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terp.asessed

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## Homework Statement

The UV/visible spectroscopy of linear conjugated molecules, particularly 1,3-butadiene in this problem, can often be modeled with the Particle-in-a-box of the electrons. Assume that we are interested in the pi electrons ONLY. A molecule with N double bonds = 2N pi electrons.

(1) To obtain the ground state, put 2N electrons in the LOWEST possible energy levels. Assume the box is of length L, what is the energy of the highest energy electron, as function of L and N?

(2) When 1,3-butadiene absorbs photon energy (hv), it uses the energy to promote 1 electron to a higher level. What is the lowest energy light the molecule can absorb, again, as a function of L and N?

## Homework Equations

C-C bond length = 1.54 Angstrom

C=C length = 1.35 Angstrom

The angle between the bonds = 120 degree

L(in Angstrom) = (2.50)N - 1.32

## The Attempt at a Solution

For butadiene, it has 2 double bonds...so N = 2 and 2(2) = 4 pi electrons, where 2 pi electrons compose energy level n=1 and second couple at n=2, with the possibility of one electron jumping to n=3 level with energy...

L = 2.50(2) - 1.32 = 3.68 Angstrom

...Could someone hint me as how to progress from here? I am stuck. Also, I know from the Schrodinger Equation that E

_{n}= n

^{2}h

^{2}/(8mL

^{2}), from H wavefunction(x) = E wavefunction (x), where V(x) = 0 b/c of confinement inside a box...so, I wonder if I should use this equation too?