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
and here’s a molecule, molecular weight 3.
(laughter).
You are all familiar with the fact that elementary molecules are normally diatomic,
like molecular hydrogen in H2 molecule or 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.
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.
Four atomic hydrogen has been discussed in connection with Van der Waals interactions
and indeed very good evidence has been produced by Professor Welsh at the university of Toronto
for the existence at very low temperatures of a molecule that consists of two hydrogen molecules, that is H22, or H4.
Until last year H3, triatomic hydrogen was not known and hardly discussed.
However H3+ is a very well known system in any discharges, the molecular ion,
H3+ 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.
Now, H3+ turns out to be a very stable system.
And no structure determinations have been made until about a month and a half ago, I mean experimental structure determinations.
But theoretical work, which is possible for as simple a molecule as H3+, have been made,
and they have fairly clearly shown that H3+ is a system like this, that is a system consisting of an equilateral triangle.
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 the system H3+, I’m not talking yet about H3, in H3+ the binding energy,
that is the energy that is required to remove one proton from the system,
is almost the same as the disassociation energy of the hydrogen molecule.
So it’s a very substantially dissociation energy.
It’s a very stable system, the H3+.
Now, in order to get experimental information about H3+ and other molecules,
one of the ways that the spectroscopist uses is to study the spectrum of such molecules.
And if we want to determine the structure of H3+ experimentally, then we should try and find, try and find a spectrum of H3+.
We have looked, my own sub group shall I say, in Ottawa for the emission spectrum of H3+.
And independently of us, Takeshi Oka in our lab also, has looked for the obstruction spectrum
and he has very recently been successful, about six weeks ago I would say,
in finding such a spectrum for the first time, 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, 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 molecular structure is determined from molecular spectra.
I’m showing you a few slides, 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 and we work in term values
which are given in units of centimetre to the minus 1, or reciprocal centimetres.
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.
But I would like to emphasise the rotational term value or the rotational energy,
which is the same except for constant factor, namely the rotational energy is given by a constant,
the so called rotational constant times J, times J plus 1.
Where J is the quantum number of the total angular momentum of the molecule.
J takes the value 012 and so on.
Now, this constant B, B sub V is given here, is simply a combination of fundamental constants
and the average of 1 over R2, the distance of the two nuclei in the particular molecule.
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.
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 RE2 .
Where mu is the reduced mass and R sub E is the internuclear distance, the distance between the two nuclei in the molecule.
This is the important relation.
So in the moment we get B from the spectrum, we get the internuclear distance or, if you like, the moment of inertia.
Now, the next slide.
Well, this repeats what I have 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.
We may have a change in vibrational energy giving mu B 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 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.
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, therefore the energy increases quadratically, J times J plus 1.
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.
And all the manyfold 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, I have just one or two very simple examples of such spectra.
This is a very old spectrum of the CN, free radical CN, where you can see particularly clearly here the regular structure,
with a gap here 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.
Now, the point that I would like to emphasise is the fact that the spacing of these lines here,
in the neighbourhood 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 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 inter nuclear distance in the molecule.
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, aluminum hydride, the aluminum hydride molecule,
in which we have here again an R branch, here a Q branch now, which starts right at the band centre,
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, 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 has been studied for about 100 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.
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, 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 great complication of the spectrum.
But this spectrum has been analysed, in the visible region for example and other parts of the spectrum,
by a succession of workers, some 50 years ago, Richardson, Finkelnburg, 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 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 ultra violet there is a very similar kind of spectrum,
which when analysed gives you the structure of H2 in its ground state.
For example you find that the internuclear distance in the hydrogen molecule H2 is .741 Å units
and that piece of information can be obtained 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 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, 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 the diatomic molecule, if this were the diatomic molecule here, the moment of inertia about this axis is zero.
So the polyatomic molecule is somewhat more complicated.
But there are some polyatomic molecules which we call symmetric top molecules
in which two of these so called principle moments of inertia are equal.
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 axis.
And that means any axis in the plain of the molecule gives the same moment of inertia,
and that is the characteristic of a symmetric top.
The moment of inertia about the axis perpendicular to this plain is different.
In the simple molecule like this, it would simply be the sum of the moment of inertia in the other two directions.
And we distinguish in this case here oblate symmetric tops from prolate symmetric tops.
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.
And since we are interested in this molecule, 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 - 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 plain of the molecule or at right angles to it,
it may have any direction to the plain of the molecule.
And then there is an additional term here which is due to this fact that we have all 3 moments of inertia different from zero
and this K that you see here,
K is the quantum number of the component of J in the direction at right angles to the plain 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 and C is a very similar constant,
only it refers to the moment of inertia about this axis, the axis perpendicular to the plain of the molecule.
So B and C correspond to the two different moments of inertia of this system.
