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#### riboswitch

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##### Particle in a Ring: Problem
« on: September 21, 2014, 06:14:51 PM »
I'm trying to solve a very simple problem from my Physical Chemistry book. I know it's about the particle in a one-dimensional ring problem, but I do need some clarifications...

Problem:

The particle on a ring is a useful model for the motion of electrons around the porphine ring, the conjugated macrocycle that forms the structural basis of the heme group and the chlorophylls. We may treat the group as a circular ring of radius 440 picometers, with 20 electrons in the conjugated system moving along the perimeter of the ring. Assume that in the ground state of the molecule quantized each level is occupied by two electrons.
• Calculate the energy of an electron in the highest occupied level.
• Calculate the frequency of radiation that can induce a transition between the highest occupied and lowest unoccupied levels.

Question:

What is the meaning of "highest occupied level"? I'm a bit lost here...
I do know that I'm dealing with 20 electrons and "each level is occupied by two electrons".
Does that mean that I have 10 levels overall?
« Last Edit: September 21, 2014, 06:44:30 PM by riboswitch »

#### Enthalpy

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##### Re: Particle in a Ring: Problem
« Reply #1 on: September 21, 2014, 06:50:38 PM »
Can you find the solutions for electrons on a ring?

They have different energies. In the ground state, provided that the energy levels are significantly more spaced than kT, the levels of lowest energy will be filled, up to the number of available electrons, and the levels of higher energy will be empty.

Don't forget to put 2 electrons per state, and remember the phase when counting the states.

#### riboswitch

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##### Re: Particle in a Ring: Problem
« Reply #2 on: September 21, 2014, 07:29:20 PM »
Can you find the solutions for electrons on a ring?

They have different energies. In the ground state, provided that the energy levels are significantly more spaced than kT, the levels of lowest energy will be filled, up to the number of available electrons, and the levels of higher energy will be empty.

Don't forget to put 2 electrons per state, and remember the phase when counting the states.

If I'm correct (and I hope so), the quantized energies of an electron in a ring are:
$$E_{m_l} = \frac{m_l^2 \hbar^2}{2mr^2}$$
with
$$m_l = 0, \pm 1, \pm 2, ...$$
According to the problem, I have to assume that in the ground state, each level occupies 2 electrons.
I have 20 electrons, so I'm guessing that I have to occupy 10 levels.
So the levels occupied are:

$$m_l = 0, +1, -1, +2, -2, +3, -3, +4, -4, +5$$
Have I interpreted the problem correctly?

Since ml=+5 and ml=-5 are degenerate levels, I assume that I can arbitrarily choose between these two the level in which I'll put the last pair of electrons.
« Last Edit: September 21, 2014, 07:58:08 PM by riboswitch »

#### Enthalpy

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##### Re: Particle in a Ring: Problem
« Reply #3 on: September 23, 2014, 04:41:07 PM »
I'm pleased with the energy equation (unless I'm wrong too!).

-----

Apparently (from +1 -1...) you've written the solution for a propagating wave, which is just fine. A different set of solutions, which is strictly equivalent, would be standing waves. A standing wave is a sum of two propagating ones, a propagating wave is a sum of two standing ones - with the proper phase as the weights in the sum.

Chemists use to prefer standing waves, because chemical bonds make these more natural. Take carbon's 2p:you can use 2px 2py 2pz (the peacock or rather dumbbell shaped orbitals) as the base to write any valid 2p orbital as a weighed sum. These are the standing waves. Or you can take the doughnut-shaped base, where a doughnut with z axis is a sum of x and y peacocks with 0° and 90° phase; this is the propagating wave. Equally good. Or you could even take an elliptical base. But as soon as you draw a C-C bond along x, the peacock is more natural, to make the first bond from 2px and the pi bond(s) from 2py and 2pz.

In your ring, the standing wave with ml=1 for instance would instead have two amplitude maximums, at geometric 0° and 180° on the ring, or at 90° and 270°: two solutions as good as the waves of uniform amplitude that propagate in clock and counterclock directions.

-----

With standing waves, the occupation of the level ml=5 is easier to represent. Because of electrostatic repulsion, the electrons won't pair as the same wave. Instead, they stay as far apart as they can, that is, one as the 0° standing wave and the other as the geometric 18° standing wave.

It's the same as for electrons occupying one 2p each in atomic carbon or nitrogen, or two unpaired electrons in molecular oxygen making it paramagnetic.

#### Corribus

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##### Re: Particle in a Ring: Problem
« Reply #4 on: September 23, 2014, 06:44:23 PM »
@riboswitch

Your approach to the problem is correct. Unfortunately the problem is wrong. Particle-on-a-ring treatments of porphyrins only use 18 electrons because this is the accepted size of the conjugation path. So actually only E up to level 4 is occupied in reality. (If you use 20 electrons, you get a triplet paramagnetic ground state like oxygen, which is certainly not accurate.) The transition energy is therefore n = 4 n = 5. Using 20 electrons, I guess it'd be n = 5 n = 6, but it's not really clear at all.

