3.3 The principle of the femtoscope explains that low energy X-rays produce resonance in K-shell

A resonance region is created in a natural way at the K-shell between the nucleus and the electrons at S-level. The condition for the photons to enter in the resonance region is given by ra ≥ rn þ λ. This resonance region gives us a new way to understand the photoelectric effect. There is experimental evidence of the existence of resonance at K-level due to photoelectric effect, represented by the resonance cross section provided by NIST and calculated with GEANT4 for each atom. In the present work we focus on the resonance effects but not on the origin of resonance region.

The resonance cross section is responsible for large and/or abnormal variations in the absorbed radiation ð Þ I<sup>2</sup> � I<sup>1</sup> :

$$\frac{I\_2 - I\_1}{\frac{r + I\_1}{2}} = -\frac{\rho r}{\mu A} (\sigma\_2 - \sigma\_1) \tag{43}$$

Theorem 10 Resonance region. The resonance cross section is produced by interference between the atomic nucleus and the incoming X-rays inside the resonance region, where the boundaries are the surface of the atomic nucleus and K-shell.

The cross section of the atomic nucleus is given by:

$$
\sigma\_{r\_n} = 4\pi r\_n^2 = 4\pi A^{2/3} r\_n^2 \tag{44}
$$

The photon cross section at K-shell depends on the wave length and the shape of the atomic nucleus:

$$
\sigma\_{r\_n + \lambda} = 4\pi (r\_n + \lambda)^2 \tag{45}
$$

Subtracting the cross sections (13) and (14) we have:

$$
\sigma\_{\lambda} = \sigma\_{r\_n + \lambda} - \sigma\_{r\_n} = 4\pi \left( 2r\_n \lambda + \lambda^2 \right) = 4\pi \left( 2r\_p \lambda + 2\left( r\_n - r\_p \right) \lambda + \lambda^2 \right) \tag{46}
$$

The resonance is produced by interactions between the X-rays, the K-shell electrons and the atomic nucleus. The cross sections corresponding to the nucleus is weighted by probability pn and should have a simple dependence of an interference term. This last depends on the proton radius rp or the difference between the nucleus and proton radius (rn � rp) according to the following relation

$$\text{Max}(\sigma\_2 - \sigma\_1) = \left(\frac{\sigma\_1}{\sigma\_2}\right)^{-b} (\sigma\_2 - \sigma\_1) = 4\pi \left(2r\_p \lambda\right) \tag{47}$$

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

We note that left hand side of Eqs. (46) and (47) should have a factor larger than one due to resonance. The unique factor that holds this requirement is <sup>σ</sup><sup>1</sup> σ2 <sup>b</sup> Where a, b constants.

$$\frac{8\pi\overline{r}\lambda}{(\sigma\_2-\sigma\_1)} = a\left(\frac{\sigma\_1}{\sigma\_2}\right)^b \tag{48}$$

After performing some simulations it is shown that the thermal a represents the dimensionless Rydberg constant <sup>a</sup> <sup>¼</sup> <sup>R</sup><sup>∞</sup> <sup>¼</sup> <sup>1</sup>:<sup>0973731568539</sup> <sup>∗</sup> 107 .

$$\frac{80000\pi\overline{r}\lambda}{(\sigma\_2-\sigma\_1)} = R\_{\approx} \left(\frac{\sigma\_1}{\sigma\_2}\right)^{2.5031} \tag{49}$$

We use σ1ð Þ Z , σ2ð Þ Z for represent cross section in resonance, R<sup>∞</sup> is the generalized Rydberg constant for all elements of periodic table, and Z atomic number.

Last equation with a <sup>R</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup>:9935 was demonstrated and constructed using the elements of the solution of the Navier Stokes equations.

$$
\left(\frac{\sigma\_1}{\sigma\_2}\right) = 0.0021Z + 0.0696
$$

This equation was obtained with a <sup>R</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup>:9939, and indicates that the ratio of the effective sections fully explain each element of the periodic table.

$$E^\* = 2 \ast 10^{-5} Z^2 - 0.0003 Z + 0.004$$

This equation obtained for a <sup>R</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup>:9996, complements the system of equations that allow to know the simulation values as a function of Z for σ1ð Þ Z , σ2ð Þ Z and <sup>E</sup><sup>∗</sup> ð Þ <sup>Z</sup> , where <sup>σ</sup>1ð Þ <sup>Z</sup> <sup>&</sup>lt;σ2ð Þ <sup>Z</sup> : The Femtoscope equations further demonstrate that energy <sup>E</sup><sup>∗</sup> ð Þ <sup>¼</sup> min <sup>E</sup> is minimum and Shannon entropy <sup>S</sup> <sup>∗</sup> ð Þ <sup>¼</sup> max <sup>S</sup> is maximum in resonance, because in equilibrium σ1ð Þ¼ Zi σ2ð Þ Zi .

