Fusion of halo nucleus 6 He on 238 U : evidence for tennis-ball (bubble) structure of the core of the halo (even the giant-halo) nucleus
aa r X i v : . [ phy s i c s . g e n - ph ] S e p Fusion of halo nucleus H e on U : evidencefor tennis-ball (bubble) structure of the coreof the halo (even the giant-halo) nucleus Syed Afsar Abbas
Centre for Theoretical Physics, JMI University, New Delhi-110025, IndiaandJafar Sadiq Research Institute, T-1 AzimGreenHome, NewSirSyedNagar,Aligarh - 202002, Indiaemail: [email protected]
Abstract
In a decade-and-a-half old experiment, Raabe et al.(Nature 431 (2004)823), had studied fusion of an incoming beam of halo nucleus He with thetarget nucleus U . We extract a new interpretation of the experiment,different from the one that has been inferred so far. We show that their ex-periment is actually able to discriminate between the structures of the targetnucleus (behaving as standard nucleus with density distribution describedwith canonical RMS radius r = r A with r = 1.2 fm), and the ”core” of the halo nucleus, which surprisingly, does not follow the standard densitydistribution with the above RMS radius. In fact the core has the structure ofa tennis-ball (bubble) like nucleus, with a ”hole” at the centre of the densitydistribution. This novel interpretation of the fusion experiment provides anunambigous support to an almost two decades old model (Abbas, Mod. Phys.Lett. A 16 (2001) 755), of the halo nuclei. This Quantum Chromodyanamicsbased model, succeeds in identifyng all known halo nuclei and makes clear-cutand unique predictions for new and heavier halo nuclei. This model supportsthe existence of tennis-ball (bubble) like core, of even the giant-neutron halonuclei. This should prove beneficial to the experimentalists, to go forwardmore confidently, in their study of exotic nuclei. Keywords : Triton clustering, halo nuclei, giant-halo nuclei, exotic nu-clei, bubble nuclei, tennis-ball nuclei, fusion, Quantum Chromodynamics,colour confinement hypothesis, quark model
PACS : 21.10.Gv, 21.60.-n, 21.60.Pj, 21.85.+p1ight from the first completely unexpected appearance of the two-neutronhalo stucture in Li in 1985, the exotic nuclei have been focus of intenseinterest in nuclear physics. Hence, really not unexpectedly, the theoreticalpredictions of new halo nuclei, and their subsequent experimental searches,have had a tortuous history. Suffice to quote the example of O (withZ=8, N=16), predicted and expected to have a neutron halo structure, waseventually and again shockingly, found to be actually a strongly doubly-magicnucleus! Clearly there is something amiss in our theoretical understandingof the nuclear halo phernomenon. What is it? Here we show that the answerlies in an internally consistent study of a decade-and-a-half old nuclear fusionexperiment.Within the framework of the studies of neutron rich nuclei, it is of greatimportance to know whether fusion of nuclei involving weakly bound particlesis enhanced or not. We look at the experimental data in this connection, andtry to extract some basic structures which these ingenious experiments aretrying to point out to us.Raabe et al. [1] fired both He and 2-neutron halo nucleus He onto thetarget nucleus U . In agreement with an earlier experiment [2], they didobtain a much increased product with the above neutron halo beam. Theselarge yields due to fission, may be attributed to fusion of He on the target-nucleus. However, the same may be due to a transfer of neutrons from He ,first onto U , and thereafter a fission from this fattened nucleus. The ear-lier experiment [2] was unable to distinguish between these two possibilities.Raabe et al. [1] cleverly, were able to distinguish between the above twophysically possible occurrences. If a fission event was detected in coincidencewith an He , it was identified with the transfer case, while one without anaccompanying He , was attributed to complete fusion. Remarkably, theydemonstrated clearly, that the large fission yields do not result from fusionwith