(Unlike the other problem you looked at (retinal), this simple treatment of the porphyrin is not particularly good because it doesn't account for spectral intensities or configuration interaction. In a real porphyrin spectrum, the lowest energy transition is partially forbidden due to destructive self-interference of the doubly-degenerate HOMO-LUMO transition. There's also a higher-energy transition originating from the same pair that is quite strong. These are referred to as Q- and B-band transitions, respectively. The Particle-on-a-Ring treatment doesn't yield any of this.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### riboswitch

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##### Re: Particle in a Ring: Problem
« Reply #5 on: September 26, 2014, 06:06:11 AM »
@riboswitch

Your approach to the problem is correct. Unfortunately the problem is wrong. Particle-on-a-ring treatments of porphyrins only use 18 electrons because this is the accepted size of the conjugation path. So actually only E up to level 4 is occupied in reality. (If you use 20 electrons, you get a triplet paramagnetic ground state like oxygen, which is certainly not accurate.) The transition energy is therefore n = 4 n = 5. Using 20 electrons, I guess it'd be n = 5 n = 6, but it's not really clear at all.

(Unlike the other problem you looked at (retinal), this simple treatment of the porphyrin is not particularly good because it doesn't account for spectral intensities or configuration interaction. In a real porphyrin spectrum, the lowest energy transition is partially forbidden due to destructive self-interference of the doubly-degenerate HOMO-LUMO transition. There's also a higher-energy transition originating from the same pair that is quite strong. These are referred to as Q- and B-band transitions, respectively. The Particle-on-a-Ring treatment doesn't yield any of this.)

I'll assume that the energy transition is n=5  n=6. The radius is 440 picometers and I'll use the equation for the quantized energies of a "particle on a ring".

$$\Delta E = \big(2m_{l} + 1 \big) \ \frac{ \hbar ^2}{2mr^2}$$

$$\Delta E = (11) \times \ \frac{(1.0546 \times 10^{-34} \ J \times s )^{2}}{2 \times \ (9.11 \times 10^{-31} Kg) \times (4.4 \times 10^{-10} m)^{2}}$$

$$\Delta E = \ 3.47 \times \ 10^{-19} \ J$$

Then to determine the frequency of radiation that can induce a transition between the highest occupied and lowest occupied levels, I'll use the following equation:

$$\nu \ = \ \frac{ \Delta E}{h} \ = \ \frac{3.47 \ \times 10^{-19} \ J}{6.626 \times \ 10^{-34} \ J.s} \ = 5.24 \times 10^{14} \ Hz$$

The corresponding wavelength is ≈ 572.46 nm (green light, I guess).

The "particle on a ring" treatment is incorrect, as all of you stated before.

I do want to admit that I'm struggling with many Phys. Chem. concepts. I have never heard of Q- and B- band transitions before, for example. Although I do heard about the "triplet state" in my "Fluorescence and Phosphorescence" lecture, I still don't know if we're dealing with the same concept here. You said that 20 electrons is unacceptable if we're using the "particle on a ring" energy equation because only levels up to n=4 is allowed in reality. So I'm guessing that a ring with 16 electrons is still acceptable, or a ring with 12 electrons.

The only thing I somewhat understand is that a green light won't make the porphin electron "jump" from n=5 to n=6. Right?
« Last Edit: September 26, 2014, 06:32:46 AM by riboswitch »

#### Corribus

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##### Re: Particle in a Ring: Problem
« Reply #6 on: September 26, 2014, 10:20:16 AM »
@riboswitch

I apologize for using terminology in my previous post that you had no reason to know. Also, there were one or two errors in my post that arose because I was working off of memory rather than actually taking the time to work through the math from the beginning. So here I'll set the record straight.

Using the information provided, you did the problem correctly.

Commentary:

The free electron model of the porphyrin (particle-on-a-ring) was first solved by William Simpson in 1949 (J Chem Phys 17, 1218-21).  Actually, porphyrins play a very important historical role in the development of quantum chemical theory. Simpson defined a porphyrin as an 18 member ring with 18 electrons.

(You will see that the ring of the porphyrin actually has 20 carbons, and 4 nitrogens. When considering the actual conjugation pathway, it is common to exclude four of the carbons and include 2 of the nitrogens, which is a shorter conjugation path. There are two mutually perpendicular ways to do this. I do not remember the reason that this is done, but it's probably because of the problem encountered in the free electron model when using 20 electrons.)