The radius of the neutron can be obtained using Eq. (49) in the following way.

$$
\overline{r} = \frac{R\_{\infty}}{8000\pi a}
$$

$$
r\_N = \frac{A}{N}\overline{r} + \frac{Z}{N}r\_P
$$

#### 3.4 Navier Stokes equation and cross section in nuclear physics

The speed needs to be defined as <sup>u</sup> ¼ �2<sup>ν</sup> <sup>∇</sup><sup>P</sup> <sup>P</sup> , where P xð Þ , y, z, t is the logistic probability function P xð Þ¼ , <sup>y</sup>, <sup>x</sup>, <sup>t</sup> <sup>1</sup> <sup>1</sup>þekt�μ<sup>r</sup> , <sup>r</sup> <sup>¼</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>y</sup><sup>2</sup> <sup>þ</sup> <sup>z</sup><sup>2</sup> ð Þ<sup>1</sup>=<sup>2</sup> defined in ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup> , t≥0 This P is the general solution of the Navier Stokes 3D equations, which satisfies the conditions (50) and (51), allowing to analyze the dynamics of an incompressible fluid.

$$\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u}, \nabla) \mathbf{u} = -\frac{\nabla p}{\rho\_0} \qquad \qquad \left( (\mathbf{x}, y, z) \in \mathbb{R}^3, t \ge 0 \right) \tag{50}$$

Where, u∈ <sup>3</sup> an known velocity vector, ρ<sup>0</sup> constant density of fluid and pressure p ¼ p0P∈ .

We use data of NIST and simulations with GEANT4 for elements Z ¼ 11 to <sup>Z</sup> <sup>¼</sup> 92 and photon energies 1:<sup>0721</sup> � <sup>10</sup>�<sup>3</sup> MeV to 1:<sup>16</sup> � <sup>10</sup>�<sup>1</sup> MeV, and have been

<sup>ρ</sup> <sup>¼</sup> <sup>σ</sup>pe <sup>þ</sup> <sup>σ</sup>coh <sup>þ</sup> <sup>σ</sup>incoh <sup>þ</sup> <sup>σ</sup>trip <sup>þ</sup> <sup>σ</sup>ph:<sup>n</sup>

The attenuation coefficient μ of a low energy electron beam 10, 100 ½ �eV will essentially have the elastic and inelastic components. It despises Bremsstrahlung

3.3 The principle of the femtoscope explains that low energy X-rays produce

we focus on the resonance effects but not on the origin of resonance region.

I<sup>2</sup> � I<sup>1</sup> <sup>I</sup>þI<sup>1</sup> 2

the boundaries are the surface of the atomic nucleus and K-shell. The cross section of the atomic nucleus is given by:

Subtracting the cross sections (13) and (14) we have:

Maxð Þ¼ σ<sup>2</sup> � σ<sup>1</sup>

σλ <sup>¼</sup> <sup>σ</sup>rnþ<sup>λ</sup> � <sup>σ</sup>rn <sup>¼</sup> <sup>4</sup><sup>π</sup> <sup>2</sup>rn<sup>λ</sup> <sup>þ</sup> <sup>λ</sup><sup>2</sup> <sup>¼</sup> <sup>4</sup><sup>π</sup> <sup>2</sup>rp<sup>λ</sup> <sup>þ</sup> <sup>2</sup> rn � rp

σrn ¼ 4πr

A resonance region is created in a natural way at the K-shell between the nucleus and the electrons at S-level. The condition for the photons to enter in the resonance region is given by ra ≥ rn þ λ. This resonance region gives us a new way to understand the photoelectric effect. There is experimental evidence of the existence of resonance at K-level due to photoelectric effect, represented by the resonance cross section provided by NIST and calculated with GEANT4 for each atom. In the present work

