# Question regarding Energy of a Particle in a Box is Quantized

1. Sep 10, 2006

### luckymango

I'm not sure how should I ask the question, so I'm gonna go ahead and quote from the book.

"The model of a particle in a one-dimensional box has been applied to the pi electons in a linear conjugated hydrocarbons. Consider butadiene, H2C=CHCH=CH2, which has four pi electrons. Although butadiene, like all polyenes, is not a linear molecule, we will assume for simplicity that the pi electrons in butadiene move along line who length can be estimated as equal to two C=C bond lengths (2 x 135pm) plus one C-C bond (154pm) plus the distance of a carbon atom radius at each end (2 x 77pm = 154pm), giving a total distance of 578pm. According to equation E = h^2 n^2 /(8m a^2), where n=1,2...

But the Pauli Exclusion Principle says that each of these states can hold only two electrons and so the four pi electrons fill the first two levels. The energy of the first excited state of this system of four pi electrons is that which has one electron elevated to the n=3 state and the energy to make a transition from n=2 state to the n=3 state is Delta E = h^2 (3^2 - 2^2) /(8m a^2)."

My first question is how did they calculate the four pi electrons in the first place? Also, how did they know that the electron elevated from n=2 state to the n=3 state?

Hopefully someone could help me with my question. I'm stuck here reading the book and I can't move on... Thanks in advance!!

2. Sep 10, 2006

### Epicurus

pi orbitals have 3-fold degeneracy, px, py and pz. Thus 4 electrons will fill these orbitals, none being the same state.