Back when I was writing my PhD dissertation, I pretty much reproduced every theoretical model there was on porphyrin electronic structure. My dissertation includes a 150 page introduction to this topic. When I reproduced the free electron method using 18 electrons, I calculated an energy separation for the lowest energy transition of 3.43 x 10-19 J, which is very close to what you calculated. The agreement is coincidental. You are using a radius of 440 pm. I used a radius of 400 pm. These values derive from an estimate of the C-C bond length multiplied by the number of bonds (circumference) divided by 2 pi to get the radius. For instance, an 18 center porphyrin has 18 bonds. Taking benzene as a model with an average bond length of 139 pm, you arrive at a radius of 398 pm. Likewise, a 20 center ring gives a radius of about 442 pm. Back when Simpson wrote his paper, I presume the bond lengths of various organic molecules were not know with much precision. To estimate a radius to use in the free electron energy formula, Simpson "calibrated" his value with the absorption wavelength of benzene. You can read his paper if you want to know more about how he did this, but the end result is that he calculated an energy separation of 615 nm. The actual experimental absorption energy of porphyrin depends quite a bit on what substituents are on the macrocycle, and "plain" (unsubstituted) porphine isn't soluble in anything, so it's hard to know what to compare these values to. Simpson used bacteriochlorophyll, which has a transition "center of gravity" (more on that in a minute) of 549 nm, and he was pretty happy with the level of agreement.

Regardless of the precision of the calculated transition wavelength, using 20 carbons/electrons is very wrong for a more important reason, and that is that it predicts the HOMOs contain two unpaired electrons. If this were the case, porphryin would be magnetic, and it isn't. 18 electrons gives rise to a fully filled m = 4 level. All electrons paired: no magnetism. Before I used the formalism "triplet" and "singlet". This is called the multiplicity, which is a formalism of spectroscopy. The multiplicity (M) is just equal to 2S + 1, where S is the total electron spin. If you have two unpaired electrons, each with a spin of 1/2, the multiplicity is 3 (2*0.5 + 1), which we call a triplet. No unpaired electrons is a multiplicity of 1 (2*0 + 1), which we call a singlet. These designations are very important for fluorescence and phosphorescence, as you've indicated. The ground state of a porphyrin (with a closed shell metal center) is a singlet, meaning all electrons are paired, so the 20-electron system cannot be right, at least in the free electron model. In fact, virtually all organic molecules are ground state singlets. If they weren't, they'd very readily react with oxygen (a ground-state triplet) and life probably wouldn't exist. So if a model predicts a common organic molecule is a triplet in its ground state, the model probably isn't right.

Here's something else interesting about the free-electron model of porphyrin. You'll notice that there are two degenerate HOMOs, one with a m = 4 and one with m = -4. Both of these have the same energy, and they are both fully filled with two electrons. The m = 5 (and -5) level is empty. This means that the lowest energy transition between levels 4 and 5 is actually four fold degenerate. You can have a transition from m = 4 to m = 5, m = 4 to m = -5, m = -4 to m = 5 and m = -4 to m = -5. Each of these transitions has the same energy equal to what was calculated above.

(Quick note: using 20 electron system means in your ground state you have 1 electron in the n = 5 orbital and one electron in the m = -5 orbital. Here it becomes unclear what the lowest lying transition is. Is it a transition from the filled m = 4 orbitals to the partially filled m = 5 orbital? Or is it from the partially filled m = 5 orbital to the fully unfilled m = 6 orbitals?)

Let's define the electron configurations as follows: {w,x;y,z}, where the four electrons are in orbitals w, x, y and z. So the ground state configuration is { +4,+4;-4,-4} because we have two electrons in the m = 4 orbital and two electrons in the m = -4 orbital. There are, as I've said, four possible excited state configurations: {+4,+5;-4,-4}, {+4,-5;-4,-4}, {+4,+4;-4,-5}, and {+4,+4;-4,+5}. (I have a nice figure that illustrates this graphically, but I can't access image sharing sites from my ISP here; I'll upload later if I get a chance.) In the Zero Order approximation, each of these excited states have the same energy, because the energy only depends on the orbital energy, which only depends on the n value. Heretofore what we've basically been doing is calculating orbital energies using one electron, and then filling in the other electrons after the fact, assuming that putting all these extra electrons in the orbitals doesn't change anything. That's completely wrong! In reality, when you have multiple electrons in a system, the electrons interact with each other, and certain configurations will have more favorable interactions than other.

You may have heard by now of Hund's Rules, possibly even in general chemistry. According to these rules, states with the highest spin multiplicity are the lowest energy (all things being equal, spins like to be unpaired), followed by those with higher orbital angular momentum. Thus we can conclude that the four excited-states we predict in the free electron model don't in reality all have the same energy. Can we say more about this?

