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NEUTRON DIFFRACTION AND STRUCTURE OF
THE ADSORBED PHASES
C. Marti, P. Thorel
To cite this version:
C. Marti, P. Thorel. NEUTRON DIFFRACTION AND STRUCTURE OF THE ADSORBED
JOURNAL DE PHYSIQUE Colloque C4, supplkment au no 10, Tome 38, octobre 1977, page C4-26
NEUTRON DIFFRACTION AND STRUCTURE OF THE ADSORBED PHASES
C. MARTI and P. THOREL
Institut Laue-Langevin B.P. 156X, 38042 Grenoble Cedex France
R&um6.
-
L'Ctude de la physisorption par diffraction neutronique est discutCe, spkcialement par rapportB
nos prCcCdents rbsultats pour le krypton sur graphite. On montre que la position du pic d e surstructure donne peu d'information ; mais sa faible intensit6 r6sulterait de la coexistence d'antiphases & une 6chelle infkrieureB
100 h;. Les variations des pics du graphite sont difficilement conciliables avec les isothermes d'adsorption : elles impliquent une persistence de la congruence m2me pour la bicouche. On indique aussi comment on pourrait r6soudre les difficultds soulevCes par les fautes d'empilement du substrat.Abstract.
-
Study of physisorption by neutron diffraction is discussed critically, especially in relation to our previous results for krypton or graphite. It is shown that the position of the superstructure peak gives little information ; but its low intensity can be related to existence of antiphases on a scale smaller than 100 h;. Variations of graphite peaks are difficult to conciliate with adsorption isotherms : they imply persistence of registry even in the bilayer. Means to overcome difficulties coming from stacking faults in the substrate are indicated.The more I think over our experiments in Grenoble with neutrons on surfaces, the less I get the impression of understanding. I take the opportunity of a small conference like this one, to put our results to question, and' to ask you for answers.
lo The first one concerns the adsorption isotherm (Fig. 1) of our most liked substrate, papyex (a French trade name for a substance very much like grafoil). I must confess that we did not take this isotherm with great care, thinking that the work was not worth a lot of time ; I confess it as a great sin, but the neutronist who has not ever sinned in that
manner can throw the first stone. Nevertheless I ask: where does the krypton lay when the adsorption rises from 50 cm3 (completion of the monolayer) to 100 cm3 (before the bilayer phase transition) ? Well, I have heard many times people well acknowledged in adsorption push the question away, raising their shoulders with
-
that's intergranular condensation, something. which has to do with capillarity,-
that's adsorption on lateral faces, on kinks, on corners, on impurities (then we would have in our case as many of these cc defects >> as of basal sites !)Now, in figure 2, the variation of the 002 graphite peak is linear with the amount of adsorbed krypton ;
the linearity, except for the first point, seems even too good, but there is no ad hoc correction in it. I always took the variation of the 002 peak as a very good measure of the adsorption on the basal plane of graphite, because it does not depend on the structure in the adsorbed layer
cm3 STP
1
I = (nb,
+
mb,)'L
pressure
FIG. 1. -Adsorption isotherm of krypton on papyex at 79 K.
I do not see how intergranular condensation or adsorption on lateral faces can give variations of the 002 peak of graphite ; perhaps you do. I am left with the idea of an unbearable number of impurities etc.
2" Moreover this variation is too high. A 1 % variation (for about the monolayer) corresponds to
NEUTRON DIFFRACTION AND STRUCTURE OF THE ADSORBED PHASES C4-27 % resp. graph. peak
FIG. 2.
-
Integrated intensity of the graphite peaks (Background subtracted) as a percentage of the intensity of the respective graphite peak : (0) 002, ( X ) 100, (+
) 101. The arrow indicatesthe full monolayer from isotherm measurements.
epitaxial ~ r ( f l x f i x 30") on graphite lamellas 55 atoms thick covered on two faces ; if the layer completion is at 50 cm3 and not at 63 cm3 and if krypton monolayer is 8 % compressed relatively to epitaxy, we find lamella 70 carbon atoms thick. This number led to a specific area of 40 m2/g, at least two times too high (at the beginning of our experiments we found 18-20 m2/g for our sample ; I am afraid that one year later, we found 10 m2, but that is another story I do not want to confess now). It must be 'recalled that in our experiments the variation of the 002 is taken on those basal planes which are 90" out of the principal orientation. I could say that preferentially oriented lamellas are 140 carbon atoms thick, but the misoriented ones are only 70" atoms thick ; I could say that preferentially oriented lamellas adsorb only on one basal plane ; I could say
.
