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Applications of Particle in a Box model (Chemistry)

I've been reading about the one dimension particle in a box model, and from some websites, I found that it can be applied in a conjugated system. However, the system is for a single particle in a box, so I don't quite get how they assume all the electrons can be taken to be one particle. Anyone has any idea why?

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Reply 1
Lifeisnice
I've been reading about the one dimension particle in a box model, and from some websites, I found that it can be applied in a conjugated system. However, the system is for a single particle in a box, so I don't quite get how they assume all the electrons can be taken to be one particle. Anyone has any idea why?


You take each electron as an individual particle - so the first in the system occupies ψ1\psi_1, the second ψ2\psi_2 etc.
Reply 2
cpchem
You take each electron as an individual particle - so the first in the system occupies ψ1\psi_1, the second ψ2\psi_2 etc.


Yeah, but the application assumes a single particle in a box which causes the conjugated system; the conjugated system is caused by many electrons 2 of each occupying each orbital.:confused:
Yes, it's an approximation. You can treat the electron individually and ignore things like electron-electron correlation. It's not a great model for such systems but it is simple.
Reply 4
EierVonSatan
Yes, it's an approximation. You can treat the electron individually and ignore things like electron-electron correlation. It's not a great model for such systems but it is simple.


Hmm... It appears to me that most of the applications are to do with the absorption spectras, and energy levels, which seems simple enough to understand.

However, how do you interpret the wave function solution? From what I understand, the square of the wave function has something to do with the probabilities of finding the electron there, but I can't quite grasp how to interpret the graphs properly.

Also, isn't the particle in a box model assuming that each individual electron acts by itself, and is spread over the WHOLE molecule, and not localized in a specific orbital? Electrons are supposed to occupy orbitals in pairs.
Lifeisnice
However, how do you interpret the wave function solution? From what I understand, the square of the wave function has something to do with the probabilities of finding the electron there, but I can't quite grasp how to interpret the graphs properly.


A wavefunction tells you everything about a physical system, when you solve a system (by applying a given operator onto a wavefunction) it spits out a set of states in which the particle can now behave.

Yes the square of the wavefunction is related to the probability of the particle being within a defined area. By definition the integral of the square of the wave function over all space is equal to unity (effectively the particle must be somewhere).

Also, isn't the particle in a box model assuming that each individual electron acts by itself, and is spread over the WHOLE molecule, and not localized in a specific orbital? Electrons are supposed to occupy orbitals in pairs.


The assumption here relies on the electron being trapped in a long thin molecule and pretty much nothing else. things like nuclear-electron, electron-electron interactions are ignored. Atomic orbitals are probability density solutions of the Schrodinger equation for a hydrogen atom (a much more complex model).
Reply 6
There is a very small part of me that loves quantum. I'm still semi-determined to get to the end of MQM properly - in which case I might sit an extra exam and inch a little closer to degree safety.

(well, I say inch - I of course mean 254 000 000 angstroms :tongue:)
Reply 7
EierVonSatan
A wavefunction tells you everything about a physical system, when you solve a system (by applying a given operator onto a wavefunction) it spits out a set of states in which the particle can now behave.

Yes the square of the wavefunction is related to the probability of the particle being within a defined area. By definition the integral of the square of the wave function over all space is equal to unity (effectively the particle must be somewhere).



The assumption here relies on the electron being trapped in a long thin molecule and pretty much nothing else. things like nuclear-electron, electron-electron interactions are ignored. Atomic orbitals are probability density solutions of the Schrodinger equation for a hydrogen atom (a much more complex model).


Oh so basically, you're assuming each electron contributes to every single bond!

What about the quantum numbers? I saw some graphs of wavefunctions of n=1 to n=4 being plotted of a conjugated system. What do these quantum numbers correspond to?
Lifeisnice
Oh so basically, you're assuming each electron contributes to every single bond!

What about the quantum numbers? I saw some graphs of wavefunctions of n=1 to n=4 being plotted of a conjugated system. What do these quantum numbers correspond to?


It would be unwise to listen to me as it has been so very long since I did QM:

Spoiler



Lot probs wrong - been too long :frown:
Reply 10
Lifeisnice
Oh so basically, you're assuming each electron contributes to every single bond!

What about the quantum numbers? I saw some graphs of wavefunctions of n=1 to n=4 being plotted of a conjugated system. What do these quantum numbers correspond to?


n=1 to n=4 will correspond to the states of the particle in a box (and the molecular orbital). This quantum number is just the n in ψn=2LsinnπxL\psi_n = \sqrt{\frac{2}{L}}\sin{\frac{n \pi x}{L}}.

