actually know what's supposed to be the right answer, but I know mine is wrong.

So yeah, if someone could point out what I'm doing wrong, I'd be really grateful!

$$E = \frac{n^2h^2}{8mL^2}$$
Your mistake is that you have $L$ not $L^2$ in the denominator so your answer is a factor of $L$ too small.