In the free electron model, all angular momentum is in the z-direction (perpendicular to the ring) and all wavefunctions are eigenfunctions of the angular momentum operators.

$$\hat{L}^2 = -\hbar^2 \frac {d^2}{d \phi^2}$$

$$\hat{L_z} = i\hbar \frac {d}{d \phi}$$

The wavefunctions from which the energy formula derives are

$$\psi(\phi) = (2 \pi)^{-1/2} e^{im\phi}$$

If we take the electrons a functionally independent, the ground state configuration(s) then are found to have a total orbital angular momentum of 0, and the excited-state configurations have angular momentum of ± 1 and ± 9 (in factors of ħ; basically, just adding up all the possible m values).

Although it's difficult at this stage to quantify the effects of electron-electron interaction in the excited-states of the free-electron model, but we can order them qualitatively using Hunds rules. Accordingly, the states with high orbital momentum (Lz = ± 9ħ) are predicted to be lower in energy (to some degree) than the states with low orbital momentum (Lz = ± ħ). Therefore instead of having a single 4-fold degenerate transition, we actually will have two 2-fold degenerate transitions: a lower energy transition with an orbital momentum change of (Lz = ± 9ħ) and a higher energy transition with an orbtial momentum change of Lz = ± ħ.

Now, take a look at an experimental porphyrin spectrum.  You can find one here:

http://www.thno.org/ms/getimage.php?name=thnov02p0916g03.png

You will see that the spectrum includes a very intense peak at around 400 nm (5 x 10-19 J) and a series of weak transitions around 600 nm (3.3 x 10-19 J). The strong band is called the Soret or B-band and the weak transitions are collectively called the Q-bands. "Q" and "B" are historical designations; "Soret" is named after a person. Let's ignore the fact that there are multiple Q-band transitions (these are vibronic in origin and also due to the fact that two nitrogens in the free base form of porphyrin have hydrogens, which lowers the symmetry). The important thing is that there are TWO transitions in porphyrin spectrum and the free base model, along with Hunds' rules, predicts that there will be two transitions with different energies. We can conveniently define the Q-band as the transition to orbitals with high orbital angular momentum and the B-band to the transition to orbtials with low orbital angular momentum. The different orbital momentum of the states has been verified experimentally (Barth et al, Ann. New York Acad. Sci, 1973, 206, 223-246). Importantly, the transition energy you calculated is the "center of mass" or average of these two transitions, and this seems to be right as well. We estimated ~570 nm, and we do have one transition on the low energy side of that and one on the high energy side of that. The real center of mass is actually somewhere in the 450-480 nm range, suggesting that the free electron model actually doesn't do a great quantitative job of predicting where the center of mass is, despite William Simpson's claims to the contrary, but given its simplicity, it's not too bad.

Ok, one more thing, and this is where I have to correct myself. The free-electron model actually can account for the intensity disparity between the Q- and B-bands, although it takes a few steps to get there. I won't take you through the math, but you can show in simple approximation that for an electron transition to be allowed in the free electron model, the orbital angular momentum has to change by ± ħ. Any transition with a different change in momentum would be strictly forbidden. Well, as it happens the B-band, which is very intense, involves a transition from a state with momentum = 0 to a state with momentum of ± ħ. And the Q-band, which is very weak, involves a transition from a state with monentum = 0 to a state with moentum of ± 9ħ. The Q-band transition should be strictly forbidden. (In reality, it's a little bit allowed, because this determination is based on a first-order approximation.)

I know that was very long-winded but maybe it helps you see some of the value in these models. The free electron model is very simple; the fact that it can predict, at least qualitatively, so many features of such a complex molecule is pretty amazing. Of course, there are better models out there that can be used to quantify these effects. We have also neglected here that the excited state has higher order electron spin multiplicities, which also complicates things further.

Also, answering this question of yours:

Quote
You said that 20 electrons is unacceptable if we're using the "particle on a ring" energy equation because only levels up to n=4 is allowed in reality. So I'm guessing that a ring with 16 electrons is still acceptable, or a ring with 12 electrons.
There is nothing special about 18 electrons specifically. It's just that for porphyrin, 18 electrons gives the best fit to empirical observation. For benzene, you'd use a ring of 6 electrons.
« Last Edit: September 26, 2014, 10:32:13 AM by Corribus »
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#### Corribus

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##### Re: Particle in a Ring: Problem
« Reply #7 on: September 27, 2014, 12:47:48 AM »
If you have two unpaired electrons, each with a spin of 1/2, the multiplicity is 3 (2*0.5 + 1), which we call a triplet.
LOL, more corrections: that should be 2*1 + 1 = 3.

Quote
The important thing is that there are TWO transitions in porphyrin spectrum and the free base model,
That should be "free-electron model" not "free base model".
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#### riboswitch

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##### Re: Particle in a Ring: Problem
« Reply #8 on: September 29, 2014, 05:44:29 PM »
Can I ask stupid questions about the concepts you explained in your latest posts? Just to make sure I interpreted them correctly.

