Polyhedrons at the nuclear structure
The space disposition of nucleons at the light nuclei is given with the help of polyhedrons. At many cases it is the collection of embedded each other regular polyhedrons with nucleons at vertexes. But at the construction one meets other polyhedrons also. We give veri cation of construction by extre...
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irk-123456789-1360632018-06-16T03:03:48Z Polyhedrons at the nuclear structure Aminov, Yu.A. Ядерная физика и элементарные частицы The space disposition of nucleons at the light nuclei is given with the help of polyhedrons. At many cases it is the collection of embedded each other regular polyhedrons with nucleons at vertexes. But at the construction one meets other polyhedrons also. We give veri cation of construction by extremal properties of regular polyhedrons. Розташування нуклонів легких ядрах у просторі описано за допомогою багатогранників. У значній кількості випадків це є множина вкладених одного в інший правильних багатогранників з нуклонами у вершинах. Але в цій конструкції теж можливо помітити і інші багатогранники. Мы приводимо обгрунтування конструкції за допомогою екстремальних властивостей багатогранників. Пространственное расположение нуклонов в легких ядрах описано с помощью многогранников. Во многих случаях это есть набор вложенных друг в друга правильных многогранников с нуклонами в вершинах. Но в этой конструкции тоже можно заметить и другие многогранники. Мы приводим обоснование конструкции с помощью экстремальных свойств правильных многогранников. 2017 Article Polyhedrons at the nuclear structure / Yu.A. Aminov // Вопросы атомной науки и техники. — 2017. — № 3. — С. 21-25. — Бібліогр.: 13 назв. — англ. 1562-6016 PACS: 21.60-n, 21.45.+v, 98.80.ft http://dspace.nbuv.gov.ua/handle/123456789/136063 en Вопросы атомной науки и техники Національний науковий центр «Харківський фізико-технічний інститут» НАН України |
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The space disposition of nucleons at the light nuclei is given with the help of polyhedrons. At many cases it is the collection of embedded each other regular polyhedrons with nucleons at vertexes. But at the construction one meets other polyhedrons also. We give veri cation of construction by extremal properties of regular polyhedrons. |
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Polyhedrons at the nuclear structure |
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Polyhedrons at the nuclear structure / Yu.A. Aminov // Вопросы атомной науки и техники. — 2017. — № 3. — С. 21-25. — Бібліогр.: 13 назв. — англ. |
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Вопросы атомной науки и техники |
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POLYHEDRONS AT THE NUCLEAR STRUCTURE
Yu.A.Aminov∗
B.Verkin Institute for Low Temperature Physics and Engineering of NAS of Ukraine, Kharkiv, Ukraine
(Received March 13, 2017)
The space disposition of nucleons at the light nuclei is given with the help of polyhedrons. At many cases it is the
collection of embedded each other regular polyhedrons with nucleons at vertexes. But at the construction one meets
other polyhedrons also. We give verification of construction by extremal properties of regular polyhedrons.
PACS: 21.60-n, 21.45.+v, 98.80.ft
1. INTRODUCTION
At the work one extends the consideration of
space disposition of nucleons at light nuclei which be-
gin at the work [1]. This question close connects with
regular and almost semi-regular polyhedrons. Under
construction we use data from the book [3, 4]. At
framework of unlimited nuclear matter from [3] we
consider the possibility of application of the polyhe-
dral theory to calculation of nuclear spectrum. The
construction of space nucleon disposition justified by
extremal properties of regular polyhedrons.
2. THE NUCLEAR CONSTRUCTIONS OF
Ca40 AND Fe56
Chemical elements Ca40 and Fe56 have special
places between other one. At the L.Aller book [2]
on the page 254, figure 21 or at the I.P.Selinov book
[4] on the Table III a the graphs of abundance of
the elements in the Solar system on these elements
have spade. It mean that these elements have larger
abundance between close to it. The nucleus of Ca40
is twice ”magic”: it has 20 protons and 20 neutrons.
