4. Discussion of results

#### 4.1 Calculation of the pressure relation of proton and neutron

The fine-structure constant, α, has several physical interpretations, we use the most known.

Exergy: Mechanical Nuclear Physics Measures Pressure, Viscosity and X-Ray Resonance… DOI: http://dx.doi.org/10.5772/intechopen.95405

The fine-structure constant, <sup>α</sup> <sup>¼</sup> <sup>1</sup> <sup>137</sup>:035999174 35 ð Þ is the ratio of two energies: the energy needed to overcome the electrostatic repulsion between two protons a distance of 2r apart, and (ii) the energy of a single photon of wavelength λ ¼ 2πr (or of angular wavelength r; according to Planck relation).

$$\alpha' = \frac{e^2}{4\pi\varepsilon\_0(2r)} / \frac{hc}{\lambda} = \frac{e^2}{4\pi\varepsilon\_0(2r)} \frac{2\pi r}{\hbar c} = \frac{e^2}{4\pi\varepsilon\_0(2r)} \frac{r}{\hbar c} = \frac{1}{2} \frac{e^2}{4\pi\varepsilon\_0\hbar c} = \frac{1}{2}a \tag{65}$$

We can find the relationship of energies between the proton and the atomic nucleus. Knowing that the two occupy the same nuclear volume V: This relationship is identical to <sup>1</sup> <sup>α</sup><sup>0</sup> , since the atomic nucleus interacts with the proton through the electromagnetic field and the nuclear force.

$$\frac{p\_p}{p\_N} = \frac{p\_p V}{p\_N V} = \frac{2}{a} = 274.07\tag{66}$$

Thus, we already know the pressure relation between the proton and the atomic nucleus pp pN ¼ 2 α :

Now we find the relation of pressures between the neutron and the atomic nucleus, which are under the action of the same nuclear force, F.

$$\frac{p\_n}{p\_N} = \frac{F/\pi (0.84184)^2}{F/\pi \left(1.2A^{1/3}\right)^2} = \frac{\left(1.2A^{1/3}\right)^2}{\left(0.84184\right)^2} \tag{67}$$

For the chemical element with maximum nuclear pressure, <sup>62</sup> <sup>28</sup>Ni we have pn pN ¼ <sup>1</sup>:2 62 ð Þ1=<sup>3</sup> ð Þ<sup>2</sup> ð Þ <sup>0</sup>:<sup>84184</sup> <sup>2</sup> <sup>¼</sup> <sup>31</sup>:830:

If we divide Eq. (67) for Eq. (66) we have.

$$\frac{p\_p}{p\_n} = \frac{p\_p}{p\_N} / \frac{p\_n}{p\_N} = 274.07 / 31.830 = 8.6104$$

## 4.2 An action on the nuclear surface produces a reaction in the nuclear volume and vice versa

It is created by the friction between the layers of nucleons.

Theorem 13 An action on the nuclear surface produces a reaction in the nuclear volume and vice versa.

Proof. The volume and the nuclear surface are connected through the Gaussian divergence theorem and the Navier Stokes equations.

For an incompressible fluid, whose velocity field u ð Þ x, y, z is given, ∇ :u ¼ 0 is fulfilled.

Logically, the integral of this term remains zero, that is:

$$\int \int \int \overleftarrow{\nabla} \vec{u} \, d\mathbf{x} dy dz = \mathbf{0}$$

Writing the Divergence theorem. Ð Ð u :n dS <sup>¼</sup> ÐÐÐ <sup>∇</sup> :u dxdydz <sup>¼</sup> 0, the first term must be equal to zero, that is:

Applying the mean value theorem of integrals, we know that there is a mean value of the nuclear pressure p and the volume of the atomic nucleus V of the integral (60)

> ðr<sup>0</sup> 0

that according to Quantum Mechanics, pð Þζ ,V rð Þ<sup>0</sup> are the observable values. By

fm, from where:

<sup>2</sup><sup>η</sup> <sup>¼</sup> <sup>1</sup>

the Navier Stokes 3D equations, Eqs. (58) and (59), it is possible to find the average or most probable values of the variables involved pressure, nuclear viscosity and

2η : Explicitly we find the nuclear viscosity as a function of the nuclear decay constant <sup>k</sup> <sup>¼</sup> <sup>k</sup> � � and the average value of the nuclear pressure <sup>p</sup> as follows:

<sup>η</sup> <sup>¼</sup> <sup>p</sup>

ρ0 m2 s h i

So the second way to find the nuclear viscosity has the form:

4.1 Calculation of the pressure relation of proton and neutron

particles and the radius from which these particles leave.