He , but from neutron transfer. Thus the ”core” of the projectile nu-cleus sees a larger nucleus which has ”eaten and digested” the halo neutrons[3]. So what is amazing here, is the new phenomenon of fusion only of theprojectile ”core” with a neutron-fattened target nucleus, which may itself bein an excited state [3].What is this experiment trying to tell us? Using Ockham’s razor, whatit is telling us is that though the 2-neutron halo is weakly bound with the core He in He , it is strongly attracted to the target nucleus. Hence min-imal requirement is that the ”core” of the halo, and the ”target” itself,should differ from each other, is some minimal significant manner. Now we2now that the density distribution of the standard nuclear medium is givenby the RMS radius r = r A with r = 1.2 fm). This is definitely true of thetarget nucleus U . And as the two neutrons (from the projectile nucleus)feel strong nuclear attraction with it, we would expect that the neutron fat-tened target nucleus would be a standard (though perhaps excited) nucleuswith density distribution conforming to the above standard nuclear RMS ra-dius. This means that, therefore, the ”core” of the projectile nucleusshould be different from the initial target nucleus , in some funda-mental manner. Note how this fusion experiment allows us to talk of thedensity distribution of the ”core” of the whole halo nucleus. The beauty ofthis experimental analysis by Raabe et al. [1], is that it is allowing us toseparate out the structure of the core-nucleus itself, which is sitting withinthe whole halo nucleus. But what does it mean?I would like to draw the reader’s attention to the density distribution aswas extracted from the classical electron scattering on nuclei. This is plottedin Fig. 1 here. These are well known figures. What is most significant isthat the central density of He is about 2.5 times higher than that of heavynuclei like bismuth, lead etc.. In fact, He has similar density distributionas that of He . Not only that, as determined from meticulous electron scat-tering experiemts, both He and He have a ”hole” (the central significantdepression near r →
0) at the centre. This is plotted in the inset of Fig. 1[4,5]. It is known that light nuclei are basically ”all surface”. Because of thecentral hole, this is more pronounced for these two nuclei. Note that as tomatter distribution of H , it is very much the same as that of He .Now to understand the fusion process of neutron halo beam, forced usto conclude logically, using the Ockham’s Razor based argument, that thecore of the projectile nucleus should be different from the target nucleus insome fundamental manner. This can now be extracted from a study of Fig.1. The much-bigger-in-magnitude ”surface-nature” of density distribution of He plus its hole at the centre, should be the reason of this ”fundamental-manner-difference”, between the target and the core of the projectile nucleus.Thus the density distribution of the core of the halo nucleus, here He , butin general any neutron halo nucleus, has a tennis-ball (bubble) like structure.This is the novel interpretation extracted from this halo fusion experiment,and which has not been understood in this manner, as of now. This newinterpretation provides an unambiguous suppport to an almost two-decades-old Quantum Chromodynamics based model predictions by the author [6].But what has been the standard interpretation of the density distribution3 -13 cm)00.511.522.5 C h a r g e d e n s it y , ρ (r) ( C / c m ) -13 cm)00.20.40.60.8 ρ (r) BiHe H x H xHe