.
.
.
I must add that the maximum intensity of diffraction (for well-oriented basal planes) is about 100 times the minimum intensity (for 90" out of the principal orientation) but that it means indeed that only one half of the graphite is preferentially oriented. Take 40 m2/g for misoriented planes and you are left with nearly zero adsorption sites for preferentially oriented lamellas !
I regret to have not been interested enough, up to now, in these quantitative aspects and I would like to
see measurements of 002 variations in the good geometry (out of plane).
3" The quantitative aspect was also neglected in the study of the surstructure ring. Up to 60 cm3 (coverage unity), there is no problem with the slope of this ring and its interpretation as coming from a registered surstructure f i x
fix
30". The coherence length is about 110hi,
like argon on grafoil at low temperatures. Its intensity seems to be proportional to the amount of adsorbed krypton during the monolayer building, but it is about 5 times smaller than expected.I have only one explanation for that : graphite monocrystalline basal faces are 120-150
hi
wide ;krypton can lay on three types of carbon hexagon centres, A, B, or C ; or it can build A, B or C 2D crystals or antiphases, which are smaller than the underlying basal graphite perfect plane ; for a mixture of antiphases on a piece of substrate, the shape of the diffraction peak is not changed (i.e. the coherence length appears to be that of the substrate) but the intensity is strongly reduced. Now a great many questions arise :
-
is this cc polycrystallinity >> general, does it depend on the quality of the substrate, on its physical perfection and its impurities,-
does it vary with growing conditions (it must be recalled that at 79 K 2D-Kr exchanges rapidly with 3D-gaseous krypton) but nevertheless one can imagine that some imperfections are dynamically trapped,- or is this polycrystallinity thermodynamically stable, like a density of dislocation depending only ori' temperature.
I was told that in LEED, people look for a good part of the surface, where spots appear clearly. Does it mean that some parts of their crystal are perfect enough to support a monocrystalline adsorption (this perfection is not a chance, it is repeatable in a series of experiments on the same place), but that other parts (the majority ?) have imperfections which trap one type of antiphase around them, leading to polycrystallinity on the 120-150
hi
scale (properly the electron beam coherence length) and hence to spots much less bright ?4" Now I come to the problem of epitaxy. We worked with 2.4
hi
neutrons ; the drawback is a large background ; the advantage is that we can observe the 100 and 101 graphite peaks. The variations of these peaks are closely related to registry of the adsorbed speciessin ?rMh sin r N k sin rPl iwp,
a =
--
e A +sin n-h sin ~k sin ?rl
C4-28 C. MARTI AND P. THOREL
where the origin is on the substrate absorbing
/--di---, ,---mdl--,
d l ! I , t , 8 , , d , , I , , t t
,
surface, A = b, e x w a is the structure factor of aS
substrate cell, hkl are indices in this cell ; t m'd I
r
=C
b, exwA is the structure factor of an adsorbate4
cell ; h ' k ' are indices in this cell.
In the simple case of krypton in registry and 10 1 peak, one can then choose h = h ' and k = k ' (note a very interesting peculiarity : these variations are not sensitive to antiphases)
sin vP61 AI = MNPM'N12
1
A (1
r
[---
sin n6lx cos (nP61
+
6) cos ywhere 6 = 2 rrlz
/
c (Z the height of krypton on=
x1
z graphite, c the cell height)y is the difference between the arguments of A and
re-'? It does not appear mixed with rrPS1, because (hkl) and ( h a occur at the same angle and their variations have to be added together. It is related to the lateral position of the adsorbate.