3.7 Calculation of nuclear pressure and viscosity

dynamic nuclear viscosity <sup>η</sup>, as follows: <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup>

4. Discussion of results

most known.

40

<sup>η</sup> <sup>¼</sup> j j <sup>u</sup> <sup>ρ</sup><sup>0</sup>

There is another experimental way of determining nuclear viscosity, through the fuzziness of alpha particles, protons or neutrons ejected in a nuclear decay using the fluid velocity equation u ¼ �2νμð Þ 1 � P ∇r, in which modulo u ¼ �2νμð Þ 1 � P ∇r,

<sup>2</sup>μð Þ <sup>1</sup> � <sup>P</sup> <sup>¼</sup> j j <sup>u</sup> <sup>ρ</sup>0rP

Eq. (64) has a more complicated form and depends on the speed of the fluid

The nuclear decay constant k is determined by the nuclear pressure p0, and the

The fine-structure constant, α, has several physical interpretations, we use the

2η :

this restriction we can equalize Eq. (60), (61) and find the value of the initial nuclear pressure as a function of the experimentally measured nuclear pressure:

> <sup>p</sup>ð Þ¼ <sup>ζ</sup> <sup>Δ</sup>Emass�excessð Þ <sup>Z</sup>, <sup>A</sup> 4 <sup>3</sup> πr<sup>3</sup> 0

ðπ 0 ð2<sup>π</sup> 0

> <sup>¼</sup> <sup>3</sup>p<sup>0</sup> μr<sup>0</sup>

r0 :

pdV ¼ pð Þζ V rð Þ<sup>0</sup> (61)

: (62)

<sup>T</sup>1=<sup>2</sup> obtained from the resolution of

<sup>2</sup><sup>k</sup> (63)

ρ0 � �

4 1ð Þ � <sup>P</sup> (64)

μð Þ 1 � P :

, we obtain: j j <sup>u</sup> <sup>¼</sup> <sup>2</sup> <sup>η</sup>

W ¼ ΔEmass�excessð Þ¼ Z, A

<sup>μ</sup> with <sup>r</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup>:2A1=<sup>2</sup>

For Yukawa's potential, it is often assumed μ≈ <sup>1</sup>

Using the fundamental expression <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup>

nuclear decay constant as follows: <sup>k</sup> <sup>¼</sup> <sup>p</sup>

replacing the dynamic viscosity <sup>ν</sup> <sup>¼</sup> <sup>η</sup>

<sup>p</sup>ð Þ<sup>ζ</sup> <sup>4</sup><sup>π</sup> <sup>3</sup> r<sup>3</sup>

<sup>0</sup> ¼ 4πp<sup>0</sup>

Nuclear Materials

r2 0

3.6 Model of nuclear viscosity

$$\int \left\{ \overleftarrow{u} . \overleftarrow{n} d\mathcal{S} = \int \left\| \left| \left| \overleftarrow{u} \right| \right| \left| \overleftarrow{n} \right| \right\| \cos \left( a \right) d\mathcal{S} = 0 \to a = \frac{\pi}{2} \right\}$$

The only possible trajectory is circular, because in this case the vector n is perpendicular to the surface of the sphere. In this way the equation of the outer sphere corresponding to the surface is: <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>y</sup><sup>2</sup> <sup>þ</sup> <sup>z</sup><sup>2</sup> <sup>¼</sup> <sup>1</sup>:2A1=<sup>3</sup> .

Within the nuclear fluid there are layers of nucleons that move in spherical trajectories. ▪