Figure 1: Schematic density distributon of nuclei as determined by elec-tron scattering. Inset shows the same from a better experiment - showing amarked ”hole” at the centre. 4igure 2: Schematically drawn from Fig. 1 of Ref. [3]of the core of the halo nucleus in the above fusion experiment of He on U ? This is best explained by looking at Fig. 1 of ref. [3], which we displayhere schematically as Fig. 2. The density distribution on the left-inset,as that the stable nucleus, is clearly that of the target nucleus U . Theright-inset shows density distribution of the halo nucleus, here He , but ingeneral of any other putative halo incoming beam. Note the identical densitydistribution structure of the core of the halo nucleus with that of the left insetfigure. The only difference there, is due to the additional extension of thehalo neutrons. The same point is also emphasized by Tanihata [7], ”Thedensity distribution of the core is assumed to be the same as the bare groundstate nucleus”. Above, we saw that this assumption is actually quitewrong . However, this is the picture of the halo nuclear density distribution,which ubiquitously pervades all the theoretical model structures available atpresent; except , of course, the 2001 model as pointed above [6].Arguments originating from Quantum Chromodynamics, have allowed usto provide an understanding of the unusual phenomenon of the halo struc-ture as part of the overall nuclear phenomenon. The author thus arrivedat a model which could explain the existence of all known halo nuclei, andprovide clearcut predictions for many more halo nuclei - which were actually5iscovered later, and thus validating this model [6]. For the sake of bettercomprehension, now onto a brief exposition of that model [6].As per Quantum Chromodynamics, the physically observed hadrons cor-respond to the colour singlet representation. So for a baryon in 3 ⊗ ⊗ ⊕ ⊕ ⊕
10, of all the representations, it is only the singlet which providesobservable baryons. All the other colour representations seem to be spuriousand unnecessary. But this may not be true for multiquark systems, where8 ⊗ .... and hence in 6-quarks this colour singlet representation maybe present [8]. But in nuclear physics, we treat the sysytem as made uponly of individual colour singlet protons and neutrons, with the commonlyheld belief that no quarks would show up in low energy nuclear physics; andonly at sufficiently high energies, these may manifest themselves in terms ofa Quark-Gluon-Plasma. However, this naive view is not correct. We showhere, that even at low energies, quarks do place their identifiable imprints ina nucleus.Deuterons should have configurations where the two nucleons overlapstrongly in regions of size ≤ f m to form 6-quark bags. Why is deuteronsuch a big and loose system? The reason has to do with the structure ofthe 6-q bags formed, had the two nucleons overlapped strongly. As per thecolour confinement hypothesis of QCD, the 6-q wave function looks like: | q > = 1 √ | ⊗ > + 2 √ | ⊗ > (1)where ’1’ represents a 3-quark cluster which is singlet in colour space and ’8’represents the same as octet in colour space. Hence | ⊗ > is overall coloursinglet. This part is called the hidden colour because as per confinementideas of QCD, these octets cannot be separated out asymptotically, and somanifest themselves only within the 6-q colour-singlet system. Group theo-retically the above hidden colour part of 80% was determined by Matveev andSorba [9]. This large hidden colour part would prevent the two nucleons tocome together and overlap strongly [8,9]. Therefore the hidden colour wouldmanifest itself as short range repulsion in the region ≤ f m in deuteron. Sothe two nucleons though bound, stay considerably away from each other.For the ground state and low energy description of nucleons, we assumethat the group SU (2) F , with u- and d-quarks in the fundamental trepresen-tation, is what is required. Hence we assume that 9- and 12-quarks belong tothe totally antisymmetric representation of the group SU (12) ⊃ SU (4) SF ⊗ SU (3) C where SU (3) c is the QCD group and SU (4) SF ⊃ SU (2) F ⊗ SU (2) S SU (12) ⊃ SU (4) ⊗ SU (3), which becomes quitecomplicated for a large number of quarks [10]. The relevant group theoreti-cal techniques were developed by So and Strottman [11] and independentlyby Chen [12], Thereafter, the author, group theoretically determined [8,13]that the hidden colour component of the 9-q system is 97 .
6% while the 12-qsystem is 99 .