And in terms of what's really going on with bonding, often the concepts can be more complicated than 2 electrons forming one bond between 2 atoms. You can have many atoms contributing partially to bonds.
Reply 11
Kyle_S-C
n=1 to n=4 will correspond to the states of the particle in a box (and the molecular orbital). This quantum number is just the n in ψn=2LsinnπxL\psi_n = \sqrt{\frac{2}{L}}\sin{\frac{n \pi x}{L}}.

And in terms of what's really going on with bonding, often the concepts can be more complicated than 2 electrons forming one bond between 2 atoms. You can have many atoms contributing partially to bonds.


so which molecular orbital would n=3 represent then?

Don't each electron from each other go into specific molecular orbitals, rather than all of them contributing to the bonds?
Reply 12
Lifeisnice
so which molecular orbital would n=3 represent then?

Don't each electron from each other go into specific molecular orbitals, rather than all of them contributing to the bonds?


Yes, they go into specific MOs. Remember that the MO model of bonding is different to the valence bond model.
Reply 13
cpchem
Yes, they go into specific MOs. Remember that the MO model of bonding is different to the valence bond model.


If so, how do you justify using the particles in a box method? The molecular orbitals are in specific places, not along the entire length of the molecule. Also, how do you interpret those graphs of n=1 to n=4.:confused: :confused: :confused:
Reply 14
Lifeisnice
If so, how do you justify using the particles in a box method? The molecular orbitals are in specific places, not along the entire length of the molecule. Also, how do you interpret those graphs of n=1 to n=4.


The MOs aren't in specific places, they're spread over the entire molecule (and beyond, what we usually draw is an isosurface at which point the integral has some fixed value). Take a look at benzene's MOs. The MO approach can explain some things which valence bond theory (i.e. electrons sit in bonds) cannot, like conjugation and aromaticity. It also has limitations (e.g. it can't do dissociation).

The interpretation of the graphs is that the MO wavefunctions look similar to the particle in the box wavefunctions, where L is the length of the molecule and n = 1 to x/2 where x is the number of electrons in the conjugated system.
Reply 15
Kyle_S-C
The MOs aren't in specific places, they're spread over the entire molecule (and beyond, what we usually draw is an isosurface at which point the integral has some fixed value). Take a look at benzene's MOs. The MO approach can explain some things which valence bond theory (i.e. electrons sit in bonds) cannot, like conjugation and aromaticity. It also has limitations (e.g. it can't do dissociation).

The interpretation of the graphs is that the MO wavefunctions look similar to the particle in the box wavefunctions, where L is the length of the molecule and n = 1 to x/2 where x is the number of electrons in the conjugated system.


Oh...

So basically this particle in a box model assumes that each electron acts independently and contributes to EVERY observed bond in the system? That's the only way I can think of how a (single) particle in a box can model the molecule.

Also, the graphs of say n=1 and n=4 look quite different, they have different nodes. How do these correspond to specific instances of the molecule?
Reply 16
Try looking at the lectures 'Particles in Boxes' and 'Spectroscopy in Boxes' here. That's a reasonable introduction.
Reply 17
cpchem
Try looking at the lectures 'Particles in Boxes' and 'Spectroscopy in Boxes' here. That's a reasonable introduction.


Thanks for the link!

I read through it, and while I can understand the mathematics of solving the Time Independent Schroedinger Equation, I don't understand what the n in the eigenfunctions correspond to. It has something to do with energy being quantized, but what physical reality is n supposed to correspond to?
Lifeisnice
Thanks for the link!

I read through it, and while I can understand the mathematics of solving the Time Independent Schroedinger Equation, I don't understand what the n in the eigenfunctions correspond to. It has something to do with energy being quantized, but what physical reality is n supposed to correspond to?


Yes, n is the quantum (~quantisisation) number for the system and is a variable (n = 1, 2, 3... (n is an element of the positive integers)) the physical realities i.e. the energy levels and wavefunctions.

By only allowing discrete values of n (and not continuous ones) we get a quantum system. Energy levels have an n2 dependence and the wavefunctions nodes is affected by the value of n, number of nodes = where the wavefunction = 0 and so the particle can not be found there, here nodes = n - 1. Again we can related the wavefunction back to the probability density.
Reply 19
EierVonSatan
Yes, n is the quantum (~quantisisation) number for the system and is a variable (n = 1, 2, 3... (n is an element of the positive integers)) the physical realities i.e. the energy levels and wavefunctions.

By only allowing discrete values of n (and not continuous ones) we get a quantum system. Energy levels have an n2 dependence and the wavefunctions nodes is affected by the value of n, number of nodes = where the wavefunction = 0 and so the particle can not be found there, here nodes = n - 1. Again we can related the wavefunction back to the probability density.


What physical reality does n=2 represent, and what does it mean when compared against n=3?

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