Hund's rule and Orbital Angular Momentum

According to my Chemistry textbooks, Hund's rule states that when assigning electrons in orbitals, each electron will first fill all the orbitals with similar energy ("degenerate orbitals") before pairing with another electron in a half-filled orbital. Atoms at ground states therefore tend to have as many unpaired electrons as possible. Thanks to the "multiplicity" formality concept you described earlier, I can now appreciate what the higher-level books tell me about Hund's rules. What I don't understand is the concept of "orbital angular momentum". Is it the same "orbital angular momentum quantum number" that "describes the shape of the orbital" according to what little I can recall from my previous Chemistry classes? Or are they not related whatsoever?

Can I determine the orbital angular momentum just by looking at the HOMO electron configuration (like multiplicity)? Or is determining the angular momentum only possible using the "angular momentum operators" you wrote previously?

Q and B- bands

Quote
Although it's difficult at this stage to quantify the effects of electron-electron interaction in the excited-states of the free-electron model, but we can order them qualitatively using Hunds rules. Accordingly, the states with high orbital momentum (Lz = ± 9ħ) are predicted to be lower in energy (to some degree) than the states with low orbital momentum (Lz = ± ħ). Therefore instead of having a single 4-fold degenerate transition, we actually will have two 2-fold degenerate transitions: a lower energy transition with an orbital momentum change of (Lz = ± 9ħ) and a higher energy transition with an orbtial momentum change of Lz = ± ħ.

You said that according to Hund's Rules, states with higher orbital angular momentum posess the lowest energy. This means that states with Lz = ± 9ħ have lower energies than states with Lz = ± ħ. You also said that Q-bands represent transitions to orbitals with higher orbital angular momentum and the B-bands represent transitions to orbitals with low orbital angular momentum. Does that mean the transition from ground state to the excited state with orbital angular momentum Lz = ± 9ħ was represented in the Q-band of the porphyrin spectrum?

What does the y-axis of the spectrum graph represent? It says "ε/l mol-1 cm-1

By the way....

Thanks for explaining the concept of multiplicity here. You saying that virtually all organic molecules are ground state singlets was a great revelation for me (together with the fact that molecular oxygen is a ground-state triplet). It's like you're implicitly saying that organic molecules can react with molecular oxygen when they receive enough energy to make the transition to the excited "triplet" state. Am I right in this one?

(P.S.: Sorry for the late reply. I was studying other non-chemistry stuff )

#### Corribus

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##### Re: Particle in a Ring: Problem
« Reply #9 on: September 29, 2014, 08:48:26 PM »
No such thing as stupid questions!

1. Orbital angular momentum in the context of the examples above is related to the orbital shapes you've learned in general chemistry. The s, p, d, and f atomic orbitals are distinguished by the quantized angular momenta of the electrons. Each electron has it's own quantized momentum. Hund's rule of orbital momentum is predicated on the hypothesis that in multielectron systems, the electrons are less likely to interact in an unfavorable way when they are moving more or less in the same direction, and higher total orbital momentum values are correlated with electrons moving in the same direction. (If electrons are moving in opposite directions, their vector magnitudes will cancel out to some degree, giving lower total momentum values.) Thus, it's not the momentum values themselves that determine the energy of the system - it's the idea that electrons don't like to share the same space, and states characterized by higher total momenta are also usually those in which electrons are farthest apart on average.

In the case of the free-electron model of the porphyrin, we're not dealing with atomic orbitals, but rather molecular-type orbitals that are also the solutions to the particle-on-a-ring problem. The general idea is the same, though.

In simple molecular and atomic systems, you can determine the total orbital angular momentum values, and therefore the relative state orderings. This is almost the entire purpose of Term Symbols.  Of course, these are ultimately derived from solutions to the wavefunctions.

http://en.wikipedia.org/wiki/Term_symbol

2.

Quote
Does that mean the transition from ground state to the excited state with orbital angular momentum Lz = ± 9ħ was represented in the Q-band of the porphyrin spectrum?
Yes.

Quote
What does the y-axis of the spectrum graph represent? It says "ε/l mol-1 cm-1.
Epsilon is the molar extinction coefficients and the l mol-1 cm-1 is the unit. The classic convention used by many journals is to report axes as unitless numbers, which is why there is the "/" divide by symbol.

3.

Quote
Thanks for explaining the concept of multiplicity here. You saying that virtually all organic molecules are ground state singlets was a great revelation for me (together with the fact that molecular oxygen is a ground-state triplet). It's like you're implicitly saying that organic molecules can react with molecular oxygen when they receive enough energy to make the transition to the excited "triplet" state. Am I right in this one?
More or less. You might notice that most combustion reactions (octane with oxygen, for example) are very thermodynamically favored. Yet gasoline don't just spontaneously combust. A significant amount of energy is required. Part of the reason is that the reaction is kinetically slow due to the fact that the reaction of oxygen with most organic substrates is not spin-allowed.