8% i.e. almost completely coloured.What is the relevance of these 9- and 12-quark configurations in nuclearphysics? The A=3,4 nuclei H , He and He have sizes of 1.7 fm, 1.88 fmand 1.674 fm respectively. Given the fact that each nucleon is itself a ratherdiffuse object, quite clearly in a size ≤ f m at the centre of these nuclei, the3 or 4 nucleons would overlap strongly. As the corresponding 9- and 12-qare predominantly hidden colour, there would be an effective repulsion atthe centre keeping the 3 or 4 nucleons away from the centre. Hence it waspredicted by the author [13] that there should be a hole at the centre of H , He and He . And indeed, this is what is found through electron scattering[4,5]. This is shown as inset in Fig. 1 here. Hence the hole, i.e. significantdepression in the central density of H , He and He , is a signature of quarksin this ground state property.This understanding of the hole, within the above QCD based arguments,leads us further to provide a consistent understanding of the halo structurepheneomenon as well [6].Due to the significantly higher density at the boundary and very smallat the centre, He is like a ”tennis-ball”. The word tennis-ball is used toemphasize the predominance of the ”surface-ness” property in the densitydistribution in the corresponding nuclei. Add two more neutrons to He tomake it He , a bound system. As the two neutrons approach the surface,they will bounce off. As the two neutrons are bound, these will ricochet onthe compact tennis-ball like nucleus. A neutron halo would be manifested asthese neutrons shall be kept significantly away from the core.How do we understand this effect? Macroscopically, we understand thisas: the density of the He core is high on the boundary, any extra neutronswould not be able to penetrate it, as this would entail much larger density on He surface than the system would allow dynamically. Microscopically, weunderstand this as: any penetration of extra neutron through the surface of He would necessarily imply the existence of five or six nucleons at the centre.7s already indicated, due to the relevant SU(12) group, only 12-quarks cansit in the s-state, which already is predominantly hidden colour. Any extraquarks hence would have to go to the p-orbital; and in the ground state ofnucleus, there is not sufficient energy to allow this. Hence, the two boundneutrons are consigned to stay outside the He boundary. In addition, if atany instant the two neutrons come close to each other while still being closeto the surface, locally the system would be like three nucleons overlapping,and which would look like a 9-q system. This too would be prevented dueto the local hidden colour repulsion. Hence as found experimentally, the twoneutrons in the halo would not come close to each other, resulting in theneutron halo in He [6].Going through the binding energy systematics of neutron rich nuclei, onenotices that as the number of α ’s increases along with the neutrons, each He + 2n pair tends to behave like a cluster of two H nuclei. Remember thatthough H is somewhat less strongly bound (ie. 8.48 MeV ), it is still verycompact (ie. 1.7 fm ), almost as compact as He (1.674 fm). In addition ittoo has a hole at the centre. Hence H is also a tennis-ball like nucleus.Hence Li which is He + H ; with two more neutrons it becomes Li ,which can be treated as made up of 3 H clusters. Let us treat tritons assitting at the vertices of an equilateral triangle. Because of the tennis-ball likestructure, the 3 H particles cannot come too close to each other. Firstly,the surface of the ball would prevent it from doing so, and secondly if somepart of the 3 H still overlap at the centre, it would look like a 6- or 9-quarksystem. Therein the hidden colour components would repel, ensuring thatthe 3 H clusters do not approach too closely at the centre. This too wouldimply a depression in the central density of Li . Be treated as 4 H sittingat the vertices of a regular tetrahedron would, for the reasons stated above,too have a central density depression. Thus Li and Be would appearmore surface-like or tennis-ball-like as well. Other evidences like the actualdecrease of radius as one goes from Be to Be [7, see Fig. 4] supports theview that it ( i.e. Be ) must be made up of four compact clusters of H .Next, the tennis-ball like nature of H and He has a unique structuralproperty which even He does not have. The nuclei H and He along withdeuteron, are the only known nuclei which have no excited state. Either thesenuclei are there, or are not there, as a single rigid entity. Due to quantummechanics, right upto their binding energy of 8.48 MeV, tritons would beimmune to any excitations; and thus their rigidity/elasticity of a tennis-balllike nature, would be more explicitly exhibited.8hat we are saying is, that the neutron rich nuclei, which are made up ofa number of tritons, each