Review
J Phys Astr, Volume: 6( 1)

Model of Nuclear

*Correspondence:
Buchakchiiskiy FF, University of St.Gagarin Severodonetsk, Ukraine, Tel: 33662922853; E-mail: [email protected]

Received: October 03, 2017; Accepted: March 21, 2018; Published: March 28, 2018

Citation: Buchakchiiskiy FF. Model of Nuclear. J Phys Astron. 2018; 6(1):141.

Abstract

It is assumed that basis for formutation of all nuclei is a nucleus of helium. All subsequent nuclei of elements consist of chain of nuclei of helium. They are constrained inter se to binding energy equal energy of separation of nucleus of helium. The chain of nuclei of helium coagulates in a ball.

Keywords

Nuclonous; Helium; Electrolysis; Calcium; Electrode

Introduction

It is known that the number of nuclonous in a nuclei is multiple four plays a large role at determination of properties of nucleus.

Foremost at the nuclei of containing an even number protons and neutrons spin of nucleus equal to the zero. To this group of elements belong helium equation ,carbon equation , oxygenequation. Because the nucleus of heliumequation is the simplest, then it serves as basis for the constriction of all anther nuclei. We will consider the chart of formation of nuclei. We will bild a chart for calcium. For this purpose we will define the chain of values of energy of separation of nucleus of helium equation [1-10].

Experimental

Energy of separation equation

equation

equation

equation

equation

equation

equation

equation

equation

Let us now construct a chain of ten helium nuclei to obtain core the calcium equation

Results and Discussion

In table (picture 1) is presented the variant all even nuclei, since the nucleus of helium equation , ending uraniumequation

Gonrned, used for the constraction of this table following:

Elements consist of chain of nuclei of helium or isotopes of helium, that enter into co-operation. The value of this cooperation is equal to energy of separation of nucleus of helium or isotope of helium from an element.

This energy links the chain of nuclei.

The constructon of nuclei with even number is begun with nucleus of helium, with an odd number a construction is begun with deuterium equation or tritiumequation Therefore there are two chains of nuclei.

In lines the numbers against the name elements stand mass of isotopes.

For example:

Element equation appears from a nucleus of helium addition of another nucleus,

equation

or addition of isotope of helium,

equation

For carbon is brought four isotope equation

The chart of formation of there isotopes is such,

equation

\

equation

equation

Another example of formation of isotopes of calcium,

equation

equation

equation

equation

equation

equation

equation

equation

On a picture 2 a chain is presented from the isotope of helium equation to uraniumequation ,This chain is also distinguished in the table of formation of nucleus (picture 1). Criterion of choice of method of formation of nucleus is a value of energy of separation of element. We find the value of energy of separation (ESα) for all elements from beryllium equation to uranium. If this value is positive, then of nucleus of helium equation becomes formative, if this value becomes negative then the isotope of helium equation becomes formative.

In a right side Figures 1-4 the value of energy of separation (ESα) equation and isotope heliumequation is presented from Coo responding clements (Table 1).

physics-astronomy

Figure 1:equation 9 ESα(1 – 9)=10 * 28.296-+(-0.092)+7.367+7.162+4.725+9.322+9.985+6.947+6.641+7.050=342.067 MeV.

physics-astronomy-helium-coagulate

Figure 2: This ten nuclei of helium coagulate in a ball.

physics-astronomy-formative-element

Figure 3: For elements from Z=2 to Z=50 this formative element is α-nucleus as and further from as and further from Z=52 to Z=92 this formative element is the isotope of helium equation, because using equation are caused negative value of energy of separation. On a picture 3 the chart equationof dependence of energy of separation is presented from Z.

physics-astronomy-chain-nucleus

Figure 4: Division of nucleus of uranium on two parts making 2/3 basic nucleus devides the break of chain near a nucleus.

A Element Z ES α ES (α+2n) ES {a+1n} ES {a+3n}
5 He 2        
9 Be 4 2.468      
14 C 6 12.011   19.716  
18 O 8 6.226      
22 Ne 10 9.668      
26 Mg 12 10.517      
30 Si 14 10.65      
36 S 16 9.009 23,820    
40 Ar 18 6.801      
46 Ca 20 11.137 25,960    
50 Ti 22 10.717      
54 Cr 24 7.931      
58 Fe 26 7.65      
65 Ni 28 8.634     29.163
69 Zn 30 5.757      
73 Ge 32 5.304      
77 Se 34 5.72      
81 Kr 36 5.52      
85 Sr 38 6.833      
89 Zr 40 6.191      
93 Mo 42 4.301      
97 Ru 44 1.734      
101 Pd 46 1.741      
105 Cd 48 1.357      
109 Sn 50 0.734      
115 Te 52 -1.46 17.11    
121 Xe 54 -0.199 18.007    
127 Ba 56 -0.009 17.938    
133 Ce 58 -0.217 17.203    
139 Nd 60 -0.209 17.163    
145 Sm 62 -1.115 16.262    
151 Gd 64 -2.653 11.132    
157 Dy 66 -1.036 13.293    
163 Er 68 -1.574 13.241    
169 Yb 70 -1.733 12.791    
175 Hf 72 -2.404 11.708    
181 W 74 2.211 11.266    
187 Os 76 -2.724 10.56    
193 Pt 78 -2.083 10.854    
199 Hg 80 -0.824 12.665    
205 Pb 82 -1.465 6.815    
211 Po 84 -7.534 6.258    
217 Rn 86 -7.887 1.497    
223 Ra 88 -5,973 3.92    
229 Th 90 -5.167 5.253    
235 U 92 -4,678 6.26    

Table 1: In a right side pictare 2 the value of energy of separation (ESα) and isotope helium is presented from COO responding clements.

Conclusion

Chain of nuclei convolves in a ball. Similar below. Division of nucleus of uranium on two parts making 2/3 basic nucleus devides the break of chain near a nucleus where ESα goes across through a Zero. Placing of the superfluous no included in a?-nuclei neutrons (n=A-2Z).

Appendix

equation

To find a radius, where potential energy of coulomb forces is equal to 10 MeV.

q=l=1,6*10-19 Cl

1 MeV=1,6*10-13 gl

k=9*109

Wc=10 Mev=1.6 *10-12 gl

1 MeV=1,6*10-13 gl

Z1=2 Z2=50|

equation

equation

References