On the nucleus of Fe56 the nuclear synthesis is almost
finished. (As the Table III a shows that synthesis is
going to Zn inclusively).
At the H.A.Bethe book [3] on the page 138 the
shell structure is given for nucleus of Ca40. Every
shell 0s and 1s has 2 protons, the shell 0p has 6
protons and the shell 0d has 10 protons. From an-
other side on the figure 25 of [3] the graphs of density
spreading of protons and separately of neutrons al-
most coincide with each other.Hence we can suppose
that the spreading of neutrons by shells is same as for
protons, t.e. the shells 0s and 1s have two neutrons,
the shell 0p has 6 neutrons and the shell 0d has 10
neutrons.Summing the numbers of protons and neu-
trons at every shell we obtain the sequence of shells
with the following numbers of nucleons: 4,4,12,20.
Supposing that nucleons at every shell lie at the
vertexes of regular polyhedrons (see chapter 6) we
obtain the following sequence of polyhedrons which
included each other with common center O:
tetrahedron ⊂ tetrahedron ⊂
icosahedron ⊂ dodecahedron. (1)
The another possible construction we obtain by union
of two first shells
cube ⊂ icosahedron ⊂ dodecahedron. (2)
Remark that icosahedron and dodecahedron are
dual figures. Therefore these two polyhedrons will
take the symmetric positions relatively each other if
the vertexes of icosahedron project from the center
O at the centers of the dodecahedron faces. This
precise position of icosahedron relatively dodecahe-
dron gives the possibility to indicate all connections
between vertexes of these figures: the icosahedron
vertex connects with those vertexes of dodecahedron,
which belong to face where lies the projection of the
considering icosahedron vertex.
At the construction (1) the tetrahedron positions
can be by two kinds: 1) the vertexes of inner tetra-
hedron project at vertexes of exterior tetrahedron, 2)
the vertexes of inner tetrahedron project at the faces
centers second one.
Remark that a tetrahedron is geometrical realiza-
tion of α- particle. But under some extremal condi-
tions as we suppose it can has form of two triangles
with common side. Under this assuming in the ar-
ticle [1] we give the constructions of a icosahedron
and dodecahedron by union of α- particles without
intersections.
Keeping in the mind the α-particle construction
we can write for Ca40 the following expansion
(2 + 3 + 5) · 4 = 40. (3)
Pass now to consideration of Fe56. This nucleus has
26 protons and 30 neutrons. Suppose that nucleus is
composed by union of α-particles and write the fol-
lowing expansion
(2 + 3 + 4 + 5) · 4 = 56. (4)
∗Corresponding author E-mail address: aminov@ilt.kharkov.ua
ISSN 1562-6016. PROBLEMS OF ATOMIC SCIENCE AND TECHNOLOGY, 2017, N3(109).
Series: Nuclear Physics Investigations (68), p.21-25.
21
The numbers in the round brackets denote the num-
bers of α-particles at shells which participate at con-
struction of the nucleus. Hence, at construction of
nucleus of Fe56 at compare with the nucleus of Ca40
appears one new shell with 4 α-particles. We put at
correspondence to that shell some polyhedron with
16 vertexes, which we denote Λ16. It is possible to
construct polyhedron Λ16 with 10 faces at forms of
squares and 8 isosceles triangles which close to equi-
lateral triangles. Therefore, we can say that Λ16 is
almost semi-regular. By analogy with (1) we give the
following construction for nucleus of Fe56
tetrahedron ⊂ tetrahedron ⊂ icosahedron ⊂
Λ16 ⊂ dodecahedron. (5)
But there is one important peculiarity at con-
structions of nucleus of Fe56. If that nucleus was
been forms precisely by all α-particles then the num-
ber of it protons will be 28. But the atom number of
this element is 26. Hence, at the process of construc-
tion two protons were converted into two neutrons.
It possible due to proton radiation of positron β+.
Consider the formation process of the nucleus of
Fe56. It is natural to suppose that it was been forms
non by exterior association α-particles with nucleus
of Ca40, because between icosahedron and dodecahe-
dron must lie the polyhedron Λ16. Therefore address
to previous chemical elements. Let us consider the
nucleus of 18Ar36. According to Table III a 18Ar
36
has larger abundance that another isotopes. We can
write the expansion for this element similar (3)
(2 + 3 + 4) · 4 = 36. (6)
The correspondent sequence of polyhedrons for
18Ar36 is following
tetrahedron ⊂ tetrahedron ⊂ icosahedron ⊂ Λ16.
Consequently we have the beginning of nucleus Fe56
construction. Later with the help of α-particle ad-
dition one forms exterior cover as dodecahedron. At
first we have the reaction
18Ar
36 + 32He4 →22 Ti48 + 2β+.
Exactly at this reaction 2 protons converted into 2
neutrons. The question why it take place at that
moment is remain open. Later we have
22Ti
48 +2 He4 →24 Cr52,
24Cr52 +2 He4 →26 Fe56.
Hence the nucleus of Fe56 can be form from the nu-
cleus of 18Ar36 by sequential associations with α-
particles.
At the conclusion of that chapter we can formu-
late the supposition that larger abundance of elements
Ca40 and Fe56 at comparison with nearest elements
justified by the circumstance that exterior shell of
their nuclei is regular polyhedron, i.e. dodecahedron.
3. THE SEQUENCE OF LIGHT NUCLEI
WITH EVEN NUMBERS OF PROTONS
Write the chain of nuclear constructions of chemi-
cal elements by sequential associations of α-particles.
One can suppose that the nucleus of nonstable isotope
of 4Be8 has form: tetrahedron ⊂ tetrahedron or
cube. Later
32He4 →6 C12,
icosahedron,
6C
12 +2 He4 →8 O16, (1 + 3) · 4 = 16,
tetrahedron ⊂ icosahedron,
8O
16 +2 He4 →10 Ne20, (2 + 3) · 4 = 20,
tetrahedron ⊂ tetrahedron ⊂ icosahedron
or dodecahedron,
10Ne20 +2 He4 →12 Mg24, (1 + 2 + 3) · 4 = 24,
tetrahedron ⊂ cube ⊂ icosahedron,
12Mg24 +2 He4 →14 Si28, (2 + 5) · 4 = 28,
cube ⊂ dodecahedron,
14Si
28 +2 He4 →16 S32, (3 + 5) · 4 = 32,
icosahedron ⊂ dodecahedron,
16S
32 +2 He4 →18 Ar36, (2 + 3 + 4) · 4 = 36.
Remark, that elements O16, Ne20, Si28 and S32 have
almost same abundance as Fe56. At that time Mg24
and Ar36 have same abundance as Ca40. It is inter-
esting to note on the Table III a from the I.P.Selinov
book [4], 1990, the abundance of Ar36 is considerable
larger that Ar40 , although at many early books the
abundance of Ar40 is on the first place.
Can we continue the sequence in the round brack-
ets in (4) ? Really , we can prolong it with indication
of corresponding elements
(1 + 2 + 3 + 4 + 5) · 4 = 60 → Ni60,
(1 + 2 + 3 + 4 + 5 + 6) · 4 = 84 → Kr84,
(2 + 3 + 4 + 5 + 6 + 7) · 4 = 108 → Ag108.
Later the sequence has prolongation, but some inde-
termination appears which connects with existence of
numerous isotopes.
4. THE SEQUENCE OF LIGHT NUCLEI
WITH ODD NUMBERS OF PROTONS
Remark, that at [1] we give representation for nu-
cleus of 3Li
7 at form of octahedron with one neu-
tron at the center.Such representation gives symmet-
ric disposition for all nucleons.
