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Pode abrir um tópico sobre isso... Se é Lagrangeano ou não. Aqui o tópico é QRT.
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tks
Agnostico
a seguir estao os calculos para o momento quadrupolar eletrico dos isotopos 8O18 excitados, conforme está na pagina 138 a 142 do livro Quantum Ring Theory.
Confira os calculos:2. ELECTRIC QUADRUPOLE MOMENT OF 8O17
The nuclei with A=odd have Q(b)¹0 even when they are stable (Ex = 0). From the nuclear experiments, we know that all nuclei have approximately the same diameter R = 10F , and we will calculate Q(b) by starting up from this information. The neutron (fermion No. 17) takes in the 8O17 the position shown in the Fig. 2.1-B. In the Fig. 2.1-A it’s shown the center of mass, which coincides with the gravitational center G.C. of the [hexagon+2He4] , but without considering the neutron No. 17. With the addition of the neutron No. 17, the new G.C. is shown in the Fig. 2.1-B.
The nucleus 8O17 turns about many chaotic axis. Suppose that it is turning about many of these axis, and it never turns about the z-axis. In this case we would have Q(b) = 0. But there is no reason why it cannot turn also about the z-axis, and in reality it does it. So, when it turns about the z-axis, shown in the Fig. 2.2-A, the geometrical center of the [hexagon+2He4+neutron-No.17] shall turn about the G.C. point, with radius “x” (see also Fig. 2.3-A) . If the position of the hexagon with regard to the 2He4 should be inflexible , the radius x would be:
From the Fig. 2.4: R.cos30 = R - a , R = 10F ==> a = 1,34F ( 2.1)
From the Fig. 2.2:
• There are 4 protons and 4 neutrons in the CD-axis => momentum = 8.x
• There are 2 protons and 2 neutrons in the AB-axis => momentum = 4.[ (R - a)+ x ]
• There are 2 protons and 3 neutrons in the EF-axis ==> momentum = 5.[ R - (x+a) ]
Thus : 8x + 4.[ R - a + x ] = 5.[ R - ( x + a ) ] (2.2)
x = 0,509F (2.3)
Therefore, the nucleus 2He4 would be turning about the G.C. point, with a radius x = 0,509F , but only if the position of the hexagon should be inflexible with regard to the 2He4 in the center of the nucleus 8O17. However the hexagon is not tied rigidly to the 2He4, because it is tied by a flux (created by the 2He4) . Therefore the hexagon has a displacement with regard to the point determined by x = 0,509F . Let’s call x’ this displacement, and let’s calculate it. Now the 2He4 does not have influence in the equilibrium, and therefore in the CD-axis of the Fig. 2.2-A only 2 protons and 2 neutrons will be considered:
4x’ + 4.[ R - a + x’ ] = 5.[ R - ( x’ + a ) ] (2.4)
x’= 0,67F (2.5)
CALCULATION OF r’ : The Fig. 2.3 shows the expressions for r’ in dependency upon the rigidity of the connection of 2He4 into the hexagon, as follows:
FIG. 2.3-A : the 2He4 is tied perfectly rigid to the center of the hexagon: r’=x .
FIG. 2.3-B : the 2He4 does not have influence in the motion due to the not rigid connection: r’=x+x’
But the real situation is not like that shown in the Figure 2.3-B , because there is a small participation of the 2H4 in the motion with radius x’ (since the conection is not 100% flexible).
Therefore, the real situation is: x <r’<x+x’, and it is very reasonable to assume the Fig. 2.3-C:
r’ = ( 2x + x’ )/2 ( 2.6 )
and thus we have r’ = ( 2x0,509 + 0,67 )/2 = 0,844F ( 2.7 )
Of course the oscillation shown in the Fig. 2.4 will be responsible for Q(b)< 0 , and the protons of the central 2He4 have no participation in the value of Q(b). By this reason:
• only 6 protons will be responsible per the value of Q(b): ( Z = 6 protons ) (2.8)
• We know that Q(b) = ò r.[3.z2 - (x2 + y2 + z2 )].dt
• But z = 0 and x2 + y2 = (r’)2 (2.9)
• Then Q(b) = 6. [ - (0,844F)2 ] = -4,27x10-30m2 (2.10)
and from experiments, Q(b) = -0,026 barns (1b=10-30cm2) = -2,6x10-30m2
But the calculation is not ready yet. Obviously the difference D=4,27-2,6=1,67 is due to the fact that it’s not exact the relation r’=(2x+x’)/2 . It’s necessary to consider the radius r’=(2x+k.x’)/2 . Lets calculate “k”:
From the experimental result, Q(b) = -2,6x10-30m2, then the radius r’ must be:
-6.(r’)2 = -2,6x10-30 m2 , ==> r’ = 0,658F.
Thus, 0,658 = (2x0,509 + 0,67.k )/2 ==> k = 0,445 (2.11)
We will apply the value k = 0,445 in the calculation of Q(b) for the 8O18.
