A particle in a box model can be applied in a simplistic manner to the considera

Question

A particle in a box model can be applied in a simplistic manner to the consideration of mobile electrons in a conjugated system. Consider vitamin A2 depicted below with 12 electrons. If these electrons fill the first 6 particle in a box energy levels by pairing spin up/spin down then the absorption of a photon of light can occur from n=6 to the vacant n-1 level. If we consider the box containing the n electrons to have infinite potential beyond the 1210 pm length of the conjugated chain and zero potential inside: What are the energies of n=l-7 levels? What is the energy difference between n=6 and n=7? What would be the wavelength of light for a photon affecting this transition? How does this very approximate calculation compare with experimental values?

Explanation / Answer

Ans:- the quantum mechanical solution for energy levels of this model is given by:- En= n2h2/8ma2

h=planks constant, m=mass of an electron, a=lenght of the one-dimensional box.

En=n2(6.6*10-34)2/8 * (9.1*10-31)* (1210*10-12)

for n=1, En=4.0868*10-20J

for n=2, En=1.634*10-19 J

for n=3,En=3.678*10-19 J

for n=4,En=6.5388*10-19 J

for n=5, En=1.02*10-18J

for n=6,En=1.47*10-18J

for n=7,En=2.0025*10-18J

b) Vitamin A2 has 12 electrons. let us denote the no of double bond in entire chain be "i". then "N" will be the ground state with N electrons with N/2 lowest enenrgy filled. since we have i=6,(12 elelctrons paired up in 6 pi orbitals ), we have N=(2i+2) electrons. now the bond lenght = (2i+2)a.

therefore, energy difference= h2((n1)2-(n2)2)/ 8ma2

energy difference=h2(2i+3)/8ma2(2i+2)=3.1276*10-21 J

C)wavelenght lambda=8mca2(2i+2)/h(2i+3)=6.05 *10-5 m

d) obtain an absorption spectrum, calculate the wavelenght of maxium absorption and compare with the experimental results.

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