The resonance cross section is responsible for large and/or abnormal variations

Theorem 10 Resonance region. The resonance cross section is produced by interference between the atomic nucleus and the incoming X-rays inside the resonance region, where

<sup>n</sup> <sup>¼</sup> <sup>4</sup>πA<sup>2</sup>=<sup>3</sup>

The photon cross section at K-shell depends on the wave length and the shape of

r 2

<sup>σ</sup>rnþ<sup>λ</sup> <sup>¼</sup> <sup>4</sup>πð Þ rn <sup>þ</sup> <sup>λ</sup> <sup>2</sup> (45)

¼ � <sup>ρ</sup><sup>r</sup>

2

The resonance is produced by interactions between the X-rays, the K-shell electrons and the atomic nucleus. The cross sections corresponding to the nucleus is weighted by probability pn and should have a simple dependence of an interference term. This last depends on the proton radius rp or the difference between the nucleus and proton radius (rn � rp) according to the following relation

> σ1 σ2 �<sup>b</sup>

=μA (41)

σtot ¼ σcoh þ σincoh (42)

<sup>μ</sup><sup>A</sup> ð Þ <sup>σ</sup><sup>2</sup> � <sup>σ</sup><sup>1</sup> (43)

<sup>n</sup> (44)

<sup>λ</sup> <sup>þ</sup> <sup>λ</sup><sup>2</sup> (46)

ð Þ¼ <sup>σ</sup><sup>2</sup> � <sup>σ</sup><sup>1</sup> <sup>4</sup><sup>π</sup> <sup>2</sup>rp<sup>λ</sup> (47)

calculated according to:

Nuclear Materials

μ

emission and Positron annihilation.

resonance in K-shell

in the absorbed radiation ð Þ I<sup>2</sup> � I<sup>1</sup> :

the atomic nucleus:

36

With speed and pressure dependent on r and t: We will write the condition of incompressibility as follows.

$$\nabla.\mathbf{u} = \mathbf{0} \qquad\qquad \left( (\varkappa, \boldsymbol{\jmath}, \boldsymbol{z}) \in \mathbb{R}^3, \boldsymbol{t} \ge \mathbf{0} \right)\tag{51}$$

Replacing the respective values for the terms: ∇<sup>2</sup>

<sup>¼</sup> <sup>μ</sup><sup>2</sup>

<sup>¼</sup> <sup>μ</sup><sup>2</sup>

<sup>μ</sup><sup>r</sup> ¼ 0.

<sup>1</sup> <sup>þ</sup> ekt�<sup>μ</sup> <sup>x</sup>2þy2þz<sup>2</sup> ð Þ1=<sup>2</sup> <sup>¼</sup> <sup>2</sup>

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

1

<sup>2</sup><sup>η</sup> <sup>t</sup>�<sup>μ</sup> <sup>x</sup>2þy2þz<sup>2</sup> ð Þ1=<sup>2</sup> <sup>¼</sup> <sup>2</sup>

<sup>P</sup> <sup>¼</sup> <sup>μ</sup>ð Þ <sup>1</sup> � <sup>2</sup><sup>P</sup> <sup>∇</sup>P:∇<sup>r</sup> <sup>þ</sup> <sup>μ</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �∇<sup>2</sup>

Exergy: Mechanical Nuclear Physics Measures Pressure, Viscosity and X-Ray Resonance…

ð Þ <sup>1</sup> � <sup>2</sup><sup>P</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �j j <sup>∇</sup><sup>r</sup>

Using gradient <sup>∇</sup><sup>P</sup> <sup>¼</sup> <sup>μ</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �∇r, modulus j j <sup>∇</sup><sup>P</sup> <sup>2</sup> <sup>¼</sup> <sup>μ</sup><sup>2</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �<sup>2</sup>

∇2 P <sup>P</sup> � j j <sup>∇</sup><sup>P</sup> <sup>2</sup> P2

" #

Replacing Eqs. (55) and (56) in (57) we obtain the main result of the Navier Stokes equations, the solution represents a fixed point of an implicit function f tð Þ ,r

An important result of the Navier Stokes 3D equation, applied to the nuclear fluid of an atom, allows us to advance our understanding of nuclear dynamics and nuclear force. Corollary 12 The nuclear decay constant k is determined by the nuclear pressure p0,