If 6 = krr, AI is like a height variation of the peak, rr
if 6 =
-
+
krr AI is close to the derivative of the 2peak as if it was slightly displaced.
We observed such a displacement on the 001 peak of MnI, covered with 36
A.
An intermediate case was shown this morning by Kim Carneiro for A on graphite at 77 K.Is it now possible to get some A 1 when the adsorbate lattice has a parameter slightly different from that of an epitaxial lattice ? I study here a one dimensional problem : non-epitaxial islands which have nevertheless one atom registered (Fig. 3). I take the case of compressed lattice ( d , < d ) and I ask for some increase of coverage when the lattice is compressed (not a too strong demand), i.e. m' = m, ( m
+
1) d l<
md, which means a relation between the island size m and the lattice incompatibility d - d L a--
dsin mxd,, sin x m 1 d N / m ' The amplitude is
sin xd, sin Xm'd the first 2 r r . .
term has its maximum at X , =
-
In figure 4d r ' '
I
'
sx
=-=A
md, 2 m '
The second term has maximums at
FIG. 3. - Non-registred islands, but connected by registred beginnings d : substrate parameter.
f--
-
range of range of
k = m k = m + l
FIG. 4. -Diffraction by an mdl island and positions of the peaks of an m'd superstructure.
For k = m the maximum (which is sharp) is outside
the first term peak ; this would correspond to the diffraction peak of the substrate lattice, but for k = m
+
i,
the maximum occurs at1
Now
-
# a ( m is the nearest superior integer to mL),
i.e. it occurs near X , ; its amplitude is about arn
.-
= N ; its width corresponds to the large net. mThe .pinning ,-of non-epitaxial islands to the substrate lattice gives a response nearly that of a lattice that is non-epitaxial but as large as the substrate.
Well, you can imagine
-
that non-epitaxial islands are really equal sized . .(then the secondary maximums of the first term are not blurred into a Gaussian)
-
that m = 5 / 4 athen the maximum for k = m occurs on a secondary maximum of .the first term which is still, in amplitude, 13 % of the principal maximum ; i.e. the substrate peak can vary 13 % as much as if the adsorbed lattice was registered (a surstructure peak would have only less than 2 % of its value),
NEUTRON DIFFRACTION AND STRUCTURE OF THE ADSORBED PHASES C4-29 so, I offer myself an answer to my question: a
variation of 100 or 101 graphite peak is indubitably a proof for registry
...
at least if this variation is greater than 10 % of the theoretical expectation for a registered lattice. In figure 2 the variation of the 100 peak should be, in a simple model, as high as the 002 variation. The former is 25 % of the latter up to coverage one.5" But I am not satisfied with that because the bare 100 graphite peak is three times smaller than it should be comparatively to the 002 ; the 101 is still smaller ; and two subpeaks can be indexed a s 102/3 and lO4I3, i.e. as if coming from rhombohedral graphite ! Our graphite has plenty of stacking faults along the c-axis. It is the ransom of lamellar structure which gives high homogeneous specific area.
I studied the following model :
Call A, B, C the three possible lateral location of carbon planes. The series ABAB
...
is the perfect hexagonal graphite, ABCABC is a perfect rhombohedral lattice. Let us imagine a growing processABA led to ABAB with the high probability 1 - r and to ABAC with the low probability r
(it measures the probability of apparition of rhombohedral parts).
ABC led to ABCA with probability u (persistence of rhombohedricity) greater than r
or to ABCB with probability 1 - u.
It is a Makhov process of second order (depending from two stages of events) and is relevant to linear algebra.
I have not yet finished the calculus but I conjecture that, if there are many stacking faults in one grain (a very essential hypothesis), then
1) the 10 1 intensity is that of a crystal of thickness L, smaller than its real thickness
2) the 10 1 variation is
2 fL f , cos y cos 6 where 6 = X , z again
but does not appear mixed with ~ P 6 1 : there is no place for shape change,
3) the adsorbed species on each face of a lamella scatter independently.
L is some coherence length for a type of stacking (hexagonalicity for 1 = 0 or 1, but rhombohedricity
2 4
for 1 =
3
or-)
and is of the order of 3A1 But L does not appear indeed in
-
.