Now, what I have written down here 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,
the electrons go around like that or they go around like that.
In 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 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 and that is given here.
And this additional term is this here minus or plus 2C 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 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
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 Tisza.
Now the next side.
Here are the energy levels, 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 01234 and so on.
And here we have the values of J, so for each value of K 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 I neglected, 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.
But when we have a degenerate electronic state, because we have these two possible directions of the angular momentum,
it 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 2C zeta K, 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.
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.
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,
the change of J is either zero or +1 or -1.
And in the spectrum that gives rise to three kinds of so called branches, the Q branches where delta J equals zero,
the R branch where delta J equals +1 and the P branches where delta J equals -1.
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 is perpendicular to the symmetry axis in this direction,
then the selection rule is delta K, is equal to + or -1.
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 am going to show presently.
The next slide, here are just two of the sets of rotational levels that I mentioned earlier here, K = 0, K = 1, K = 2.
And now we have to apply the rule that we can either have delta J is equal to plus 1 like here,
delta J is equal to minus 1 like here, this is the upper state, the upper electronic state, presumably.
And the Q branch doesn’t occur here, but here you have J = 1 here, J = 1 here,
you get Q 1 line and more lines for higher J values, and the same here.
Now, you can see that because these levels coincide with these levels and these levels coincide with these levels,
except for a 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 sub bands will be super imposed and that is shown in the next slide.
Here are the, for K equals zero, you have one sub band K = 1, K = 2, K = 3.
And you always see this, here the R branch, here the P branch and here the Q branch.
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 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 about an axis like that.
Now, you have to imagine that all these sub bands are super imposed 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, lines in quotation mark,
and here and what you will see is then what you see in the next slide.
Simple case that was found some 20 years, almost 20 years ago, no, it was more than 20 years ago, CD3.
CD3 unlike CH3, which has just a broad spectrum here that there’s nothing to analyse.
Here you have this fine structure, and this is exactly the P branch
where each of these lines really consists of a number of lines 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 and that is the intensity alternation.
And in a symmetric top molecule of this particular symmetry, B3H symmetry, you find that when K is = 0,
alternate lines are missing, in H3 and they have the ratio 10:1 in D3.
And that applies whether it’s H3 or NH3 or CH3, in all such molecules
we have this intensity alternation for K = 0 and we have another intensity alternation in K,
namely that we have strong, weak, weak, strong 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, for K equals zero you see this intensity alternation,
there’s a strong line and a very weak line and 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 alternately,
that there will be a slight intensity alternation 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 that is absent or weak or strong.
But it was in that way that for the first time in CD3 it was established that CD3 is a plainer molecule,
a plainer molecule, I don’t want to go any further into that.
Now, let us go to the next slide.
What I have 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.
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 sub bands
which no longer coincide because you don’t keep K the same in the upper and lower state.
But you get a series of sub bands 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.
The spacing between these sub bands is given by this expression 2A - B,
I should have written here C – 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.
Now, I come 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+.
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 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, positive column.
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,
we found several spectral lines that were candidates for H3+.
But we weren’t sure and fortunately we didn’t publish that
because they were not the lines that were six or seven weeks ago found by Oka in our lab
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+ in absorption.
But at the time we didn’t know that 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, you saw in a previous slide,
the many line spectrum of ordinary hydrogen or deuterium,
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 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,
we found some of these broad diffuse features 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 realised what really this spectrum means.
Now, I like to remind you of what I said earlier in connection with diatomic molecules,
that the spacing between successive lines in the neighbourhood of the so called band origin,
this spacing here is very close, is 2B, where B is this rotational constant
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 centimetres,
it came out to be, I think it was, well, it doesn’t really quite matter, 44 reciprocal centimetres, 44.
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.
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 artefact, how the wrong order had come through or some foolish thing like that.
But when I got, finally we repeated and repeated and always came up again and then I looked at this distance,
44 centimetre minus 1, I tried to see what could it be.
Now, the most natural thing of course would have been to say well, it’s some unknown spectrum of ordinary hydrogen,
ordinary molecular hydrogen.
In this case deuterium, of course.
But we knew the B values of deuterium, in the ground state, the B value of deuterium is sixty.
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.
In deuterium, they would be half that, in other words 15.
Here we have 2B = 44, therefore B = 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.
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.
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 ab-initio theory,
were 43 for H3+ and correspondingly the half of that, 21.5 for D3+.
And I noticed that here we measure a B value of 22 and predicted for D3+ is 21.5.
Well, if we had been foolish we could have said, 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, well, I shouldn’t say ‘we know’
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 2 in D3+, out of the lowest orbital and put it into an excited orbital,
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 right hand side of this central part of the band.
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,
that is an orbital with a principle quantum number higher than in the ground state, somewhere round here.
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, 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 state are similar.
