5. A particle in a box model can be applied in a simplistic manner to the consid

Question

5. 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-7 level. If we consider the box containing the electrons to have infinite potential beyond the~1210 pm length of the conjugated chain and zero potential inside: a) What are the energies of n=1-7 levels? b) What is the energy difference between n-6 and n=72 c) What would be the wavelength of light for a photon affecting this transition? d) How does this very approximate calculation compare with experimental values? (See: Fieser, L. F., "ABSORPTION SPECTRA

Explanation / Answer

a)

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 length = (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 maximum absorption and compare with the experimental results.

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