Consider the sequence 3Li
7 + n2He4. We have
3Li
7 +2 He4 →5 B11,
5B
11 +2 He4 →7 N15,
22
7N
15 +2 He4 →9 F 19,
9F
19 +2 He4 →11 Na23,
11Na23 +2 He4 →13 Al27,
13Al27 +2 He4 →15 P 31,
15P
31 +2 He4 →17 Cl35,
17Cl35 +2 He4 →19 K39,
19K
39 +2 He4 →21 Sc43.
At that sequence are represented isotopes which have
larger abundance, except elements N and Sc , for
which the largest abundance have 7N
14 and 21Sc
45.
The curve of the abundance of chemical elements has
form of saw with cogs directed up. The tops of cogs
correspond to elements with even atomic numbers.
And abysses correspond to elements with odd num-
bers. Write two lines:
5B
11,7 N
15,9 F
19,11 Na23,13 Al
27,15 P
31,17 Cl35;
6C
12,8 O
16,10 Ne20,12 Mg24,14 Si
28,16 S
32,18 Ar
36.
We see that nucleus at the first line has on one proton
less than nucleus on the second line, which stand un-
der it. The number of neutrons is same. Therefore,
we can suppose that from exterior shell of nucleus
from the first line is taken one proton. Under this
operation the stability of shell is disordered and the
abundance of elements falls. For example, the abun-
dance of C12 is enough high, but for near standing
element B11 it is exceptionally low.
However it is difficult to imaginer the removal re-
action of one proton from nucleus. Therefore let us
consider the formation of light nuclei by join of α-
particles.
We can represent the construction of nucleus of
3Li
7 at form of octahedron with one neutron at the
center. It is symmetric construction with one neutron
at distinguished place.
The addition to nucleus of 3Li
7 of two α- particles
gives the nucleus of 7N
15, to which corresponds the
following construction
n ⊂ octahedron ⊂ cube.
The addition of three α-particles gives the nucleus
of 9F
19 and the possible construction
n ⊂ octahedron ⊂ icosahedron.
The addition of 4 α-particles to nucleus of 9F
19
gives nucleus of 17Cl35 and the construction
n ⊂ octahedron ⊂ icosahedron
⊂ polyhedron with 16- vertexes.
Then the addition to nucleus of 9F
19 of five α-
particles gives the nucleus of 19K
39 with the following
construction
n ⊂ octahedron ⊂ icosahedron
⊂ dodecahedron.
We remark that the element 19K
39 has almost
same abundance as element 20Ca40.
Thus the the difference at this construction of nu-
clei with odd numbers of protons consists at using of
octahedron with neutron at the center at beginning
of construction.
At the book [2] on the p. 348 author write that
”the formation problem of visible quantity of light el-
ements Be,Li,B remains as principal unsolvable dif-
ficulty of modern theory of element origin”. On these
elements the curve of abundance falls deeply down,
that tails on singularity of its origin. At that time for
the next elements with odd numbers of protons in nu-
clei the corresponding points on the curve are close to
points of elements with even numbers of protons. The
second deep cavity of the curve lies between points of
Ca40 and Fe56.
5. POLYHEDRONS AND SPECTRUM OF
NUCLEI AT FRAMEWORK OF THE
NUCLEAR MATTER THEORY
The density of heavy nuclei ρ is approximately
constant and at the center ρ = 0, 17nucl.
fm3 . At the
H.Bothe’s book [3] one considers the theory on nu-
clear matter which infinitely stretched and has con-
stant density. For the volume energy Ev of such mat-
ter included at some domain with volume V will be
Ev = ρV .
Consider the polyhedron M as convex hull of all
nucleons centers of nucleus. By analogy we can con-
struct convex hull of inner shell also. If all faces of
M are triangles then by I.Sabitov’s theorem [6] the
volume V (M) is a root of some polynomial equation
V k + ak−1V
k−1 + ... = 0, (7)
where the coefficients ai are determined by squares
of edge lengths. If some face is non triangle , then we
divide it on triangles by diagonals of this face. The
equation (6) has some number of real roots V1, V2, ...