We chose to calculate the Q(b) of 8O17 because only the oxygen supplies isotopes with eccentric distribution of nucleons 1H2 but without distortions in the hexagonal floor. If we take another nuclei , where the second hex. floor is beginning , for example the 9F18, the introduction of one more nucleon 1H2 (to be added to the structure of the 8O16), produces a distortion in the hexagonal floor, and the calculation of Q(b) for 9F18 is not practicable. This distortion is shown in the Fig. 2.5: the nucleon 1H2 added in the second incomplete floor has repulsion with the first complete floor, and so the first floor is distorted.
3. Q(b) OF EXCITED ISOTOPES OF THE OXYGEN
Table I exhibits the nuclear properties of some oxygen's isotopes. The unity of half-time T1/2 are ns=10-9s and ps=10-12s. For the understanding of the Table I , first of all we must understand three behavior of the neutrons when they are introduced within a nucleus:
Fig. 3.1: It shows the distribution of spins of nucleons 1H2 in the structure of 8O16. By looking at, for example, the nucleons 3 and 4, we note that their spins have contrary directions. From this distribution it is minimum the conflict between repulsions and attractions.
Fig. 3.2: 1- When one neutron is introduced into the 8O16, performing the 8O17, the nucleon 3 is constrained to change its spin (Fig. 3.3), in order to avoid the conflict of magnetic fluxes (Fig. 3.2-B) into the stable 8O17.
2- When two neutrons are introduced, they can take (or not) two opposites places, like shown in the Fig. 3.4: it depends on the excitation of the 8O18
Then now let us try to understand the Table I.
• 8O16 with Ex = 6130 (see Fig. 3.1) : Although the nucleus is excited, it has Q(b) = 0. And we can understand why. Indeed, the energy of excitation Ex only increases the pulsation of the nucleus. It does not changes the distribution of loads. Therefore the 8O16 behaves like a perfect boson, at least concerning to its symmetry: its Q(b) continues zero. The energy of excitation is very high, and it changes the spin of one of the protons of the central 2He4 of the 8O16. As we will see in the paper No. 18, within an excited 2He4 one of the protons can change its spin, in spite of the direction of the gravitational flux into the 2He4 does not change, because what actually changes is the electron's spin(*). In the whole Raghavan's table only the following nuclei are submitted to energy highest than 6000keV: 6C14 (spin 3-) ,12Mg24 , 46Pd96, and isotopes of 86Rn . One of the protons of 2He4 changes its spin only into the 8O16 and into the 6C14 (in the Raghavan’s table).
• 8O17: With the addition of one neutron to the 8O16, it loses its perfect distribution of nucleons. It has Q(b)¹0 even without excitation. As it is impossible to add one neutron in the structure of 8O16 of the Fig. 3.1 without having a conflict between magnetic fluxes (see Fig. 3.2 and 3.3), then the introduction of one additional neutron performing the stable 8O17 constrains the nucleon No. 3 to get a spin up, in order to avoid the conflict of magnetic fluxes. The total spin of 8O17 is i = 4 - 2 + 1/2 = 5/2+ .
• 8O18 : The two neutrons have contrary spins, and the structure tries to have the maximum symmetry, like it is shown in the Fig. 3.4.
a) 8O18 with Ex = 0 and with Ex = 3555: The point G.C. is not eccentric, the reason why Q(b)=0.
b) 8O18 with Ex = 1982: The 8O18 can have two structures, as shown in the Fig. 3.4 , with two different eccentricities.
• Fig. 3.4: 8O18 with Ex = 0: Q(b) = 0 , i = 0
Symmetry:
1- Conjunct A contrary of that of B
2- Conjunct C: nucleons 5 and 2 with contrary spins.
3- The conjunct C is symmetrical with regard to A and B.
• Fig. 3.4: 8O18 with Ex = 3555: Q(b) = 0 , i = 4+
• Fig. 3.4: 8O18 with Ex = 1982: Q(b) = 0 , i = 2+
Pay attention that the neutron B does not take a place in a position opposite to that of the neutron A.
• (Fig. 3.5)- 8O19 with Ex = 96: Q(b)=0 , i= 3/2+ . The symmetrical distribution of the three neutrons explains why Q(b)=0 even for the excited nucleus 8O19: although A= odd, it is almost a perfect boson (of course not considering the problem of spin i ¹ 0 ). It is of interest to note that it is impossible to explain why it has Q(b)= 0 from the models of current Nuclear Theory, because with A=19= odd it would be expected that Q(b) ¹ 0, even for the stable 8O19 without excitation.
• (Fig. 3.6)- 8O20 with Ex =1674: Q(b)=0 , i= 2+ . The symmetrical distribution of the four neutrons explains why Q(b) = 0.