of which is tennis-ball like and compact, should becompact as well. These too would develop tennis-ball like property. This is,because the surface is itself made up of tennis-ball like clusters. Hence whenmore neutrons are added to this ball of triton clusters, these extra neutronswill ricochet on the surface. Hence we expect that one or two or more boundneutrons outside these compact clusters, would behave like neutron halos.Let Li be treated as made up of 3 H clusters and which should have holeat the centre. Therefore Li with Li + 2 n should be two neutron halo nuclei- which it is. So should Be be. It turns out that internal dynamics of Be is such that it is a cluster of α − t − t , with one extra neutron halo aroundit [6]. Similarly e.g. B, C, etc. would be neutron halo nuclei and so on.These specific predictions of 2001 have been confirmed in later years [7,14].Still more heavy nucleus N e , predicted to be one neutron-halo as perour model, found experimental confirmation in 2009 [14]. Still heavier nu-cleus M g , was found to be a one neutron halo nucleus in 2014 [15], onemore confirmation of our model. All this provides clear and unambiguossupport to our model. M g remains the heaviest halo nucleus discoveredso far. However our model predicts many more heavier halo nuclei; and theexperimentalists are urged to look for those.The proton halo nuclei can also be understood in the same manner. Here,another nucleus with a hole at the centre He (binding energy 7.7 MeV, size1.88 fm), would play a significant role.Thus all light neutron rich nuclei ZZ A Z are made up of Z H clusters.Recently we used the RMF model with the best interactions, to study thepredictions of this model [16]. That [16] too has confirmed our model [6].Note that we prefer to use the word tennis-ball, vis-a-vis the word bubble.Firstly, because while the word bubble, connotes ephemeral nature of theentity, the word tennis-ball signifies its rigidity/elasticity/stability. Next,the surface of the tennis-ball is more resilient, so that a neuton coming fromoutside, can at best, just ricochet on top. Thus the stabilty of the core,and the halo nature of the extra neutrons, is connoted more naturally in it.However we use both the words here, due to the fact that in the nuclearphysics literature, it is more common to come across the word bubble, ratherthan the word tennis-ball. However, in connection with the core of the halo,it is clearly better to use the word, tennis-ball like, rather than bubble like.This brings into focus the recently discovered [17] bubble nature of thenucleus Si . Our model predicts that Si = 14 H , be a bound system with9 hole at the centre. Hence the bubble nature of Si , as observed by them[17], should be just a remnant effect of the more strongly bound tennis-balllike state of Si . We urge the expeimentalists to look for it.We have treated all neutron rich nuclei ZZ A Z , as made up of Z number of H clusters. This is easily understood geometrically, as say 3-tritons sittingat the vertices of an equilateral traingle for li , and 4-tritons sitting at thevertices of a regular tetrahedron for Be . In Table 1, we show severalneutron rich nuclei which may be treated as being composed of n-clusters of H . We write the binding energy of these nuclei as [6], E B = 8 . n + Ck (2)where 8.48 MeV is the binding energy of H , with n-cluster of tritons formingk bonds and with C being inter-triton-bond energy. Here we take the samegeometric structure of clusters in these nuclei as conventionally done for α -clusters in A=4n nuclei. Thus this model seems to hold out well with inter-triton cluster bond energy of about 5.4 MeV, which continues to work foreven heavier neutron rich nuclei, e.g. for Si ,Table 1: Neutron-rich nuclei - inter-triton cluster bond energyNucleus n k E B -8.48n(MeV) C(MeV) Li 3 3 19.90 6.63 Be 4 6 34.73 5.79 B 5 9 45.79 5.09 C 6 12 64.78 5.40 N 7 16 79.43 4.96 O 8 19 100.64 5.30Hence the tennis-ball (bubble) like structure seems to hold good for evenheavy nuclei. We can understand this as per QCD as follows. As alreadyindicated, due to the relevant SU(12) group, only 12-quarks can sit in thes-state, which already is predominantly hidden colour. Hence if two tritonscome close to each other in a heavy triton-rich nucleus, then it would neces-sarily imply overlap of six nucleons at the centre of mass of these overlappingtritons. Any extra quarks above the number twelve, relevant for the groupSU(12), hence would have to go to the p-orbital; and in the ground stateof that nucleus, there is not sufficient energy to allow this. Hence, any two10ound tritons are consigned to stay away from each other. Hence, this wouldresult in the tennis-ball (bubble) like structure in say, Si [6].However, in this context, interesting is the concept of ”giant-halo-nucleus”,as proposed by Meng and Ring [18] for halo in Zr nuclei (Z=40, A >
A >