The other point is, and that was important, 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 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,
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 linewidth,
the lifetime that is connected with this radiationless transition is of the order of 10 to the -11 or 10 to the -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,
in which a transition takes place, shall I say from N = 3 to N = 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.
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.
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.
That’s not all that we have as evidence.
Here’s the same spectrum you’ve seen a moment ago, but here’s the corresponding spectrum in H3.
And you just, can just see that there’s one line here, much weaker line here, just this background here,
not the sharp lines, the sharp lines are all due to H2 or D2, whichever the case may be.
And there’s another one here.
Now, the important thing that is an important confirmation 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 ordination.
But the more important thing still is that in H3 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,
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 is such that alternate levels for K = 0 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, if this is really a symmetric top molecule,
as I think it is, each of these lines really consist of a number of components.
And only the components with K = 0 are the ones that are missing or have low intensity.
So that you cannot expect that alternate lines are completely out,
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, naturally when we first found this band at 5,600 Å units,
we tried to see whether there might be something else somewhere along the spectral regions.
And low and behold at 7,100 Å we found a very strong feature here, much stronger than the band that you saw a moment ago,
this in D3 and this in H3, that is in a hollow cathode discharge in hydrogen or deuterium.
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.
And the help of Dr. Jim Watson 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 because it would take much too long, but I think the next slide might show you...
Let us forget now about the perpendicular band because the explanation we feel is 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, 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.
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.
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, 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 and you have a P branch here and you have,
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 symmetric top is a superposition of sub bands.
In the original band that we found these sub bands 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, = 3.
You have three sublines, 0, 1 and 2, these are the K values.
For the next one you have only 1, 2, 3, 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.
Here it’s there, there it’s absent, here it’s there, there it’s absent and so on, if there were a 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 threefold axis of symmetry like this system here.
And I think, the next slide please, well, this is a corresponding thing for H3 which is not quite as clear cut,
because there 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 say, this is a photometer curve of the perpendicular band at 7,100 Å units.
The spikes here you must disregard because they correspond to deuterium.
This is what you might have expected naïvely,
and this is what Dr. Watson finally obtained by introducing all these corrections and interactions.
And you see that, then introducing the linewidth,
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.
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+ and H3+.
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
are very close to the B value of the ground state of D3+, and the same for H3 and H3+.
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.
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.
But when you now separate them, then, while in a 1S electron simply gives an A1 prime electron, according to the notation,
which I cannot explain now here, the 2P electron has 6-fold degeneracy and is resolved in an E prime and an A2 double prime.
Well, no, I should say threefold 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,
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 then give rise to the spectrum.
And so I think the next slide shows an energy level diagram and shows the transitions that we have found.
For N = 2 we have actually three states, one of them isn’t shown here, we have the 2S state,
which is a double A1 prime and a 2P state, which is the A2 double prime.
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 = 3 you have 3P here and here, these two states.
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 pre-dissociation of this state is strongly forbidden,
while the pre-dissociation 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 levels of these other transitions, and everything fits together in almost perfect ways,
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.
This is somewhat similar to what I just showed for D3,
all I want to point out is the internuclear distance observed here for these various excited states.
That is we can determine the distance between the deuterons or protons in the H3+ molecule and, as you see,
they are similar to the ones in D3+ or H3+.
Some are smaller, some are larger indicating whether the orbitals involved are slightly bonding or slightly anti bonding.
Next slide, please.
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 and H plus H2 on the other hand.
I have shown here H3 if I gradually increase the principle quantum number, eventually I come to H3+ and an electron.
And in the same way, if I ionise the hydrogen atom here, I eventually come to H plus, plus H2, plus an electron.
If I now bring H plus and H2 together, I get H3+, plus an electron.
And this is the energy I mentioned at the beginning, the dissociation energy of H3+, 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 when you have the principle 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.
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 Franck-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, 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 it’s 7.1, of NH3 it’s 8.8, of CH4 it’s 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’s states which are stable.
In one of these cases we have indeed, we believe, found such Rydberg states, and that is the system NH4 plus.
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,
and the other has been known since 1955, through the work of Schüler,
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.
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+ molecules are not only fairly abundant
but also fairly important and that is in the interstellar medium,
H3+ is present in the interstellar medium by common consent of theoretical astrophysicists.
Although nobody has seen H3+ 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+ in the interstellar medium.
But suppose that H3+ is really there, and that is generally believed as I say,
and 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, to recombination with an electron.
If there are H3+ ions, there must also be electrons, in fact for many other reasons there are.
Now, if H3+ and an electron recombine, then we have the phenomenon of dissociated recombination.
And in this process we first form a highly excited state of H3, neutral, 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 that we have found in the laboratory
in the spectra of suitable interstellar clouds.
Thank you very much.
(Applause).