Assume that M with volume Vi transforms under
emanation or irradiation with preservation of edge
lengths into polyhedron with volume Vj . Then the
corresponding change of energy
Evi − Evj = ρ(Vi − Vj)
will give some spectral lines of emanation or new ex-
cited state of nucleus under irradiation. If the length
of edges change and the polyhedron M1 transforms
to M2 then we consider the difference
Ev(M1)− Ev(M2) = ρ(V1 − V2).
For example at the work [7] some cases of volume cal-
culations are given for octahedron type polyhedrons.
For this case the equation (6) is following
V 16 + a1(l
2)V 14 + ...+ a8(l
2) = 0,
where every summand has form ai(l
2)V 16−2i and l2
denotes the edge squares. The number of monomials
for calculation of ai can be very large ( order 1010).
But if M has many edges with equal lengths then
calculation will be possible.
23
6. EXTREMAL PROPERTIES OF
REGULAR POLYHEDRONS
It is well known that for figures in 3-dimensional
Euclidean space (convex or non convex) the following
inequality has place
F 3 ≥ 36πV 2,
where F is the surface area of the figure and V is its
volume.The equality can be for a ball and only at this
case.
There are many remarkable and difficult works de-
voted to extremal properties of regular polyhedrons
at some class of convex polyhedrons. Enough detailed
exposition of this question is given at the L.F.Tóth’s
book [8] . For some kind of polyhedrons one considers
the fraction
F 3
V 2
and looks for a polyhedron of given kind for which
this fraction is smallest. That polyhedron is called
best. Lhuilier proved that between polyhedrons of the
tetrahedron kind the best is tetrahedron.J. Steiner in
[9] formulated the supposition that between polyhe-
drons which have type of some regular polyhedron the
best is the regular polyhedron. He proved his suppo-
sition for kind of octahedron. Then M.Goldberg in
[10] for polyhedrons with n faces discovered and par-
tially proved the inequality
F 3
V 2
≥ 54(n− 2)tgωn(sin
2 ωn − 1),
where ωn = n
n−2 · π
6 and equality has place only for
regular polyhedrons which angles have 3 faces.
From here it follows that cube and dodecahedron
are best. The complete proof was been given by
L.F.Tóth at [11, 12]. The case of icosahedron don’t
have proof. But L.F.Toth told that probably for con-
vex polyhedron with n vertexes the following inequal-
ity has place
F 3
V 2
≥ 27
√
3
2
(n− 2)(3tg2ωn − 1),
and equality can be only for regular polyhedron with
triangle faces.
With volume of figure one can connect the vol-
ume energy Ev = ρV , where ρ is the density of nu-
clear matter, and the surface area F corresponds the
surface energy Ef , (see [13],p 48,133 and 136). By
formula of H.Bothe-C.F.Weizsäcker the nucleus en-
ergy is given by following expression
M(A,Z)c2 = Zmpc
2 +Nmnc
2 − avA+ afA
2/3+
ac
Z2
A1/3
+ aa
(N − Z)2
A
,
wheremp,mn are the proton and neutron masses, the
third member is volume energy, the fourth member
is surface energy, the fifth member is Coulomb’s en-
ergy and the last member is symmetry energy. Here
av, af , ac, aa are positive constants. For the light nu-
clei the last two members are very small, and we re-
ject it.
The equilibrium state of a nucleus (under constant
volume) determined by minimum of the free energy
of nucleus. It means that in the considering case sur-
face energy achieves minimum and the surface area
F also. Therefore for equilibrium state of nucleus the
fraction F 3
V 2 will have the minimal value among con-
sidering kind ( and with given number of vertexes)
of polyhedrons. By stated above it leads to regular
polyhedrons.
Under some natural suppositions the combinator
kind of a polyhedron can be reconstructed by the
number of vertexes. Remember that by well known
hypothesis every nucleon contains 3 quarks. Suppose
that every quark from nucleon can be connect with
one other nucleon of the nucleus. At our polyhedron
this connection corresponds to one edge which is go-
ing from the vertex. So from every vertex can go out
three edges.
Consider, for example, the polyhedron with 20
vertexes from every which are going 3 edges. Hence
the number of all edges is 3·20
2 = 30. By Euler formula
we obtain the number m of faces: 20 − 30 +m = 2.
Consequently, the number of faces is m = 12.Suppose
that all faces are same type polygons with l sides.
Calculating the number of all edges of the polyhe-
dron we obtain l·12
2 = 30, that is l = 5. So, we have
polyhedron with 20 vertexes, 30 edges and 12 faces
every which is a pentagon.
Draw the image of the polyhedron on a plane
begin from some pentagon. At first adjoin to it 5
pentagons and then again 5 pentagons keeping the
condition that from every vertex 3 sides are going
out. Exterior 5 sides of the last pentagons form the
boundary of exterior domain which is image of the
last face. Hence we have on the drawing image of
1+5+5+1 = 12 faces. But such image the dodecahe-
dron has also. Hence our polyhedron has combinator
kind of a dodecahedron.
Similar consideration can be given for polyhe-
drons with 4 or 8 vertexes.
References
1. Yu.A.Aminov. One hypothesis on the nuclear
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don, 1961.
3. H.A.Bethe. Theory of nuclear matter. Annual
Review of nuclear science. Paulo Alto, Califor-
nia, USA, v.21, p.93-244.
24
4. I.P.Selinov. The structure and systematization of
atomic nuclei. M: ”Nauka”, 1990 (in Russian).
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perticules, 1. Approche elementaire. Paris. Her-
mann.
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Þ.À.Àìèíîâ
Ïðîñòðàíñòâåííîå ðàñïîëîæåíèå íóêëîíîâ â ëåãêèõ ÿäðàõ îïèñàíî ñ ïîìîùüþ ìíîãîãðàííèêîâ. Âî
ìíîãèõ ñëó÷àÿõ ýòî åñòü íàáîð âëîæåííûõ äðóã â äðóãà ïðàâèëüíûõ ìíîãîãðàííèêîâ ñ íóêëîíàìè
â âåðøèíàõ. Íî â ýòîé êîíñòðóêöèè òîæå ìîæíî çàìåòèòü è äðóãèå ìíîãîãðàííèêè. Ìû ïðèâîäèì
îáîñíîâàíèå êîíñòðóêöèè ñ ïîìîùüþ ýêñòðåìàëüíûõ ñâîéñòâ ïðàâèëüíûõ ìíîãîãðàííèêîâ.
ÁÀÃÀÒÎÃÐÀÍÍÈÊÈ Â ÑÒÐÓÊÒÓÐI ÀÒÎÌÍÈÕ ßÄÅÐ
Þ.À.Àìiíîâ
Ðîçòàøóâàííÿ íóêëîíiâ ëåãêèõ ÿäðàõ ó ïðîñòîði îïèñàíî çà äîïîìîãîþ áàãàòîãðàííèêiâ. Ó çíà÷íié
êiëüêîñòi âèïàäêiâ öå ¹ ìíîæèíà âêëàäåíèõ îäíîãî â iíøèé ïðàâèëüíèõ áàãàòîãðàííèêiâ ç íóêëîíàìè
ó âåðøèíàõ. Àëå â öié êîíñòðóêöi¨ òåæ ìîæëèâî ïîìiòèòè i iíøi áàãàòîãðàííèêè. Ìû ïðèâîäèìî îá-
ãðóíòóâàííÿ êîíñòðóêöi¨ çà äîïîìîãîþ åêñòðåìàëüíèõ âëàñòèâîñòåé áàãàòîãðàííèêiâ.
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