The work in the nuclear fluid is given by the variation of the energy necessary to form that atomic nucleus, that is, by the excess mass required in the process. Using

> ð<sup>r</sup><sup>0</sup> 0

sin θdθ

ðπ 0 ð<sup>2</sup><sup>π</sup> 0 p0 2 μr r

ð<sup>2</sup><sup>π</sup> 0

dϕ ¼ 4πp<sup>0</sup>

Eq. (59) we can find the explicit form of nuclear pressure p ¼ p0P ¼ p<sup>0</sup>

ð<sup>2</sup><sup>π</sup> 0

pdV ¼

rdrð<sup>π</sup> 0

ð<sup>r</sup><sup>0</sup> 0

ðπ 0

2 μ ð<sup>r</sup><sup>0</sup> 0

Proof. A nuclear decay is a reaction of degree 1, which is explained by the exponential law N tðÞ¼ <sup>N</sup>0e�kt: Rewriting Eq. (58) as a function of the initial nuclear pressure p<sup>0</sup> and the nuclear viscosity η we can prove that they are related to the nuclear decay constant, in an intrinsic way and allow to explain dynamic

<sup>μ</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>y</sup><sup>2</sup> <sup>þ</sup> <sup>z</sup><sup>2</sup> ð Þ<sup>1</sup>=<sup>2</sup> ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup>

2η :

<sup>μ</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>y</sup><sup>2</sup> <sup>þ</sup> <sup>z</sup><sup>2</sup> ð Þ<sup>1</sup>=<sup>2</sup> ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup>

ð Þ <sup>1</sup> � <sup>2</sup><sup>P</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � � <sup>þ</sup> <sup>μ</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � � <sup>2</sup>

Laplacian of P can be written as follows.

DOI: http://dx.doi.org/10.5772/intechopen.95405

∇2

in (56).

where f tð Þ¼ ,<sup>r</sup> <sup>P</sup> � <sup>2</sup>

<sup>P</sup> <sup>¼</sup> <sup>1</sup>

nuclear phenomena.

1 þ e p0

3.5 Model of nuclear pressure

W ¼ ΔEmass�excessð Þ¼ Z, A

W ¼ ΔEmass�excessð Þ¼ Z, A p<sup>0</sup>

P xð Þ¼ , y, z, t

39

<sup>P</sup> and j j <sup>∇</sup><sup>P</sup> <sup>2</sup> of Eq. (55). The

r

¼ 0 (57)

j j ∇r

, t≥ 0 � � (58)

, t ≥0 � �

2 μr :

<sup>2</sup> cos θdrdθdϕ � �

r2 0 μ

(59)

(60)

▪

(56)

P

▪

<sup>2</sup> and ∇<sup>2</sup>

r

<sup>2</sup> <sup>þ</sup> <sup>μ</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �∇<sup>2</sup>

r

Theorem 11 The velocity of the fluid given by: <sup>u</sup> ¼ �2<sup>ν</sup> <sup>∇</sup><sup>P</sup> <sup>P</sup> , where P xð Þ , y, z, t is6the logistic probability function P xð Þ¼ , <sup>y</sup>, <sup>x</sup>, <sup>t</sup> <sup>1</sup> 1þe kt�<sup>μ</sup> <sup>x</sup>2þy2þz<sup>2</sup> ð Þ1=<sup>2</sup> , defined in

ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup> , t≥0 is the general solution of the Navier Stokes equations, which satisfies conditions (50) and (51).

Proof. Firstly, we will make the equivalence u ¼ ∇θ and replace it in Eq. (50). Taking into account that ∇θ is irrotational, ∇ � ∇θ ¼ 0, we have.

$$(\mathbf{u} \cdot \nabla)\mathbf{u} = (\nabla \theta . \nabla)\nabla \theta = \frac{1}{2}\nabla(\nabla \theta . \nabla \theta) - \nabla \theta \times (\nabla \times \nabla \theta) = \frac{1}{2}\nabla(\nabla \theta . \nabla \theta),$$

We can write,

$$\nabla \left( \frac{\partial \theta}{\partial t} + \frac{1}{2} (\nabla \theta. \nabla \theta) \right) = \nabla (-p)$$

It is equivalent to,

$$\frac{\partial \theta}{\partial t} + \frac{1}{2} (\nabla \theta. \nabla \theta) = -\frac{\Delta p}{\rho\_0}$$

where Δp is the difference between the actual pressure p and certain reference pressure p0: Now, replacing θ ¼ �2ν ln ð Þ P , Navier Stokes equation becomes.

$$\frac{\partial P}{\partial t} = \frac{\Delta p}{\rho\_0} P \tag{52}$$

The external force is zero, so that there is only a constant force F due to the variation of the pressure on a cross section σ. Where σ is the total cross section of all events that occurs in the nuclear surface including: scattering, absorption, or transformation to another species.

$$\begin{aligned} F &= p\sigma\_2 = p\_0\sigma\_1\\ \Delta p &= p - p\_0 = \left(\frac{\sigma\_1}{\sigma\_2} - 1\right) p\_0 = -(1 - P)p\_0 \end{aligned} \tag{53}$$

putting (52) in (53) we have

$$\frac{\partial P}{\partial t} = -\mu k (\mathbf{1} - P)P \tag{54}$$

In order to verify Eq. (51), <sup>∇</sup>:<sup>u</sup> <sup>¼</sup> 0, we need to obtain <sup>∇</sup><sup>r</sup> <sup>¼</sup> <sup>x</sup> r , y r , z r , <sup>∇</sup><sup>2</sup><sup>r</sup> <sup>¼</sup> <sup>∇</sup>:∇<sup>r</sup> <sup>¼</sup> <sup>y</sup>2þz<sup>2</sup> ð Þþ <sup>x</sup>2þz<sup>2</sup> ð Þþ <sup>x</sup>2þy<sup>2</sup> ð Þ <sup>x</sup>2þy2þz<sup>2</sup> ð Þ3=<sup>2</sup> <sup>¼</sup> <sup>2</sup> r :

$$\nabla \cdot \boldsymbol{\mu} = -2\nu \nabla \cdot \frac{\nabla P}{P} = -2\nu \mu \nabla ((\mathbf{1} - P)\nabla r) \tag{55}$$

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

Replacing the respective values for the terms: <sup>∇</sup><sup>2</sup><sup>P</sup> and j j <sup>∇</sup><sup>P</sup> <sup>2</sup> of Eq. (55). The Laplacian of P can be written as follows.

$$\begin{aligned} \nabla^2 P &= \mu (\mathbf{1} - 2\mathbf{P}) \nabla P. \nabla r + \mu \left(\mathbf{P} - P^2\right) \nabla^2 r \\ &= \mu^2 (\mathbf{1} - 2\mathbf{P}) \left(\mathbf{P} - P^2\right) |\nabla r|^2 + \mu \left(\mathbf{P} - P^2\right) \nabla^2 r \\ &= \mu^2 (\mathbf{1} - 2\mathbf{P}) \left(\mathbf{P} - P^2\right) + \mu \left(\mathbf{P} - P^2\right) \frac{2}{r} \end{aligned} \tag{56}$$

Using gradient <sup>∇</sup><sup>P</sup> <sup>¼</sup> <sup>μ</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �∇r, modulus j j <sup>∇</sup><sup>P</sup> <sup>2</sup> <sup>¼</sup> <sup>μ</sup><sup>2</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �<sup>2</sup> j j ∇r <sup>2</sup> and ∇<sup>2</sup> P in (56).

$$
\left[\frac{\nabla^2 P}{P} - \frac{|\nabla P|^2}{P^2}\right] = \mathbf{0} \tag{57}
$$

Replacing Eqs. (55) and (56) in (57) we obtain the main result of the Navier Stokes equations, the solution represents a fixed point of an implicit function f tð Þ ,r where f tð Þ¼ ,<sup>r</sup> <sup>P</sup> � <sup>2</sup> <sup>μ</sup><sup>r</sup> ¼ 0.

$$P = \frac{1}{1 + e^{kt - \mu(\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2)^{1/2}}} = \frac{2}{\mu(\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2)^{1/2}} \qquad\qquad \left( (\mathbf{x}, \mathbf{y}, \mathbf{z}) \in \mathbb{R}^3, t \ge 0 \right) \tag{58}$$

▪ An important result of the Navier Stokes 3D equation, applied to the nuclear fluid of an atom, allows us to advance our understanding of nuclear dynamics and nuclear force.

Corollary 12 The nuclear decay constant k is determined by the nuclear pressure p0, and the dynamic nuclear viscosity <sup>η</sup>, as follows: k <sup>¼</sup> <sup>p</sup><sup>0</sup> 2η :

Proof. A nuclear decay is a reaction of degree 1, which is explained by the exponential law N tðÞ¼ <sup>N</sup>0e�kt: Rewriting Eq. (58) as a function of the initial nuclear pressure p<sup>0</sup> and the nuclear viscosity η we can prove that they are related to the nuclear decay constant, in an intrinsic way and allow to explain dynamic nuclear phenomena.

$$P(\mathbf{x}, y, \mathbf{z}, t) = \frac{1}{1 + e^{\frac{\mathbf{p}\_0}{2\hbar} - \mu(\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2)^{1/2}}} = \frac{2}{\mu(\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2)^{1/2}} \tag{59}$$


#### 3.5 Model of nuclear pressure

The work in the nuclear fluid is given by the variation of the energy necessary to form that atomic nucleus, that is, by the excess mass required in the process. Using Eq. (59) we can find the explicit form of nuclear pressure p ¼ p0P ¼ p<sup>0</sup> 2 μr :

$$\begin{aligned} W &= \Delta E\_{\text{max}-\text{excess}}(Z, A) = \int\_0^{r\_0} \int\_0^{\pi} \int\_0^{2\pi} p dV = \int\_0^{r\_0} \int\_0^{2\pi} p\_0 \frac{2}{\mu r} \left(r^2 \cos\theta dr d\theta d\phi\right) \\\ W &= \Delta E\_{\text{max}-\text{excess}}(Z, A) = p\_0 \frac{2}{\mu} \int\_0^{r\_0} r dr \int\_0^{\pi} \sin\theta d\theta \int\_0^{2\pi} d\phi = 4\pi p\_0 \frac{r\_0^2}{\mu} \end{aligned} \tag{60}$$

With speed and pressure dependent on r and t: We will write the condition of

, t≥0 is the general solution of the Navier Stokes equations, which satisfies

Proof. Firstly, we will make the equivalence u ¼ ∇θ and replace it in Eq. (50).

ð Þ ∇θ:∇θ 

ð Þ¼� ∇θ:∇θ

where Δp is the difference between the actual pressure p and certain reference

The external force is zero, so that there is only a constant force F due to the variation of the pressure on a cross section σ. Where σ is the total cross section of all events that occurs in the nuclear surface including: scattering, absorption, or

> σ2 � 1

pressure p0: Now, replacing θ ¼ �2ν ln ð Þ P , Navier Stokes equation becomes.

∂P <sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>Δ</sup><sup>p</sup> ρ0

1þe

∇ ∇ð Þ� <sup>θ</sup>:∇<sup>θ</sup> <sup>∇</sup><sup>θ</sup> � ð Þ¼ <sup>∇</sup> � <sup>∇</sup><sup>θ</sup> <sup>1</sup>

¼ ∇ð Þ �p

Δp ρ0

p<sup>0</sup> ¼ �ð Þ 1 � P p<sup>0</sup>

<sup>∂</sup><sup>t</sup> ¼ �μkð Þ <sup>1</sup> � <sup>P</sup> <sup>P</sup> (54)

<sup>P</sup> ¼ �2νμ∇ð Þ ð Þ <sup>1</sup> � <sup>P</sup> <sup>∇</sup><sup>r</sup> (55)

r , y r , z r , <sup>∇</sup><sup>2</sup><sup>r</sup> <sup>¼</sup>

, t ≥0 (51)

kt�<sup>μ</sup> <sup>x</sup>2þy2þz<sup>2</sup> ð Þ1=<sup>2</sup> , defined in

2

P (52)

(53)

∇ ∇ð Þ θ:∇θ ,

<sup>P</sup> , where P xð Þ , y, z, t

<sup>∇</sup>:<sup>u</sup> <sup>¼</sup> <sup>0</sup> ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup>

Theorem 11 The velocity of the fluid given by: <sup>u</sup> ¼ �2<sup>ν</sup> <sup>∇</sup><sup>P</sup>

Taking into account that ∇θ is irrotational, ