Using liquid drop nuclear equation, to obtain quark total binding energy of nucleons in the nuclide. (Part I)

Transcription

1 Using liquid drop nuclear equation, to obtain quark total binding energy of nucleons in the nuclide (Part I) The General Science Journal (ISSN ) No E Using liquid drop nuclear equation, to obtain quark total binding energy of nucleons in the nuclide (Part I) R. Javahery 1, A. Jahanshir 2,3 1 Associated Professor, University of Tehran Sci. & Tech Park, Islamic Republic of Iran rjavaheri@ut.ac.ir 2 Ph.D, FKN University, Faculty of physics, Department of Theoretical Physics, Almaty, Kazakhstan 3 University of Tehran Sci. & Tech Park, Islamic Republic of Iran aresuj@gmail.com Abstract An approach and new vision in the primary structure around the nuclear from past up to now are an integral part of the modern physics. Precise nuclear binding energies and nuclear structure are basic data for various fields, in particular for nuclear physics. Since the beginning of nuclear physics, nuclear binding energy played a key role in the understanding of the structure of the atomic nucleus. Studying nuclei is interesting from the point of view of their multi-body systems and structures. Lighter and heavier nuclei contain at least two or a few hundred particles and thus their nuclear structure and overlapped between few-particle systems and the large number of particles involved in nuclide therefor design and determination of nuclear structural form based on binding energy is very important. From the Bethe-Weizsäcker formula (also called semi-empirical mass formula) inspired by the liquid-drop model the development of theories of the nucleus has gone a far way. The aim of this article is to explain nuclear structural form in the framework of the total energy consumed by the quark which causes binding energy between all nucleons in the nuclide. Keywords: nuclear mass; liquid drop model; Coulomb repulsion; quark binding energy.

2 Theoretical Framwork In this paper, it will be considered the total quark binding energy which is consumed for binding the nucleons together in the nuclide. In previous articles [1-4], it was intended that quark spends some value of energy as the binding energy (BE) of the nuclide, and spends another value of energy to prevent the protons' Coulomb repulsion from each other in a nuclide. But in this article the total energy caused by quark will be considered. Besides those two effects that were mentioned above, in this article two more phases are added. One is symmetry energy and the other energy which is related to odd and even number of protons and neutrons in the nuclide. For the quark reaction, the nucleons inside a nuclide bound together strongly and they arrange themselves symmetrically distributed and preferably spherical. The maximum number of nucleons bounding together inside the nuclide is six and for the nucleons in the surface, the minimum bounding is two. Starting with binding energy equation of nuclide: = ( )+ (1) In this equation since the binding energy of electrons in the atom A Z X are very small compared to nuclear binding energy, the binding energy of electrons are neglected. On the other hand Weizseker's formula [5, 6] which is semi empirical nuclear binding energy or liquid drop model of nucleons is: = ( ) ( ) (2) The sum of is actually the total energy consumed by the quark which causes binding energy between all nucleons in the nuclide. This quantity will be shown by (QBE) with parenthesis. (This quantity (QBE) is different from = + which was used in the [4]). The equation (2) could be written as it follows: ( )= + ( ) ( ) (3) In this equation each phase belong to some effect of quark for nuclear binding energy. The first point to be obtained is the value of (QBE) for each nuclide. For this purpose, in the above equation BE could be determined easily by calculation, for each nuclide. The values of the coefficients in the above equation are as it follows: = 0.711, = 23.7 Z even and N even = ( ) Z and N if one of them is odd and another is even = 0 Z and N if both are odd = ( ) These values of coefficients have been so precisely obtained by least square calculation [7] that the total value of quark binding energy (QBE) for each nuclide could be determined. There are 130 nuclides which are considered in this paper. The second point to be found is the number of bounding between the nucleons in each nuclide. In the appendix A of this paper and previous paper the diagram for nucleons' distribution in the layers of nuclide are given. The nucleons in each layer are bound to 2

3 the neighboring nucleons in the same layer. The total number of this kind of bounding is shown by ( )and besides that the two neighboring nucleons in the two neighboring layers are bound together. The sum of the bounding of these nucleons in the nuclide is shown by (,. ) Thus, the total numbers of nucleons bounding in the nuclide are = ( ) + (,, ) for more details refer to previous articles [1-4]. Using = ( ) which represents the total binding energy caused by quark divided by the total number of quark bounding between the nucleons in the nuclide, is more perceptible and meaningful than. Conclusion Using equation (3) the total quark bounding energy is obtained and by counting the total number of nucleons' binding in the nuclide is derived. There are some interesting results found from this research as it follows: what is the origin of nuclear binding energy? The binding energy of nuclide is caused by the quark attraction between two neighboring nucleons in the nuclide. The value of = ( ) for is greater than all other nuclides: it is 3.07MeV per nucleons. That is because the nuclide is very symmetric and spherical. That is one of the important results that have been obtained. There is not any nuclide with A=5 in the nature. That is because nuclide has the strongest bounding between each nucleons and it does not accept another one nucleon. The variation of ( ) versus A for some nuclide up to A=40 are shown in the following diagram: For the following nuclide the values of = ( ) are greater than any other neighboring nuclide:,,,,,,, The value of A and Z of this nuclide show that this is constructed by. In other words, these nuclides are composed from three to ten nuclides respectively. It is another interesting result that which is obtained. There is no stable nuclide with A=8 in the nature and the nuclides with A=8 are all radioactive and finally they decay to two nuclides. The reason is that these nuclides do not have 3

4 spherical shape. For most nuclides the values of = ( ), as it show in the diagrams (Apendix), is equal to 2.7MeV per nucleon, except for nuclides that are composed by nuclides and also for nuclides that have magic number of proton or neutron or have magic number of both N and Z. Acknowledgement The authors would like to express sincere appreciation to Mr. Khosrow Karimi, Mr. Hossein Javadi, Mr. Mohammad Mahammadi-Erbati, and Mr. Masood Sanati. References [1] R. Javahery, "A new explanation of quark causing binding energy between nucleons in the nuclide- I", The General Science Journal, ISSN: (GSJ), No E, 2008 [2] R. Javahery, "A new explanation of quark causing binding energy between nucleons in the nuclide-ii", The General Science Journal, ISSN: (GSJ), No E, 2008 [3] R. Javahery, "A new explanation of quark causing binding energy between nucleons in the nuclide- III ", The General Science Journal, ISSN: (GSJ), No E, 2008 [4] R. Javahery, "A new explanation of quark causing binding energy between nucleons in the nuclide- IV ", The General Science Journal, ISSN: (GSJ), No E, 2008 [5] E. G. Segre, Nuclei and particles, Benjamin Press, p , 1964 [6] P. Marmier, E. Sheldon, "Physics of nuclei and particle", Academic Press, p , 1969 [7] Rohlf, James William, "Liquid drop model of nucleus, Modern Physics, Wiley press, sec. 11.3,

5 Appendix BE=2.224 MeV NQB=1 Equation 3 can not be used for deuteron because it is only composed of two nucleons. A=4 mass= U BE= MeV (QBE)= MeV n(i)=.. m(ij)=. NQB=6 A=6 mass= U BE= MeV (QBE)= MeV n (i)=4 m (ij)=8 NQB=12 R[ ]=3,1117 5

6 A=7 mass= U BE= MeV (QBE)= MeV n(i)=4 m(ij)=10 NQB=14 a b R[ ]= A=9 mass= U BE= MeV (QBE)= MeV n(i)=6 m(ij)=16 NQB=22 R[ ]= A=10 mass= U BE= MeV (QBE)= MeV n(i)=9 m(ij)=16 NQB=25 6

7 R[ ]= A=11 mass= U BE= MeV (QBE)= MeV n(i)=10 m(ij)=20 NQB=29 R[ ]= A=12 mass= U BE= MeV (QBE)= MeV n(i)=9 m(ij)=20 NQB=29 R[ ]=

8 A=13 mass= u BE= MeV (QBE)= MeV n(i)=10 m(ij)=24 NQB=34 A=14 mass= u BE= MeV (QBE)= MeV R[ ]=

9 n(i)=12 m(ij)=32 NQB=44 R[ ]= A=16 mass= u BE= MeV (QBE)= MeV A=15 R[ ]= mass= U BE= MeV (QBE)= MeV n(i)=12 m(ij)=32 NQB=44 9

10 n(i)=16 m(ij)=32 NQB=48 R[ ]= A=18 mass= u BE= MeV (QBE)= MeV n(i)=16 m(ij)=36 NQB=52 R[ ]=

11 A=20 mass= U BE= MeV (QBE)= MeV n(i)=23 m(ij)=40 NQB=63 R[ ]= A=23 mass= U BE= MeV (QBE)= MeV n(i)=24 m(ij)=52 NQB=76 R[ ]=

12 A=28 mass= U BE= MeV (QBE)= MeV n(i)=33 A=24 m(ij)=64 NQB=97 R[ ]= mass= U BE= MeV (QBE)= MeV n(i)=31 m(ij)=48 NQB=79 R[ ]=

13 A=31 mass= U BE= MeV (QBE)= MeV n(i)=36 m(ij)=72 NQB=108 ]= A=32 mass= U BE= MeV (QBE)= MeV n(i)=36 m(ij)=76 NQB=112 R[ ]= A=36 mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV 13

14 n(i)=44 m(ij)=84 NQB=128 R[ ]= R[ ]= A=37 mass= U BE= MeV (QBE)= MeV n(i)=40 m(ij)=92 NQB=132 R[ ]= A=39 mass= U BE= MeV (QBE)= MeV n(i)=46 m(ij)=96 NQB=136 14

15 R[ ]= A=40 mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=47 m(ij)=96 NQB=143 R[ ]= a b b c R[ ]= A=42 15

16 mass= U BE= MeV (QBE)= MeV n(i)=56 m(ij)=104 NQB=160 R[ ]= A=43 mass= U BE= MeV (QBE)= MeV n(i)=48 m(ij)=108 NQB=152 R[ ]= A=44 mass= U BE= MeV (QBE)= MeV n(i)=56 m(ij)=112 NQB= R[ ]=2.7606

17 A=46 mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=49 m(ij)=120 NQB=169 R[ ]= R[ ]= A= mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV 17

18 n(i)=55 m(ij)=132 NQB=187 R[ ]= R[ ]= A=50 mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=55 m(ij)=138 NQB=193 R[ ]= R[ ]= R[ ]=

19 A=58 mass= U BE= MeV (QBE)= MeV mass= U BE= Mev (QBE)= MeV n(i)=73 m(ij)=156 NQB=229 R[ ]= R[ ]= A=60 mass= U BE= MeV (QBE)= MeV n(i)=74 m(ij)=160 NQB=234 19

20 R[ ]= A=61 mass= U BE= MeV (QBE)= MeV n(i)=72 m(ij)=168 NQB=240 R[ ]= A=66 mass= U BE= MeV (QBE)= MeV n(i)=76 m(ij)=184 NQB= R[ ]=2.8530

21 A=70 mass= U BE= MeV (QBE)= MeV n(i)=97 m(ij)=184 NQB=289 R[ ]= A=71 mass= U BE= MeV (QBE)= MeV n(i)=104 m(ij)=192 NQB=296 21

22 R[ ]= A=73 mass= U BE= MeV (QBE)= MeV n(i)=100 m(ij)=204 NQB=304 R[ ]= Having the diagram of Nuclide A=73, by adding one nucleon on each side of its series on obtained diagram for nuclide A=75: 22

23 A=75 R[ ]= mass= U BE= MeV (QBE)= MeV n(i)=100 m(ij)=212 NQB=312 A=74 mass= U BE= MeV (QBE)= MeV mass= u BE= MeV (QBE)= MeV n(i)=105 m(ij)=200 NQB=305 [ ]= [ ]=