One of the exited states of oxygen, however, is a singlet state.  So-called singlet state is quite reactive with organic substrates and oxidizes them very readily. This is the basis of some modern cancer therapies, in which a sensitizer (often a porphyrin; this for a long time was one of the only FDA cleared photodynamic pharmaceuticals) produces singlet oxygen in response to light, and the singlet oxygen tears apart membranes and other biomolecules in the vicinity of the therapeutic agent. Singlet oxygen has quite a bit more energy available for reaction, but it's also a spin-allowed reaction, which is part of the reason its chemistry is so different from ground-state oxygen.

http://en.wikipedia.org/wiki/Singlet_oxygen

http://en.wikipedia.org/wiki/Photodynamic_therapy

The generation of singlet-oxygen from a photosensitizer is some interesting photochemistry. Porphyrin is a ground-state singlet, as we've seen. When it is photoexcited, the excited state is initially also a singlet.  This may relax back to the ground-state by fluorescence, but some porphyrins have a high triplet yet, meaning that instead of relaxing to the ground-state, a nominally forbidden transition to an excited-triplet state can occur. Normally, this would relax back to the ground state by phosphorescence (or other nonradiative pathways), but if oxygen is around, reaction between triplet oxygen and triplet porphyrin is facile and the triplets annihilate to form singlet (ground-state) porphyrin and singlet (excited-state) oxygen. Essentially, an energy transfer reaction has occurred. There are a number of photoactive nanoparticles that can do this as well, which is why a number of plastics that have these nanoparticles in them are much more biodegradeable - the singlet oxygen oxidizes and destroys the polymer in the presence of sunlight.
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#### riboswitch

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##### Re: Particle in a Ring: Problem
« Reply #10 on: October 06, 2014, 07:26:23 PM »
Angular momentum of a particle on a ring

I was asked to review the concept of angular momentum of the free particle on a ring and to apply it in our porphin problem. In fact, the textbook problem also asks the angular momentum of the electron in the highest occupied level. Since the angular momentum of the electron in the "particle on a ring" model is quantized, I should calculate it by expressing the angular momentum in terms of the de Broglie wavelength and considering the allowed wavelengths of the particle ( λ = 2pr/ml).

$$J_{z} \ = \ pr \ = \ \frac{hr}{ \lambda }$$

$$J_{z} \ = \ \hbar m_{l}$$
Suppose I calculate the angular momentum of our porphin electron in the highest occupied level. If I treat said electron as a free particle on a ring, I'll just use the equation above to solve Jz.

$$J_{z} \ = \ m_{l}\hbar \ = \ \pm 6 \hbar$$
My book says nothing beyond that. You said earlier that we don't have a single 4-fold degenerate transition, but we have two 2-fold degenerate transitions (represented in the Q- and B- bands of the porphyrin spectrum). This is because we're dealing with a multi-electronic system, so we can't just ignore the interactions between the numerous electrons that we have in our porphin system. That means my calculations don't correspond to any real experimental data. My assumption that the electron transition happens from the highest occupied level m=5 to lowest unoccupied level m=6 (m=5 m=6) is also wrong and misleading, as you have hinted previously.

According to my solution manual, the angular momentum value should be:

$$J_{z} \ = \ \pm 5.27 \ \times 10^{-34} \ Js$$
No matter how many times I use my calculator, ±6ħ is never equal to the value indicated above. The whole problem becomes more complicated the longer I analyze it. Or maybe the whole problem is wrong, as you have stated previously.

Quote
The generation of singlet-oxygen from a photosensitizer is some interesting photochemistry. Porphyrin is a ground-state singlet, as we've seen. When it is photoexcited, the excited state is initially also a singlet.  This may relax back to the ground-state by fluorescence, but some porphyrins have a high triplet yet, meaning that instead of relaxing to the ground-state, a nominally forbidden transition to an excited-triplet state can occur. Normally, this would relax back to the ground state by phosphorescence (or other nonradiative pathways), but if oxygen is around, reaction between triplet oxygen and triplet porphyrin is facile and the triplets annihilate to form singlet (ground-state) porphyrin and singlet (excited-state) oxygen. Essentially, an energy transfer reaction has occurred. There are a number of photoactive nanoparticles that can do this as well, which is why a number of plastics that have these nanoparticles in them are much more biodegradeable - the singlet oxygen oxidizes and destroys the polymer in the presence of sunlight.

So porphyrins occasionally undergo intersystem crossing to the triplet state. I'll just assume every photosensitizer in this kind of therapy undergoes this kind of transition. What I understand is that intersystem crossing is possible when the energy levels of the excited singlet state and the excited triplet state overlap. If the energy level gap between the two states is large enough, intersystem crossing will not happen.

