# Number of pi-electrons in an energy state

I am learning quantum chemistry and I am trying to understand the statement in a .pdf I'm reading which states "The energy transition associated with the spectral transitions observed is: $$\Delta E = E_{LUMO} - E_{HOMO} = \frac{(N+1)h^2}{8mL^2}$$ where $L$ is the length of the "box" (it's talking about particle in a box), $m$ is the mass of the electron, and $N$ is the number of $\pi$-electrons in the system. I know due to the solution of the Schrodinger equation that $$E = \frac{n^2 h^2}{8mL}$$ So, $$\Delta E = E_{HOMO} - E_{LUMO} = E_{k+1} - E_{k} = \frac{(2k+1)h^2}{8mL^2}$$ So, I want to figure out why $2k = N$?

• Welcome to Chemistry.SE. Is there any possibility that you could link to the pdf in question? – Ben Norris Sep 28 '13 at 1:15

Your calculation is indeed correct. However, because each electron in the $\pi$ system can have two spin states, there are two electrons per energy level of the “particle in a box” Hamiltonian. Hence, $N=2k$.
• I've said that $N$ is the number of $\pi$-electrons. That's all my source says. – Samuel Reid Sep 28 '13 at 15:37