24 A=76 mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=100 m(ij)=212 NQB=312 R[ ]= R[ ]=

25 A=77 mass= U BE= MeV (QBE)= MeV n(i)=112 m(ij)=210 NQB=322 [ ]= A=83 mass= U BE= MeV (QBE)= MeV n(i)=112 m(ij)=232 NQB=344 25

26 A=84 mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=105 m(ij)=236 NQB=341 R[ ]= R[ ]= R[ ]= A=85 mass= U BE= MeV (QBE)= MeV n(i)=112 m(ij)=240 NQB=352 R[ ]=

27 A=89 mass= U BE= MeV (QBE)= MeV n(i)=118 m(ij)=256 NQB=374 R[ ]=

28 R[ ]= A=95 mass= U BE= MeV (QBE)= MeV n(i)=122 m(ij)=272 NQB=394 28

29 29

30 A=99 mass= U BE= MeV (QBE)= MeV n(i)=124 m(ij)=284 NQB=408 R[ ]=

31 A=101 mass= U BE= MeV (QBE)= MeV n(i)=124 m(ij)=288 NQB=412 R[ ]=

32 A=113 mass= u BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=166 m(ij)=328 NQB=494 R[ ]= R[ ]= A=115 mass= u BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=162 m(ij)=344 NQB=506 R[ ]= R[ ]=

33 33

34 A=116 mass= u BE= MeV (QBE)= MeV mass= u BE= MeV (QBE)= MeV n(i)=156 m(ij)=336 NQB=492 R[ ]= R[ ]=

35 A=117 mass= u BE= MeV (QBE)= MeV n(i)=162 m(ij)=354 NQB=516 R[ ]= A=119 mass= u BE= MeV (QBE)= MeV 35

36 n(i)=168 m(ij)=356 NQB=524 R[ ]= Having the diagram of Nuclide A=119, by adding one nucleon on each side of its series on obtained diagram for nuclide A=121 g f : A=121 mass= u BE= MeV (QBE)= MeV n(i)=168 m(ij)=349 NQB=520 A=120 mass= U BE= MeV (QBE)= MeV R[ ]= mass= U BE= MeV (QBE)= MeV n(i)=180 m(ij)=351 NQB=532 R[ ]=

37 R[ ]= A=126 mass= u BE= MeV (QBE)= MeV mass= u BE= MeV (QBE)= MeV n(i)=180 m(ij)=384 NQB=564 R[ ]= R[ ]= A=128 mass= u BE= MeV (QBE)= MeV 37

38 mass= u BE= MeV (QBE)= MeV n(i)=188 m(ij)=384 NQB=572 R[ ]= R[ ]=

39 A=130 mass= u BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=188 m(ij)=392 NQB=580 R[ ]= R[ ]= R[ ]=

40 A=133 mass= U BE= MeV (QBE)= MeV n(i)=194 m(ij)=398 NQB=592 R[ ]= A=136 mass= u BE= MeV (QBE)= MeV 40

41 mass= u BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=209 m(ij)=408 NQB=617 R[ ]= R[ ]= R[ ]= A=137 mass= u BE= MeV (QBE)= MeV n(i)=196 m(ij)=404 NQB=600 R[ ]=

42 42

43 A=138 mass= u BE= MeV (QBE)= MeV mass= u BE= MeV (QBE)= MeV mass= u BE= MeV (QBE)= MeV n(i)=199 m(ij)=416 NQB=615 R[ ]= R[ ]= R[ ]= A=142 43

44 mass= u BE= MeV (QBE)= MeV mass= u BE= MeV (QBE)= MeV n(i)=201 m(ij)=432 NQB=633 R[ ]= R[ ]= Having the diagram of Nuclide A=142, by adding one nucleon on each side of its series on obtained diagram for nuclide A=144: f A=144 g mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV n(i)=201 m(ij)=436 NQB=637 R[ ]= R[ ]= A=148 mass= U BE= MeV (QBE)= MeV mass= U BE= MeV (QBE)= MeV 44

45 n(i)= m(ij)= NQB= R[ ]= R[ ]= R[ ]= R[ ]= A=150 mass= u BE= MeV (QBE)= MeV mass= u BE= MeV (QBE)= MeV n(i)=233 m(ij)=448 NQB=681 45

46 R[ ]= R[ ]= A=164 mass= u mass= U BE= MeV BE= MeV (QBE)= MeV (QBE)= MeV n(i)=241 m(ij)=508 NQB=749 46

47 A=170 mass= U mass= u BE= MeV BE= MeV (QBE)= MeV (QBE)= MeV n(i)=246 m(ij)=520 NQB=765 R[ ]= R[ ]=

48 A=172 mass= u BE= MeV (QBE)= MeV n(i)=258 m(ij)=528 NQB=786 R[ ]= A=174 mass= u mass= u BE= MeV BE= MeV (QBE)= MeV (QBE)= MeV n(i)=248 m(ij)=528 NQB=776 48

49 R[ ]= R[ ]= A=176 mass= u BE=1382,5268MeV (QBE)=2168,9935 MeV 49

50 n(i)=248 m(ij)=532 NQB=780 R[ ]=2,7 A=177 mass= u BE= MeV (QBE)= MeV n(i)=260 m(ij)=552 NQB=812 50

51 R[ ]= A=181 mass= u BE=1414,1840MeV Having the diagram of Nuclide A=177, by adding one nucleon on each side of its series on obtained diagram for nuclide A=179: A=179 mass= u BE= MeV (QBE)= MeV n(i)=236 m(ij)=560 NQB=796 R[ ]=

52 A=181 mass=180, u BE=1414,1840MeV (QBE)=2235,9809 MeV n(i)=268 m(ij)=560 NQB=828 R[ ]= A=182 mass= u BE= MeV (QBE)= MeV n(i)=264 m(ij)=568 NQB=832 52

53 R[ ]= Having the diagram of Nuclide A=182, by adding one nucleon on each side of its series on obtained diagram for nuclide A=184: f g A=184 mass= u mass= u BE= MeV BE=1434,0878MeV (QBE)= MeV (QBE)=2275,4756 MeV n(i)=264 m(ij)=572 NQB=836 R[ ]= R[ ]=

54 A=191 mass= u BE= MeV (QBE)= MeV n(i)=284 m(ij)=600 NQB=884 R[ ]=

55 A=193 mass= u BE= MeV (QBE)= MeV n(i)=300 m(ij)=600 NQB=900 R[ ]= A=195 mass= u BE= MeV (QBE)= MeV 55

56 n(i)=300 m(ij)=608 NQB=908 R[ ]= A=197 mass= u BE= MeV (QBE)= MeV n(i)=306 m(ij)=592 NQB=898 R[ ]= A=198 mass= u BE= MeV (QBE)= MeV 56

57 mass= u BE= MeV (QBE)= MeV n(i)=302 m(ij)=624 NQB=926 R[ ]= R[ ]= A=201 mass= u BE= MeV (QBE)= MeV n(i)=316 m(ij)=632 NQB=948 57

58 R[ ]= A=202 mass= u BE= MeV (QBE)= MeV n(i)=304 m(ij)=632 NQB=936 R[ ]=

59 Having the diagram of Nuclide A=202, by adding one nucleon on each side of its series on obtained diagram for nuclide A=204: f g A=204 mass= U mass= U BE= MeV BE= MeV (QBE)= MeV (QBE)= MeV n(i)=304 m(ij)=636 NQB=940 R[ ]= R[ ]= A=208 mass= u BE= MeV (QBE)= MeV n(i)=321 m(ij)=656 NQB=977 R[ ]=

60 A=235 * mass= u BE= MeV (QBE)= MeV n(i)=366 m(ij)=736 NQB=1102 R[ ]=