I love the fact that these concepts have real-world applications (like photodynamic therapy). Thanks for the additional information.

#### Corribus

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##### Re: Particle in a Ring: Problem
« Reply #11 on: October 07, 2014, 12:05:24 AM »
According to my solution manual, the angular momentum value should be:

$$J_{z} \ = \ \pm 5.27 \ \times 10^{-34} \ Js$$
No matter how many times I use my calculator, ±6ħ is never equal to the value indicated above.
Well, you're right that 6ħ never will equal the book answer, but 5ħ does. And the highest occupied orbital has an angular momentum of 5, not 6.

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So porphyrins occasionally undergo intersystem crossing to the triplet state.
More than occasionally. It depends on the porphyrin of course, and especially on the coordinated metal, but a closed shell model compounds like zinc tetraphenyl porphyrin has a triplet yield of around 80% if I recall. A free base porphyrin (like photofrin) is probably higher. I can look back through my materials and get you some hard numbers if you are really interested.

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What I understand is that intersystem crossing is possible when the energy levels of the excited singlet state and the excited triplet state overlap. If the energy level gap between the two states is large enough, intersystem crossing will not happen.
The parameters that impact triplet yield are complicated. Experimentally measuring triplet yield isn't particularly easy, either. Any one bulk photophysical property will be affected by all the various rates of excited-state deactivation to some degree, and there's no simple way to summarize the structural features that give rise to this, that, or the other.

That said, the presence of a heavy atom in the proximity of the conjugated system (what is "heavy" is arbitrary; there's no hard cut-off, but usually 3rd row or lower is heavy enough to be significant) generally will lead to high triplet yields because of the heavy atom's impact on spin-orbit coupling. Explaining why heavy atoms increase the rate of intersystem crossing isn't straightforward, though, so I won't bother unless you really want me to try. I've got a great picture of the fluorescence of porphyrin and brominated porphyrin side by side that show the effect, though. I'll see if I can't hunt it down.

Overlap of the triplet and singlet exciton wavefunctions is also important; as a result, it is often observed, all things being equal, that similar compounds with lower energy optical absorption bands have lower triplet yields. This has important effects in conducting polymer physics. As I recall, you can see this effect quite nicely in the oligothiophenes, where the triplet yield drops steadily as the oligomer length increases. (Singlet states tend to by highly delocalized, triplet states not so much, so the degree of overlap decreases as the singlet conjugation length increases.)

Finally, the triplet yield is heavily dependent on other singlet-state deactivation pathways. Since the singlet-to-triplet transition is nominally forbidden, it is also typically rate limiting. Any structural feature of a molecule that offers a convenient relaxation pathway will tend to short-circuit triplet formation. So the presence of things like open shell metals, amino or other functional groups that can facilitate the production of charge-transfer transitions, or simply the presence of good overlap of the excited and ground vibronic wavefunctions can really impact the triplet yield.  And naturally, to sensitize singlet oxygen, the excited triplet state of the photosensitizer has to have a higher energy than that of singlet oxygen, because energy only flows downhill.

So, finding good photosensitizers is a tricky business!
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### Corribus

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##### Re: Particle in a Ring: Problem
« Reply #12 on: October 07, 2014, 10:12:08 AM »
Photograph showing heavy atom effect, as promised. And just to be extra relevant, it's with a porphyrin. So you see, under UV illumination, the unsubstituted porphyrin is fairly fluorescent, but the brominated porphyrin has quenched fluorescence because the bromine facilitates rapid intersystem crossing to the triplet state (which is non-fluorescent).

What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### riboswitch

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##### Re: Particle in a Ring: Problem
« Reply #13 on: October 14, 2014, 02:25:45 PM »
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Well, you're right that 6ħ never will equal the book answer, but 5ħ does. And the highest occupied orbital has an angular momentum of 5, not 6.
Wow! No wonder I flunked the exam!
You're right. I confused the highest occupied level to the lowest unoccupied energy level.

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Overlap of the triplet and singlet exciton wavefunctions is also important; as a result, it is often observed, all things being equal, that similar compounds with lower energy optical absorption bands have lower triplet yields.

Let me restate what you have just said. I want to make sure I understand what you're saying.
According to the "Free particle in a box" energy equation, the allowed energy values are inversely proportional to the square of the "length" of the "box".

$$E \ \propto \frac{1}{L^{2}}$$

That means that lower energy values result from the increase of the value of L.

The same phenomenon happens in the "Particle on a ring" system. In this case, the allowed energy values of a free particle restricted to a ring are inversely proportional to the square of the radius of said ring.

$$E \ \propto \frac{1}{r^{2}}$$

The higher the value of r (the larger the ring), the lower the energy level of the free particle.