4. ELECTRIC QUADRUPOLE MOMENT OF 8O18
The results of the experiments with this isotope of oxygen are shown in the Table I .
Then let us see the average Q(b)medium:
-1,0 -7,3 -10,7 -5,2 = -24,2x10-30m2 ® Q(b)medium = -24,2/4 = -6,05
+2,0 -4,5 -7,6 -2,3 = -12,4x10-30m2 ® Q(b)medium = -12,4/4 = -3,1
We will calculate the theoretical values of Q(b) . In the Figure 3.4 are exhibited the three possible structures of 8O18 isotopes . The structure with Ex = 3555 has Q(b) = 0 because it is symmetrical .
Let us calculate the Q(b) of the two other structures with Ex =1982.
4.1) FIRST STRUCTURE (see Fig. 4.1)
a = R.(1-cos30) = 1,34F (4.1 )
b = (R / 2).cos30 = 4,33F (4.2)
c = R.(1-sen30)/ 2 = 2,5F (4.3)
4.1.1 ) - Calculation of the ( x , y ) coordinates of the G.C. point, with the 2He4 situated rigidly in the center of the hexagon (Fig. 4.1-A).
side left : 8.x + 4.( R - a + x ): (see Fig. 4.1-D)
side right : 5.[ R - (a + x ) ] + 1.( b - x ): (see Fig. 4.1-D)
8.x + 4.( R - a + x ) = 5.[ R - (a + x ) ] + 1.( b - x ) (4.4)
x = 0,722F (4.5)
side up : 4.y + 2.( R + y ) + 4.( R / 2 + y ) + 1.y: (see Fig. 4.1-C)
side down : 4.( R / 2 - y ) + 2.( R - y ) + 1.( R / 2 + c - y ): (see Fig. 4.1-C)
4y + 2.(R + y) + 4.(R / 2 + y) + 1.y = 4(R / 2 - y) + 2(R - y) + 1.(R / 2 + c - y) (4.6)
y = 0,417F (4.7)
From the Figure 4.1-E : d2 = x2 + y2 , d = 0,834F (4.8)
4.1.2 ) - Calculation of the ( x’, y’ ) coordinates of the hexagon, without the 2He4 :
Side left : 4.x’ + 4.( R - a + x’ )
Side right: 5.[ R - (a + x’ ) ] + 1.( b - x’ )
4.x’ + 4( R - a + x’ ) = 5[ R - (a + x’ ) ] + 1.( b - x’ ) (4.9)
x’ = 0,928F (4.10)
Side up: 2( R + y’) + 4( R/2 + y’) + 1.y
Side down: 4.( R/2 - y’ ) + 2.(R - y’ ) + 1.(R/2 + c - y’)
2(R + y’) + 4(R/2 + y’) + 1.y = 4.( R / 2 - y’) + 2(R - y’) + 1.(R/2 + c - y’) (4.11)
y’ = 0,536F (4.12)
From the Figure 4.1-E , (d’)2 = (x’)2 + (y’)2 , d’ = 1,071F (4.13)
Let us apply the coefficient k obtained in (2.11), k = 0,445 (see Fig. 4.1-F):
r’ = ( 2d + k.d’ ) / 2 = ( 2x0,834 + 0,445x1,071 )/2 = 1,072F (4.14)
And the electric quadrupole moment of the first structure is:
Q(b) = 6x [ -(1,072F)2 ] = -6,43x10-30m2 (4.15)
4.2) SECOND STRUCTURE (Fig. 4.2)
4.2.1 ) - Calculation of the ( x , y ) coordinates of the G.C. point, with the 2He4 situated rigid in the center of the hexagon:
8x + 4(R - a + x) + 1.(b + x) = 5[R - ( a + x )] (4.16)
18.x = 4,33 (4.17)
x = 0,24F (4.18)
4y + 2(R + y) + 4(R/2 + y) + 1.y = 4(R/2 - y) + 2(R - y) + 1.(R/2 + c - y) (4.19)
18y = 7,5 (4.20)
y = 0,417F (4.21)
d2 = x2 + y2 d = 0,482F (4.22)
4.2.2 ) - Calculation of the ( x’, y’ ) coordinates of the hexagon, without the 2He4:
14.x’ = 4,33 (similar to 4.17) (4.23)
x’ = 0,309F (4.24)
14y’ = 7,5 (similar to 4.20) (4.25)
y’ = 0,536F (4.26)
(d’)2 = (x’)2 + (y’)2 d’ = 0,618F (4.27)
r’ = ( 2x0,482 + k.0,618 ) / 2 = 0,619F (4.28)
Q(b) = 6x( -0,6192) = -2,3x10-30m2 (4.29)
CONCLUSION : the two theoretical values are:
Q(b) = -6,43x10-30m2 or -2,3x10-30m2 ,
and they are coherent with the Q(b)medium of experiments (-6,05x10-30m2 or -3,1x10-30m2)