I
This conjecture gives equal variation of 002 and 101 and an exactly opposite variation for 100 for the registered monolayer. For the bilayer the variation of 101 should be zero but the double for 100.
Looking back at figure 2, I would say as in [I] that registry persists up to the second layer, if not completely : one half of the bilayer is registered. Note that a non-registered second layer does not explain our results.
But I must emphasize the need for substrates crystallographically better ; it seems possible to use our present samples, but the 10 1 are weaker than in good crystals, and they are mixed in a mess ; would it be possible to get pure hexagonal papyex ? ~e same sort of things happen in heavy metal halides. Perhaps a very good candidate is alkali halides, which can be produced with a good specific area corresponding to the same type of crystallographic plane.
6" Now I must again call your attention to the fact that the angular position of a 2D superstructure ring does not appear at the Bragg position.
In the simple case (structure factor independent of X ) the displacement is well known to be A x = 2 n - x - 0'32 the line width being about
L ' 1
A x = 2 . r r x -
0.57 L ' the tail shape being given by
,
the maximum then being =I--
a x
For a very simple bilayer, the structure factor is 1
I - exp(ix, z ) giving
I
f1
= 1 -5
cos X, z (the average is on X , and - x,). It has a maximum atX, z = n- but due to the line shape, if a maximum in the ring appears, it is displaced towards the low
2 % - angles, say at about X; z =
-
.
3
The ratio of this maximum to the normal maximum is
if L = nd, X, = X, tg a it is
1
then n
-
36 ; tga-
- : the ratio is smaller than one. 3C4-30 C . MARTI AND P. THOREL
now This can easily be confused with a non-registered
2 0 lattice for a compression of about 4.5 %. and the peak has a principal maximum very much I a n a l ~ s e d previously our krypton b i l a ~ e r peak as
displaced : a bilayer epitaxial peak, because of registry in 100
ax
= I &
and 101 graphite peaks. After a more quantitative 2 X O analysis I was led t o only 50 % of registry. And my,& conclusion is that I cannot deduce anything from this
18 ' superstructure ring.
References
[I] MARTI, C., CROSET, B., THOREL, P., COULOMB, J. P., Surf.
Sci. 65 (1977) 532.
DISCUSSION
X. DWAL.
-
Over what coverage range do you observe a linear relationship between the intensity of the 002 peak and the amount of krypton, and to what precision is it actually linear ?C. MARTI.
-
Between 0.5 and 2.1 coverage. Except for low coverage, the linearity is much better than 10 %.Y . LARHER.
-
I have measured an adsorption isotherm of krypton on non-compressed exfoliated graphite (from Union Carbide, the FOAM substrate mentionned by Prof. Dash) at about 83 K, at coverages up to a monolayer. The uniformity of this substrate is comparable to that of the Thomy-Duval exfoliated graphite. It presents the advantage over grafoil that adsorption equilibrium is attained within very short time intervals, less than 10 min. For adsorption on grafoil in similar conditions, equilibrium is not attained within half an hour.J. A. VENABLES. - You find a reduction in a factor of
-
5 in the expected intensity due to an island structure. We find a factor of-
4 reduction for dislocated layers by Theed (these preceedings). This seems to be in remarkable agreement.C. M a n .
-
Note that the results we discussed for the intensity of superstructure ring are for a registered lattice.S. FAIN.
-
Would you please review your explanation of the changes in the over-layer superstructure peak that occur with increasing coverage ? In particular, how much of your sample has a compressed, out-of-registry Kr bilayer ?C. MARTI.
-
From coverage = 1 to coverage--
2, the superstructure peak shifts from 38" to 40°, but there is still 25 % of the previous intensity at 38". This shift may be attributed either to a compressed perfect bilayer, or to a polycristalline registered bilayer : the superstructure ring does not tell anything about that, at least with our weak accuracy. But from the variation of the (100) graphite peak and out-of-plane study of the superstructure ring, I infer an equal participation from both structures.R. KERN.
