# Question #45cd0

Jan 18, 2016

I'm not sure what your $K$ refers to.

This equation, derived from quantum mechanics, gives a similar relation to the energy of the hydrogen atom...

${E}_{n} = \frac{{n}^{2} {h}^{2}}{8 m {a}^{2}}$

where:

• $n$ is the principal quantum number, and $n = 1 , 2 , 3 , \ldots , N$ where $N$ is a real integer
• $h$ is Planck's constant ($6.626 \times {10}^{- 34} \text{J"*"s}$)
• $m$ is the mass of the electron ($9.109 \times {10}^{- 31} \text{kg}$)
• $a$ is the width of the box that confines the particle in the Particle in the Box model, i.e. the domain for the boundary conditions of the system's wave function

From this equation, as $n$ increases, $E$ increases.

If the size of a molecule increases, it is proportional to $a$ increasing, causing the energy levels to begin converging together. Eventually, as $a \to \infty$, the energy ${E}_{n}$, for every energy level $n$, will approach the same energy level, and the energy will distribute itself evenly throughout the molecule.

That's why larger aromatic systems tend to have smaller HOMO-LUMO gaps than smaller aromatic systems, and thus, they have better conductivity.