You said that similar compounds with lower energy optical absorption bands have lower triplet yields. That explains why longer oligothiophenes (with lower energy optical absorption bands) have lower triplet yields.

Question: So, the lower the triplet yield of the compound the higher its quantum yield? Since compounds in the excited triplet state tend to phosphoresce, I'll assume that higher triplet yields equal to lower chances that the compound will fluoresce. In fact, you said that a brominated porphyrin, which has a higher triplet yield, tends to not fluoresce.

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Any structural feature of a molecule that offers a convenient relaxation pathway will tend to short-circuit triplet formation. So the presence of things like open shell metals, amino or other functional groups that can facilitate the production of charge-transfer transitions, or simply the presence of good overlap of the excited and ground vibronic wavefunctions can really impact the triplet yield.

Wow, new terminologies! Never heard of the charge-transfer transition before. Is it the same thing as the "charge delocalization" that happens in carboxylate anions (and benzene rings), making them more stable via resonance? Why would an amino group participate in a charge-transfer transition?
« Last Edit: October 14, 2014, 02:44:14 PM by riboswitch »

#### Corribus

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##### Re: Particle in a Ring: Problem
« Reply #14 on: October 14, 2014, 06:36:08 PM »
That means that lower energy values result from the increase of the value of L.
A better way to state this is to say that an increase in the value of L results in energy levels that are spaced closer together. In the limit that L extends to infinity, there is no spacing at all between adjacent levels. In other words, when the electron is not confined at all, the quantization condition disappears for all intents and purposes, and the classical limit is reached.

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Question: So, the lower the triplet yield of the compound the higher its quantum yield? Since compounds in the excited triplet state tend to phosphoresce, I'll assume that higher triplet yields equal to lower chances that the compound will fluoresce. In fact, you said that a brominated porphyrin, which has a higher triplet yield, tends to not fluoresce.
This is often true, but not always, because there are other possible relaxation pathways. (Just a terminology note: "quantum yield" can be generically applied to any excited-state relaxation process. It is usually understood to mean "quantum yield of fluorescence", but when you are comparing multiple possible pathways, it is better to use "fluorescence yield" or "quantum yield of fluorescence" because "quantum yield" could mean "quantum yield of phosphorescence", or "quantum yield of intersystem crossing to the triplet surface" something else.) So, a molecule with a low triplet yield could also have a very low fluorescence yield if the nonradiative relaxation pathways dominate. All of the yields have to add up to 100%, though, so all things being equal, highly fluorescent molecules do tend to have low intersystem crossing yields.

Also, even if the triplet surface is populated after photoexcitation, this does not mean phosphorescence is certain. In fact, strong phosphorescence is pretty rare, even among good triplet sensitizers. Often it might only be observed at low temperature.

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Wow, new terminologies! Never heard of the charge-transfer transition before. Is it the same thing as the "charge delocalization" that happens in carboxylate anions (and benzene rings), making them more stable via resonance? Why would an amino group participate in a charge-transfer transition?
Not really. A charge transfer transition is one in which there is a large spatial change of electron density (or other charge carriers) after photoexcitation. This could be almost like an intramolecular redox reaction in some molecules, or it could only be a partial relocation of charge density, as is the case in, for example, metal polypyridyl complexes (e.g., Ru(bpy)32+, in which photoexcitation results in a partial oxidation of the ruthenium and partial reduction of the terpyridine ligand: http://en.wikipedia.org/wiki/Tris(bipyridine)ruthenium(II)_chloride). Photon-driven electron transfer reactions are important in, among other things, photosynthesis and dye-sensitized solar cells (http://en.wikipedia.org/wiki/Dye-sensitized_solar_cell).

Because a large movement of charge almost invariably results in large (though delayed by many picoseconds) structural changes, charge transfer states tend to also quench fluorescence. The nitrogen on an amino group (like the oxygen in a carbonyl) is a good temporary acceptor of an excited electron. What will often happen in the excited state is that after the nitrogen accepts the electron, it will then twist or bend to stabilize this state. This twisting motion (called a twisted intramolecular charge transfer, or TICT, state) represents a pretty massive structural change, which efficiently quenches the excited state before fluorescence can occur. The likelihood of this happening is very solvent-dependent: polar solvents are more compatible with a charged, almost ionic structure, and so they tend to stabilize the charge transfer state, resulting in red-shifted, quenched fluorescence. Nonpolar solvents are less compatible, and favor fluorescence before charge transfer can occur.

Take a look at this photograph to see how structure influences the fluorescence yield of some common coumarin laser dyes.  (I think I've shown this here before somewhere since it was already in my photobucket album.)

You can see that the dye on the left that features a diethyl amino group has quenched fluorescence in the most polar solvents. On the other hand, the other dye on the right features an amino group that can't rotate efficiently - which inhibits formation of the twisted charge transfer state, and results in high fluorescence yields even in polar solvents.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman