Spectra and Structure of Three-Atomic Hydrogen
This is a modal window.
The media could not be loaded, either because the server or network failed or because the format is not supported.
Formal Metadata
| Title |
| |
| Title of Series | ||
| Number of Parts | 340 | |
| Author | ||
| License | CC Attribution - NonCommercial - NoDerivatives 4.0 International: You are free to use, copy, distribute and transmit the work or content in unchanged form for any legal and non-commercial purpose as long as the work is attributed to the author in the manner specified by the author or licensor. | |
| Identifiers | 10.5446/52585 (DOI) | |
| Publisher | ||
| Release Date | ||
| Language |
Content Metadata
| Subject Area | ||
| Genre | ||
| Abstract |
|
00:00
ChemistryNobeliumEmission spectrumUltraviolettspektrumMoleculeAgeingBranch (computer science)AmmoniaWursthülleSea levelBearing (mechanical)Systemic therapyÜbergangszustandDipol <1,3->HydridePropylthiouracil <6-Propyl-2-thiouracil>Chemical structureDiatomaceous earthErdrutschOptische AktivitätMeatElectronProcess (computing)Complication (medicine)Electronic cigaretteAtomic orbitalSet (abstract data type)Left-wing politicsAtomic numberButcherBase (chemistry)Separator (milk)ReaktionsgleichungAluminiumZellmigrationNew Millennium ProgramCubanePill (pharmacy)TopicityChemical formulaGap junctionWine tasting descriptorsRadical (chemistry)Ice frontStructural steelDeath by burningStream gaugeLibrary (computing)Origin of replicationWhitewaterRekombinante DNSThermoformingSurface finishingNucleolusRadioactive decayCombine harvesterSunscreenSpectroscopyWaterfallStorage tankMolecularityPainStarvation responseSetzen <Verfahrenstechnik>GezeitenküsteWaterErdölraffinationProtonationGlassesCupcakeElektrolytische DissoziationWeaknessFunctional groupIceAntibodies (film)CobaltoxideSpectral lineMolecular geometryAageSong of SongsTool steelConnective tissueHoneyController (control theory)AreaHydrophobic effectNeutralization (chemistry)AbbruchreaktionHuman subject researchHuman body temperatureDischarge (hydrology)CHARGE syndromeTempehGoldCrown etherStickstoffatomTitanateGermanic peoplesHydrogenAgriculturePhosphorusGene expressionChemical reactionDevolution (biology)Binding energyChemical elementOcean currentActive siteCalculus (medicine)Flüssiger StickstoffVSEPR theorySlightingBranntweinAbsorption spectroscopyCaveCuring (food preservation)PlasticStereoselectivityArtifact (archaeology)SynergyAusgangsgesteinISO-Komplex-HeilweiseU.S. Securities and Exchange CommissionDistortionSense DistrictDiet foodAtomic orbitalDeuteriumCommon landAmmoniumMethaneAluminium hydridePhosphorescenceRedoxQuantum chemistryCryogenicsIonenbindungZeta potentialAddition reactionVakuumverpackungInfrarot-SpektrumCyanogenRydberg atomPhotodissociationRiver deltaSolution
Transcript: English(auto-generated)
00:11
Thank you very much. I feel a little out of place when we hear about molecules with molecular weights of 20,000 or 100,000,
00:24
and here's a molecule, molecular weight three. You're all familiar with the fact that elementary molecules are normally diatomic, like molecular hydrogen in H2,
00:41
molecular nitrogen into O2, F2, CO2, and so on. But there are some exceptions to this statement. You are also well aware of the fact that oxygen also forms a triatomic system, O3, and phosphorus forms a four atomic system, P4.
01:01
More recently, now already some 20 years ago, triatomic nitrogen was observed as a free radical. It is physically stable in its ground state, but is a transient molecule, and obtained, for example, by the photolysis of HN3.
01:21
Four atomic hydrogen has been discussed in connection with van der Waals' interactions, and indeed very good evidence has been produced by Professor Welch at the University of Toronto for the existence at very low temperatures of a molecule that consists of two hydrogen molecules, that is H2-2 or H4.
01:44
Until last year, H3 triatomic hydrogen was not known and hardly discussed. However, H3 plus is a very well-known system in any discharges.
02:02
The molecular ion, H3 plus, was first discovered back in 1907 by J.J. Thomson, and when the first mass spectrum of hydrogen coming from an electric discharge was obtained.
02:23
Now, H3 plus turns out to be a very stable system, and no structure determinations have been made until about a month and a half ago.
02:41
I mean experimental structure determinations. But theoretical work, which is possible for as simple a molecule as H3 plus, have been made, and they have fairly clearly shown that H3 plus is a system like this, that is a system consisting of an equilateral triangle.
03:06
The symmetry, there's a three-fold axis of symmetry in the system that has come out of these theoretical considerations. And indeed, the binding energy in this system, H3 plus, I'm not talking yet about H3.
03:24
In H3 plus, the binding energy, that is the energy that is required to remove one proton from this system, is almost the same as the dissociation energy of the hydrogen molecule. So it's a very substantial dissociation energy.
