Chemistry 351L Wet Lab
The Particle-in-a-Box Model for Describing the
Electronic Transitions in Conjugated Dye Molecules
The quantum mechanical solutions derived for a particle in a box assumes
a free particle moving within a region of zero potential inside “walls” of
infinite potential. The one-dimensional version of this solution, the
“particle-on-a-line” is attractively simple. It would be interesting to
discover whether there are any real systems that approximate this very
simple model. Real systems are, of course, inherently three-dimensional,
but this laboratory exercise is intended to demonstrate that there are real
systems that can be treated as one-dimensional problems for certain
purposes.
For example, consider the molecule ion:
This can, of course, may be written in an equivalent form:
Since the double bonds can be in either of the positions shown, it would
appear that the electrons represented by the double bonds are not
localized in one area. Rather they appear to be “spread out” over the
entire length of the double bond chain. These so-called π electrons are
free to “move” along the chain between the terminal nitrogen atoms. If it
is assumed that the potential energy influencing the electrons along the
chain due to the carbon centers is constant over the length of the chain,
this potential may, for convenience, be set at zero. Then this problem is
similar to the one-dimensional particle that is free to move along the x
direction between the limits x=0 and x=L.
In the real molecule analog, the electrons are free to move along the
chain (x) between the limits prescribed by the length of the chain. As we
learned in class, the allowed energy values and the eigenfunctions are
thus:
Unlike the class example where a single particle was confined to the 1D
box, in the molecular analogue there are a number of electrons confined
to one box. (We are assuming that the electrons in this system do not
interact with one another excepting, as seen below, through their spin).
Each of these particles must be described by an eigenfunction. Because
of spin, which we have not yet introduced, but with which you are familiar
from previous courses, there is a maximum of two electrons that can
have the same spatial eigenfunction. In the molecular ion shown above,
there are a total of 12 π-electrons, one each from the 9 carbon atoms
and three from the two nitrogen atoms. Hence these 12 electrons will
occupy first 6 wavefunctions given by ψn (n = 1, 2,….6) each holding 2
electrons (one spin up and one spin down). In the more general case of N
electrons in a “conjugated” π systems (meaning alternating single and
double C-C bonds), the first N/2 (assuming N is even) energy levels would
be occupied.
PRE-LAB Question # 1. Use Mathematica to graph the potential
corresponding to a box of length 10 where the potential outside the box
is infinite and inside the box is zero. Now superimpose on this potential
the first six energy levels for a particle moving in this box. Make your unit
of energy equal to h2/8mL2. Superimpose on each energy level the
corresponding probability distribution. Now in someway depict that the first six
energy states are occupied and the sixth is not (consider using dashed lines or
something of that sort. Finally, provide an appropriate caption for your plot to
explain its meaning to someone else. Your plot should look a bit like Figure 3.2
in your text.
In absorption spectroscopy, an electron in a given energy level absorbs a
photon of a specific energy and is consequently elevated to a higher
energy level.
PRE-LAB Question # 2. Explain why one will only see a transition if the
energy of the photon is equivalent to the energy difference between
states
For example, in our case above, one of the electrons in the 6th energy
level could be excited to, say, the 7th level if the enrgy of the incoming
photon were equal to 13 h2/8mL2.
PRE-LAB Question # 3. Explain how the preceding energy was derived.
We can write a general expression for a transition between an energy
level ni (the initial n-value of an electron) to nf (the empty
level). The energy change would be:
ΔE = [h2/8mL2](nf2
– n i2 )
If we consider transitions from the highest occupied level to the lowest
unoccupied level (as in our example above) and we let N be the number
of π-electrons in the molecule, then ni = N/2 and n f = (N/2 + 1),
hence,
ΔE = hν = hc/λ = [h2/8mL2](N+1)
where ΔE is the energy of the photon necessary to cause the transition.
PRE-LAB Question # 4. Write a Mathematica function that gives the
wavelength of a photon (in nm) necessary to cause a transition of an
electron in the nth energy state to the mth when in a box of length of L.
That is, there will be three inputs to your function—n, m, and L.
For molecules such as the one above, N is equal to the number of carbon
atoms in the chain (since each carbon atom contributes one electron)
plus 3 (each N atom would contribute 2 electrons, but one electron has
been lost to form the ion), so,
N = nC + 3.
The parameter L is just the length of the “box.” It can be expressed as
the number of C-C bonds (which is the number of carbon atoms minus
one) times the average length of the C-C bonds, plus the number of C-N
bonds times the length of the C-N bond, plus some unknown distance
representing the penetration of the wave function beyond each end of
the conjugated chain. So,
L = (nC-1)lCC
+ 2lCN + 2p
where lCC is the length of the C-C bond, lCN is the length of the C-N bond
and p is the (unknown) penetration at each end.
PRE-LAB Question # 5. Write a Mathematica function to calculate the
wavelength, λ, of an absorbed photon in terms of nC, lCC, lCN and p. Note
that these parameters give you nothing more than L, n, and m you used
in Question 4. Thus you might want to thnk about combining functions
to achieve this end.
In the equation you derived in pre-lab question #5, there are four
unknown quantities that must be determined in order to estimate the
wavelength of absorbed photons for a particular molecule. The first, the
number of carbon atoms, nC, can be determined from the molecular
structure. The other parameters, lCC, lCN, and p are difficult to determine
exactly for these molecules, but appropriate values can be estimated
from experiment, as will be shown below. The experiments involve the
comparison of experimental and predicted absorption wavelengths for
several dyes. The results will then be used to predict the absorption
wavelength for a new molecule.
The structures of the three dyes to be used in these experiments are
shown below:
As a first approximation, we will assume that the value of lCC in these
molecules is the same as in benzene, a somewhat similar delocalized
system, where lCC = 0.140 nm. The value of lCN will be assumed to be
similar to that in pyridine, where lCN = 0.134 nm. Since the value of p is
difficult to assess, we will assume a range of p values ranging from p= 0
to p = 0.140 nm. Using this range of values and your Mathematica
functions, plot the wavelength of the photon needed to cause the
transition versus p.
Having done this, you will follow the procedures outlined below to obtain
the absorption spectrum of dye # 1. From the wavelength of the
absorption maximum (λmax) observed for this dye you will use your graph
to pinpoint the value of p that gives the experimentally observed value of
the absorption maximum.
Now using this value of p, you can estimate the wavelength of the
absorption maximum for dye # 2 and for dye # 3 and compare this to
your measured value. Comment on any differences observed and propose
additional changes to the model that might give a better estimate of
(λmax). Also comment on the origin of this penetration depth in terms of
the particle-on-a-line quantum mechanical model. Is reasonable to
assume that the value of this parameter should be the same for all of the
molecules tested? Why or why not? When considering this question, take
a look at the problem given in the practice test of a particle in a finite
square well.