03:41
It's a very stable system, the H3 plus. Now, in order to get experimental information about H3 plus and other molecules, one of the ways that the spectroscopist uses is to study the spectrum of such molecules,
04:03
and if we want to determine the structure of H3 plus experimentally, then we should try and find a spectrum of H3 plus. We have looked, my own subgroup, shall I say, in Ottawa, for the emission spectrum of H3 plus,
04:27
and independently of us, Takeshi Oka in our lab also has looked for the absorption spectrum, and he has very recently been successful, about six weeks ago, I would say, in finding such a spectrum for the first time.
04:43
But this is not the subject of my talk. Now, before I can tell you about the spectrum of H3, neutral H3, I would like to give you a very brief summary of molecular spectroscopy, if I may put it that way.
05:02
That is, of course, not really possible in the matter of a few minutes, but just to remind some of you, and to introduce others who are not familiar with it, to the way in which a molecular structure is determined from a molecular spectra, I'm showing you a few slides.
05:21
So the first slide just reminds you of the fact that the energy of a molecule can be subdivided into three parts, the electronic energy, the vibrational energy, and the rotational energy. In spectroscopy, we usually divide these energies by the quantity h times c,
05:41
and we work in term values, which are given in units of centimeter to the minus one, or reciprocal centimeters. And I don't need to explain how the vibrational energy depends on the vibrational quantum numbers, because I will not need it for the special purpose of this talk.
06:01
But I would like to emphasize the rotational term value, or the rotational energy, which is the same except for a constant factor, namely the rotational energy is given by a constant, the so-called rotational constant, times j times j plus one, where j is the quantum number of the total angular momentum of the molecule.
06:25
j takes the value zero, one, two, and so on. Now this constant b, b sub v is given here, is simply a combination of fundamental constants and the average of one over r squared, the distance of the two nuclei in the particular molecule.
06:46
And we have to take the average of that. And then we find that this rotational constant depends slightly on the vibration if the vibrations have large amplitude, since they are anharmonic, you find that this bv depends on the vibrational quantum number.
07:03
But again, that is something we will not need for the further considerations. The main point is that you remember that the rotational constant in the equilibrium position is inversely proportional to the moment of inertia of the molecule, which is mu times Re squared,
07:22
where mu is the reduced mass and Re is the internuclear distance, the distance between the two nuclei in the molecule. This is the important relation. So the moment we get b from the spectrum, we get the internuclear distance,
07:41
or if you like, the moment of inertia. Now the next slide. Well, this repeats what I had just said here and adds the following point. When we have a transition from one state to another in the molecule, then we may have a change in electronic energy giving rise to this difference,
08:02
we may have a change in vibrational energy giving mu v, and we have a change in the rotational energy giving this mu r, so that the frequency of the spectral lines may be represented as a sum of three terms, of which the term mu sub e is the most important
08:20
when we are talking about an electronic spectrum, and where the term mu sub r is interesting because it gives us information about the structure of the molecule. Well, that is shown here at the bottom. Now the next slide just shows in an energy level diagram the kind of transition that we are talking about.
08:42
We have an electronic state up here with its various vibrational levels, and in each of these vibrational levels, we have a series of rotational levels, and you notice that the spacing of the rotational levels is quadratic. The spacing between succeeding levels increases linearly,
09:02
therefore the energy increases quadratically, j times j plus one. And the same for the lower state of the particular transition that we are considering, the various vibrational levels, which are nearly but not quite equidistant, and the various rotational levels in each of these vibrational levels.
09:21
And the manifold of transitions from these various levels up here to the various levels down here gives what we call a band system in the spectrum of the molecule. And now I think in the next slide,
09:41
I have just one or two very simple examples of such spectra. This is a very old spectrum of the Cn free radicals, Cn, where you can see particularly clearly here the regular structure with a gap here,
10:00
and then it goes on like that. This is what we call an R branch to the short wavelength side, and a P branch to the long wavelength side of the new zero, which is the transition which corresponds to the change of electronic state, as well as possibly vibrational state.
10:21
Now the point that I would like to emphasize is the fact that the spacing of these lines here, in the neighborhood of this so-called band origin, this spacing is simply equal to two times this constant b, 2b. That's a very rough statement, but by and large,
10:41
it helps in understanding the following considerations. This is 2b, so if we measure the spacing between these two lines, we get immediately the moment of inertia of the molecule, or if we know the reduced mass, we get the internuclear distance in the molecule.
11:01
This is one case, now I didn't mention that the R branch corresponds to a change of angular momentum by plus one, the P branch corresponds to the change of angular momentum by minus one. The next slide shows another old example, an aluminum hydride, the aluminum hydride molecule,
11:22
in which we have here again an R branch, here a Q branch now, which starts right at the band center, and then underneath a P branch, which is indicated here. These lines belong to the P branch. And again, you might say that if I take this spacing here,
11:42
divided by two, I have a rough value for the constant b, which tells me the internuclear distance in this particular molecule. Now, in the case of the hydrogen molecule, ordinary hydrogen, H2, the spectrum, of course,
12:00
has been studied for about a hundred years, and in this case, the spectrum doesn't look as simple as it does in aluminum hydride or cyanogen in the previous slide, but in the next slide, you see two small sections of spectra of, in this case, deuterium, and in this case, hydrogen.
12:23
And you see there are just many, many lines, and the regularity is by no means as obvious as it is in the case of CN or aluminum hydride. The reason is that in molecular hydrogen, the moment of inertia is very small, and therefore the value of b is very large,
12:41
therefore the separation of successive lines is large. Maybe one line is here and another line is here, and in between are all sorts of other bands, and that makes for a great complication of the spectrum, but this spectrum has been analyzed in the visible region, for example, and other parts of the spectrum by a succession of workers some 50 years ago,
13:04
Richardson, Finkelberg, Mecke, Weizel, and others have clarified the interpretation of the spectrum, and by this study, they have shown that this spectrum that you see here and adjacent parts of it
13:20
correspond to the transitions between excited electronic states in the H2 molecule. They are not related to the ground state of the H2 molecule. In order to establish the structure of H2 in its ground state, you have to go to the ultraviolet, in the vacuum ultraviolet,
13:42
there's a very similar kind of spectrum which when analyzed gives you the structure of H2 in its ground state. For example, you find that the internuclear distance in the hydrogen molecule H2 is 0.741 angstrom units, and that piece of information can be obtained
14:02
with a very great deal of precision. Now, before I can now come to the actual topic of my talk, I have to tell you a few things about polyatomic molecules and their spectra. Of course, linear molecules are very similar to diatomic molecules,
14:21
and I don't want to go into that now because I will not need it. But non-linear molecules, unlike diatomic molecules, have non-zero moments of inertia about any axis through them. For example, if you take this system here,
14:44
and any more complicated system, you can see that about any axis of rotation, this axis or this axis, any other axis, you find the moment of inertia is not zero. While in a diatomic molecule, if this were the diatomic molecule here, the moment of inertia about this axis is zero.
15:01
So, the polyatomic molecule is somewhat more complicated, but there are some polyatomic molecules which we call symmetric top molecules, in which two of the so-called principal moments of inertia are equal.
15:20
And that is indeed the case in this system here, where the moment of inertia about this axis here is the same as the moment of inertia about this axis and any of the other axes. And that means any axis in the plane of the molecule gives the same moment of inertia, and that is the characteristic of a symmetric top.
15:43
The moment of inertia about the axis perpendicular to this plane is different. In the simple molecule like this, it would simply be the sum of the moments of inertia in the other two directions. And we distinguish, in this case here, the oblate symmetric tops from prolate symmetric tops.
16:03
Prolate symmetric tops are those in which the moment of inertia, for example, this stick here, has a very small moment of inertia about this axis here, but a large moment of inertia about this axis or that axis. This is what we call a prolate symmetric top. This is an example of an oblate symmetric top. Since we are interested in this molecule,
16:22
I only give you the energy levels for an oblate symmetric top, but the formula are almost identical. You only change names a little bit. So the next slide shows, for an oblate symmetric top, the rotational energy or term value, if you like, is given by this formula here.
16:42
Forget about this because that's only a very small correction due to centrifugal force, which I won't discuss here. But you see this term here is exactly the same as in the case of a diatomic molecule. Only that J is now the total angular momentum which may not be in the plane of the molecule
17:00
or at right angles to it. It may have any direction to the plane of the molecule. And then there is an additional term here, which is due to this fact that we have three, all three moments of inertia different from zero. And this K that you see here, K is the quantum number of the component of J
17:24
in the direction at right angles to the plane of the molecule. So K is the component of J in the direction of the symmetry axis. It is multiplied by a constant C minus B where B is the same constant as this
17:41
and C is a very similar constant. Only it refers to the moment of inertia about this axis. The axis perpendicular to the plane of the molecule. So B and C correspond to the two different moments of inertia of this system. Now, what I've written down here
18:03
refers to a non-degenerate state. But in a molecule you may also have so-called degenerate states. For example, in a diatomic molecule, take away one of these, you can have an electronic angular momentum about this axis,
18:22
the electrons go around like that, or they go around like that. And this alternative, whether they go either way, this leads to a double degeneracy. But such a degeneracy arises also in the case of various states of symmetric top molecules. If you imagine this molecule now as a symmetric top
18:45
and you imagine an electron is moving around this way, it can also move around that way. And therefore, there would be a degenerate state, a doubly degenerate state. And in such a state, the energy formula has an additional term compared to this here,
19:01
and that is given here. And this additional term is this here, minus R plus two C zeta K, where K and C I have already defined here, K being the angular momentum, the component of the angular momentum in the symmetry axis, and zeta, this zeta here,
19:21
is the electronic angular momentum. Just as in a diatomic molecule, you have what is called lambda, the electronic angular momentum about this axis, so in a symmetric top molecule, we can have an electronic angular momentum about this axis. And this electronic angular momentum can interact with the rotational angular momentum,
19:43
and that gives rise to this additional term in the energy formula, which was first discussed, I believe, some 50 or 48 years ago by Teller and Tisser. Now, the next slide. Here are the energy levels of,
20:03
let's only look at the right-hand side of this diagram, of an oblate symmetric top molecule. We have a number of values of K, K equals zero, one, two, three, four, and so on, and here we have the values of J. So for each value of K,
20:20
we have a set of rotational levels, very similar to the rotational levels of a diatomic molecule. But we have, instead of one of these sets, we have a number of sets, and all these sets are similar. By just shifting them a little bit, they would come into coincidence if it weren't for the effect of these terms that are neglected,
20:43
namely the centrifugal distortion terms. So we have these different sets. Now, let's look at the next slide. This was what I just said was for the non-degenerate electronic state, when the electronic angular momentum, or the vibrational angular momentum for that matter, about this axis is zero.
21:01
But when we have a degenerate electronic state, because we have these two possible directions of the angular momentum, can either go this way or this way, we find that there will be a splitting, and this splitting is given here. It's due to this minus or plus sign in front of 2K zeta K, 2C zeta K,
21:23
and you have this splitting, which is the same for all these levels here. It's larger here because it's proportional to K, and still larger here and still larger here. I don't want to go into great detail about that because time would not permit to explain everything that one could say about the structure of this molecule.
21:42
But I did want to mention the complication that arises even in a simple system like triatomic hydrogen when you introduce degeneracy and Jahn-Teller effect and all sorts of other things that can be considered.
22:00
Now, the next slide gives you the selection rules for a symmetric top molecule. For the rotational quantum number J, you have the same selection rule as in a diatomic molecule, or for that matter in any atomic system or even a nucleus, for so-called dipole radiation, delta J,
22:21
the change of J is either zero or plus one or minus one. And in the spectrum, that gives rise to three kinds of so-called branches, the Q branches with delta J equals zero, the R branches with delta J equals plus one, and the P branches with delta J equals minus one.
22:42
In a so-called parallel band, that is to say when the dipole moment of the transition is parallel to the symmetry axis, then the selection rule for K is that delta K is equal to zero. When, however, the dipole moment of the transition
23:01
is perpendicular to the symmetry axis in this direction, then the selection rule is delta K is equal to plus or minus one. Now, let us first consider this kind of parallel band, and we have to consider it if we want to interpret the spectrum that I'm going to show presently.
23:20
The next slide. Here are just two of these sets of rotational levels that I mentioned earlier here for K equals zero, K equals one, K equals two. And now we have to apply the rule that we can either have delta J is equal to plus one, like here, delta J is equal to minus one, like here. This is the upper state, the upper electronic state, presumably.
23:42
And the Q branch doesn't occur here, but here you have J equals one here, J equals one here, you get Q one line, and more lines for higher J values, and the same here. Now, you can see that because these levels coincide with these levels,
24:02
and these levels coincide with these levels except for shift, since in this case where delta K is equal to zero, that is, we go from this stack to this stack, from that stack to that stack, from that stack to that stack. All the subbands will be superimposed, and that is shown in the next slide.
24:22
Oh, could we just try the one after that? Yes. I'll come back to the other one in a moment. Here are the, for K equals zero, you have one subband, K equals one, K equals two, K equals three. And you always see this, here the R branch, here the P branch, and here the Q branch.
24:42
Now, what has been done here is assuming that the moments of inertia, and therefore the rotational constants, are different in the upper and the lower state, and therefore there is a shading. It widens out here and narrows down this way. But even so, the same statement applies
25:04
that I made earlier for a diatomic molecule, namely that the spacing between these lines near the band origin, so-called, is roughly 2B, where B is the rotational constant that corresponds to the moment of inertia
25:21
about an axis like that. Now, you have to imagine that all these subbands are superimposed, and that is what you see here. And if you don't have enough resolution, either because the lines are broad or because your instrument is not powerful enough, then you will get just a simple series of lines,
25:41
lines in quotation marks, and here. And what you will see is then what you see in the next slide, for a simple case that was found some 20 years, almost 20 years ago. Oh, it was more than 20 years ago.
26:01
CD3. CD3, unlike CH3, which has just a broad spectrum here, that cannot be, there's nothing to analyze, here you have this fine structure. And this is exactly the P branch where each of these lines really consists
26:21
of a number of lines of different K values, of different K values. Now, if we could just go back two slides. In molecules that have identical nuclei, there is a very important phenomenon that occurs,
26:40
and that is the intensity alternation. And in a symmetric top molecule of this particular symmetry, D3H symmetry, you find that when K is equal to zero, alternate lines are missing in H3, and they have the ratio 10 to 1 in D3,
27:02
and that applies whether it's H3 or NH3 or CH3. In all such molecules, we have this intensity alternation for K to zero, and we have another intensity alternation in K, namely that we have strong, weak, weak, strong,
27:21
and the same in D3, but with another intensity ratio. I don't want to go into that further. Now, let us just look forward again to the next slide. Now, I just wanted to point out... No, no. One back. No. There. I just wanted to point out... I'm not quite tall enough.
27:41
For K equal to zero, you see this intensity alternation. There's a strong line and then a very weak line, then a strong line, very weak line, and so on. Now, if you superimpose this all, then it is clear that if you have always one line very weak here alternately, that there will be a slight intensity alternation
28:00
in this group of lines if you haven't resolved them. And that is what I think you can see in the next slide. I think it's just visible that you see here strong, weak, strong, weak, strong, weak, not terribly strong because it's only one line out of several
28:20
that is weak or... that is absent or weak or strong. And... but it was in that way that for the first time in CD3, it was established that CD3 is a planar molecule, a planar molecule. I don't want to go any further into that. Now, let us go to the next slide.
28:41
What I've just discussed were parallel bands of symmetric top molecules. You have this stack combines with this stack, this stack with this stack, and so on. If we now come to perpendicular bands, the situation becomes more complicated because we go from this stack to this stack and from this stack to that stack, and so on.
29:02
And these several subbands no longer coincide, but they rather look like shown in the next slide, which is, again, a very old slide due to Dennison many years ago. Here you have these various subbands which no longer coincide because you don't keep K the same in the upper and lower state,
29:21
but you get a series of subbands whose Q branches, which are unresolved here, form a series of lines like that. Well, one could say a lot more about this, but I think time would not be sufficient to go into all that, and I will proceed immediately to the next slide.
29:41
The spacing between these subbands is given by this expression 2A minus B. I should have written here C minus B, which is a negative quantity. I don't think there's a complication in degenerate states, but I think I will omit that complication for the purpose of this lecture. So let us go to the next slide.
30:02
Now I come really finally to the real subject of my talk. As I mentioned at the beginning, we were actually engaged in experiments that had the aim of finding an emission spectrum of H3 plus.
30:22
And in doing so, we found a number of features which seemed to agree very nicely in the infrared spectrum of a discharge through a discharge tube of this sort with what had been predicted from theory. In this discharge tube, we have a cathode, a so-called hollow cathode,
30:44
and we have an anode, and we fill this with hydrogen or deuterium, and a discharge is struck between these two electrodes. We have here the negative glow, which withdraws into the hollow, a hollow cathode glow, and then we have here an anode glow, a positive column.
31:07
And we look at the spectrum of the cathode glow through this window, and we look at the spectrum of the anode glow through this window. And we found in this way, in the hollow cathode, which is assumed to be rich in molecular ions,
31:26
we found several spectral lines that were candidates for H3 plus. But we weren't sure. Unfortunately, we didn't publish that, because they were not the lines that were six or seven weeks ago found by Oka in our lab
31:45
in the absorption spectrum of a similar kind of discharge, which was done with very great refinement, and where he actually found the spectrum of H3 plus in absorption. But at the time, we didn't know that,
32:01
and we thought we must first establish whether in this discharge there is rotational equilibrium. And so we decided we take some spectra of this discharge in the ordinary region where we know, as you saw in the previous slide, the many-line spectrum of ordinary hydrogen or deuterium,
32:24
and see whether from this spectrum of ordinary hydrogen or deuterium we can establish the rotational temperature. And when we did this, we obtained a spectrum that I think is shown in the next slide. And you see here, this is the spectrum in the same region,
32:42
as I showed in an earlier slide, of the anode glow and the normal set of spectral lines of ordinary, in this case, deuterium. Deuterium is shown. This is nothing new. But when we looked at the spectrum of the cathode glow,
33:00
we found some of these broad diffuse features here and here and here and here and here. On the first spectrum that we obtained, it wasn't as clear as it is on this, particularly when we cooled with liquid nitrogen, it came out much stronger. And it took several months before we realized what really this spectrum means.
33:24
Now, I'd like to remind you of what I said earlier in connection with diatomic molecules, that the spacing between successive lines in the neighborhood of the so-called band origin, this spacing here is very close, is 2b, where b is this rotational constant,
33:44
which is simply the reciprocal of the moment of inertia except for a constant factor. And when I looked at this separation here, I just measured it with a ruler and converted to reciprocal centimeters. It came out to be, I think it was, well, it doesn't really quite matter, 44 reciprocal centimeters, 44.
34:10
And then I tried to go backward now and try to identify the molecule that is responsible for this spectrum here from the value of this constant b that I found.
34:21
Of course, the first assumption, the very first assumption when we first saw that, was that we simply had to do with some kind of artifact, somehow the wrong order had come through or some foolish thing like that. But when I got, finally we repeated and repeated and it always came up again, and then I looked at this distance, 44 centimeter minus one, I tried to see what could it be.
34:44
Now, the most natural thing, of course, would have been to say, well, it's some unknown spectrum of ordinary hydrogen, ordinary molecular hydrogen. But, in this case deuterium, of course, but we knew the b values of deuterium.
35:01
In the ground state, the b value of deuterium is 60. And in the excited states, the only states that come into play in this spectral region, because we cannot involve the ground state here, in the excited states, the b values are all of the order of 30 in ordinary hydrogen.
35:20
In deuterium, they would be half that, in other words, 15. Here we have 2b equal 44, therefore b equal 22, and 22 is not equal to 15, not anywhere near that. And so it was rather baffling. Why we should find something with a b value of 22?
35:40
Now you could say, well, couldn't it be some hydride, shall we say, some impurity, OH or CH or some other hydride? Well, indeed, hydrides have fairly large b values, but when they are deuterated, all the b values, the maximum b value that has ever been found for a deuteride is 10, and not 22.
36:02
So that didn't work either. And then fortunately, I remembered that in the calculations of the structure of H3+, the b values that were predicted from theory, from up-in-issue theory, were 43 for H3+, and correspondingly the half of that, 21.5 for D3+.
36:29
And I noticed that here we measure a b value of 22, and predicted for D3+, it's 21.5. Well, if we had been foolish, of course, we could have said,
36:41
well, here we have the spectrum of D3+, but it was immediately clear to me that we didn't have the spectrum of D3+, because there's no way in which you can account for a spectrum of D3+, in this particular spectral region, because we know that D3+, is unstable in, well, I shouldn't say we know,
37:03
but from theory, it had been predicted that all excited states of D3+, are unstable. Moreover, if we excite D3+, to an excited state, we take one of the electrons that holds it together, and there are only two in D3+, out of the lowest orbital and put it into an excited orbital,
37:23
and then it would have a very much smaller binding energy, and this spectrum is more the spectrum between two states that have nearly the same binding energy, because it's fairly uniform spacing on the left and the right-hand side of this central part of the band.
37:40
So, it was clear to me it couldn't be D3+, and yet the b value was the same. Now, what could that be? Well, the assumption that we have confirmed by many further observations, is simply that we have here neutral D3, in which an electron is moving in a Rydberg orbital,
38:05
that is an orbital with a principle quantum number higher than in the ground state, somewhere around here, this Rydberg orbital, this Rydberg electron, does not change the moment of inertia, it does not change the binding, and therefore the b value is the same as if this Rydberg electron were not there,
38:24
that is the same as in D3+. And since Rydberg states have much the same rotational constants in various Rydberg orbitals, it is explained that the upper and lower states are similar. The other point is, and that was important,
38:40
you see that these lines, why are these lines so broad here? Well, there's a whole field in molecular spectroscopy that deals with the phenomenon of pre-dissociation, and I cannot undertake to describe that in detail here, but I can only remind you that the pre-dissociation
39:01
is caused by a radiationless transition from some excited state into the continuum of another state. And the reason why these lines are broad here, must clearly be that the lower state, not the upper state,
39:21
of this particular transition is pre-dissociated. It cannot be the upper state, because if the molecule were pre-dissociating in the upper state, it wouldn't emit light, it would dissociate, and you wouldn't see the spectrum. And that is a very strong conclusion, because when you have this kind of line width,
39:41
the lifetime that is connected with this radiationless transition is of the order of 10 to the minus 11 or 10 to the minus 12 seconds, so there is no time for the emission of light. So the assumption then was that what we see here is a spectrum of H3, or D3 rather in this case,
40:04
in which a transition takes place, shall I say, from n equal 3 to n equal 2, where n is the principle quantum number, simply analogous to the ordinary H alpha transition in atomic hydrogen. And indeed, this spectrum is not very far from the H alpha line of atomic hydrogen.
40:24
And the lower state, the lower state, pre-dissociates because the ground state of D3 or H3 is, of course, unstable. Everybody knows that if you take a hydrogen atom and bring it up to a hydrogen molecule, it will not be attracted.
40:40
We have a repulsive state, which in fact is a degenerate state, but never mind about that. Now let's look at the next slide. Now that's not all that, of course, we have as evidence. Here's the same spectrum we've seen a moment ago, but here's the corresponding spectrum in H3.
41:03
And you can just see that there's one line here, a much weaker line here, just this background here, not the sharp lines. The sharp lines are all due to H2 or D2, whichever case may be. And there's another one here. Now the important thing that is an important confirmation
41:23
of the idea that this is really H3 is the intensity alternation. Namely, you see that you have strong, weak, strong, weak, and then it fades out. The Boltzmann factor overcomes the intensity alternation. But the more important thing still is that in H3,
41:42
the first line is strong, while in D3, the first line is weak. In other words, in H3, the odd lines are strong, and in D3, the even lines are strong. And that is precisely what I showed in an earlier theoretical slide,
42:00
precisely what we expect in a molecule like this for the rotation about this axis, because we exchange two hydrogen nuclei. But it's a little more complicated because we have a triatomic system, and one finds from theory that the intensity alternation in hydrogen
42:23
is such that alternate levels for k equals zero are missing, while in deuterium, alternate levels are very weak. And on top of that, in one case, they are the odd levels, and the other, the even levels. Now, I also mentioned earlier that each of these lines,
42:44
if this is really a symmetric top molecule, as I think it is, each of these lines really consists of a number of components, and only the components with k equals zero are the ones that are missing or have low intensity, so that you cannot expect that alternate lines are completely out.
43:02
They are there, but they are weak, because one of the lines contributing is missing or is very weak. Now, let's look at the next slide. In addition, and naturally, when we first had found this band at 5600 angstrom units, we tried to see whether there might be something else
43:20
somewhere along the spectral regions. And lo and behold, at 7100 angstroms, we found a very strong feature here, much stronger than the band that you saw a moment ago, in this in D3 and this in H3, that is, in a hollow cathode discharge in hydrogen or deuterium.
43:44
Now, the structure of this is not as clear as the structure of the band that you saw on the preceding slide, and the reason for that is clearly that this is a perpendicular band. That was clear to me right from the beginning, but it took quite a long while in the help of Dr. Jim Watson
44:03
of Southampton University to really confirm unambiguously that this is a perpendicular band of the system D3 or H3, whichever the case may be. I don't want to go through the whole argument,
44:20
because it would take much too long, but I think the next slide might show you... No. Next one, please. Oh, perhaps I should omit that next slide, please. And the next one. I hope I get that. Yeah. Now, let's forget now about the perpendicular band, because the explanation,
44:41
while we feel it's completely unique and unambiguous, is somewhat involved, because it involves Jahn-Teller interaction, lambda-type doubling, lambda-type resonance, and all these things that make the spectrum rather complicated and would take an extra hour to explain these irregularities,
45:07
if you like to call them that. Anyway, it has been completely explained, but what I would like to show you very much is the spectrum that you see here. This slide that you see here is not what we directly observe in the spectrum.
45:22
We have used a trick in order to isolate that part of the spectrum that belongs to H3 or D3 from the part that belongs to H2 or D2, and we have done it by taking two pictures. We take the original, and we take a negative of the anode spectrum.
45:44
As you may recall, in the spectrum of the anode, there's no sign of H3 or D3. Therefore, if we now take the negative of that, put it on top of the cathode glow spectrum of H3 or D3, and then copy through that,
46:01
we eliminate all the lines that are either due to H2 or D2, and only the lines of H3 or D3 remain, and that is what you see here. And now you see here an additional band that is very much nicer than the first one that we found. Namely, you have here a Q branch,
46:20
and you have a P branch here, and this goes on here. This is the Q branch again, the same as that, and you have here the R branch. And the important thing is now, here the lines are sharp, or almost sharp, and here you see that the so-called K structure is resolved. You recall that a parallel band of a symmetric top
46:41
is a superposition of subbands. In the original band that we found, these subbands were not resolved because the lines were broad. But here the lines are sharp, and therefore the subbands are resolved. And more than that, you see here, for example, for J, well, it's called N here because this is the angular momentum without spin,
47:03
equal three, you have three sublines, zero, one, and two, these are the K values. For the next line, you have only one, two, three. The zero line is missing, or very weak, so we don't see it. Here the zero line is there, so you have a 100% intensity alternation here in the K equals zero subband.
47:21
Here it's there, there it's absent, here it's there, there it's absent, and so on, if there were, and so on. On the basis of this spectrum here, there can be absolutely no doubt that we have here a spectrum of a system that has a three-fold axis of symmetry, like this system here. And I think, next slide, please.
47:42
Well, this is a corresponding thing for H3, which is not quite as clear-cut because their pre-dissociation sets in that limits the number of K values. I don't want to go further into that. Next slide, please. And I think I will forget the discussion of this, too. The next slide, please. This, I think I should perhaps say, this is a photometer curve
48:01
of the perpendicular band at 7100 angstrom units. These spikes here you must disregard because they correspond to deuterium. This is what you might have expected naively, and this is what Dr. Watson finally obtained by introducing all these corrections and interactions, and you see that, and then introducing the line width,
48:23
and you see that this spectrum is a very good reproduction of this spectrum here, of this perpendicular band, but I don't want to go further into that. The next slide, please. And this is an infrared band. I think I will not go into that. Next slide, please. My time is gradually coming to an end.
48:40
These are the results in the form of the values of this constant B and C. If you just look at B, these are the B values of the various electronic states of D3 and here of H3, and for comparison, we've given here the theoretical values for D3 plus and H3 plus.
49:02
This slide was made a little before Dr. Oka got the spectrum. We have now an experimental value. He has an experimental value for this quantity, but it's only half a percent higher than this. It's 43.6, I believe, or something like that. What you do see is that the B values of all these Rydberg states of D3
49:24
are very close to the B value of the ground state of D3 plus, and the same for H3 and H3 plus. The next slide, please. I should say a few words, if I may go on for a few minutes longer, about the electronic states that we see.
49:42
The orbitals in this symmetry, D3H, of an electron, well, in the limiting case of the United Atom, if you bring these three nuclei together, then you would have a 1s electron, 2s, 2p, 3s, 3p, and so on.
50:01
But when you now separate them, then while a 1s electron simply gives an A1 prime electron, according to the notation, which I cannot explain now, here, the 2p electron has six-fold degeneracy, and is resolved in an E prime and an A2 double prime,
50:22
well, no, I should say three-fold degeneracy, and you have an E prime and A2 double prime electron or orbital that you get out of that, and the same out of this one, for 3D you get this. And if you now consider the electronic structure of this H3 or D3, then you have a total of three electrons.
50:43
The first two electrons go into this lowest orbital, 1s A1 prime. The other electron goes into any of these orbitals here, and depending on which one it goes to, we have different electronic states, and they give them rise to the spectrum.
51:01
And so I think the next slide shows an energy level diagram and shows the transitions that we have found. For n equal 2, we have actually three states. One of them isn't shown here. We have the 2s state, which is a doublet A1 prime, and a 2p state, which is doublet A2 double prime,
51:21
but there is also derived from that an E, a degenerate state, which should be shown here, but it is a continuous state. It's a repulsive state. It is the ground state of the H3 or D3 system. The excited states are shown here. For n equal 3, you have 3p here and here, these two states.
51:42
This is the upper state of the degenerate, of the perpendicular band. This is the upper state of the parallel band that was first found. And then we have this new system here, which has this upper state and the other ground state here. These lines are much sharper because the predissociation of this state is strongly forbidden
52:03
while the predissociation of this is allowed. And so these two transitions are broad. I have broad lines. These two transitions here have sharp lines. We have also observed infrared transitions, this one here, which combine upper or lower,
52:21
upper levels of these other transitions, and everything fits together in an almost perfect way so that there can be absolutely no doubt that we have this kind of energy level diagram. Let's just look at the next slide. Well, this is somewhat similar to what I just showed for D3. All I want to point out is the inter-nuclear distance
52:43
observed here for these various excited states. That is, we can determine the distance between the deuterons or protons in the H3 plus molecule. And as you see, they are similar to the ones in D3 plus or H3 plus.
53:01
Some are smaller, some are larger, indicating whether the orbitals involved are slightly bonding or slightly anti-bonding. Next slide, please. Yeah, perhaps this may be the last slide I will show. This is just to indicate the correlation between the energy levels of H3 on the one hand
53:21
and H plus H2 on the other hand. I've shown here H3, if I gradually increase the principal quantum number, eventually I come to H3 plus and an electron. And in the same way, if I ionize the hydrogen atom here, I eventually come to H plus plus H2 plus an electron.
53:41
If I now bring H plus and H2 together, I get H3 plus plus an electron. And this is the energy I mentioned at the beginning, the dissociation energy of H3 plus, which is large. And the important thing is now that all the excited states of this system lie above this limit. The first excited state arises
54:02
when you have the principal quantum number equal to two. And that is already above this level, and similarly with all the others. In other words, all these excited states of H3 have no other way to go but go up here. And it's only the ground state which is unstable which goes here.
54:20
But of course we can have pre-dissociation from any of these levels into this level here, but this pre-dissociation is strong only for the lowest ones according to the Frank-Condon principle. Now one might ask, what is the determining factor in obtaining such a spectrum? It is clearly the fact that this so-called proton affinity,
54:43
proton affinity of molecular hydrogen is large. And there are other systems in which this is the case, and maybe I can just show the next slide if it's the one I think it is. The proton affinity of molecular hydrogen is 4.4, of water 7.1, of NH3 is 8.8,
55:01
of CH4 is 5.3 eV. In all these systems, we have very stable ions, these ions here, and we can therefore expect if we add an electron to these systems, we should get Rydberg states which are stable. In one of these cases, we have indeed, we believe,
55:22
found such Rydberg states, and that is the system NH4+. If we add an electron to that, we get Rydberg states and they give rise to a spectrum, to two spectra, one of which has been known for 108 years, the so-called Schuster band of ammonia,
55:41
and the other has been known since 1955 through the work of Shuler. But I think time does not permit me to go into further detail about that. Now there's one further remark that I am anxious to make because I'm naturally always asked, well, what is it good for? Where can it be applied, this sort of thing?
56:03
Well, I don't know whether it can be applied to any chemical reactions. That is for the future to show. But there is one place where it has long been assumed that H3 plus molecules are not only fairly abundant, but also fairly important, and that is in the interstellar medium.
56:22
H3 plus is present in the interstellar medium by common consent of theoretical astrophysicists, although nobody has seen H3 plus yet in the spectrum, because it's difficult to see, but Dr. Oka has now supplied the basis for establishing by the spectrum the presence of H3 plus in the interstellar medium.
56:43
But suppose that H3 plus is really there, and that is generally believed, as I say, and it is considered to be the origin of the formation of all these fancy molecules that have been found in the interstellar medium. If it is there, it is subject to formation,
57:01
to recombination with an electron. If there are H3 plus ions, there must also be electrons, in fact, for many other reasons there are. Now, if H3 plus and an electron recombine, then we have the phenomenon of dissociative recombination, and in this process, we first form a highly excited state of H3 neutral.
57:23
It cascades down, emits the spectrum that I have been reporting, and then dissociates into H2 plus H by this phenomenon of dissociative recombination. And I have every hope that in the not-too-distant future, my astronomical friends will observe the spectrum
57:41
that we have found in the laboratory in the spectra of suitable interstellar clouds. Thank you very much.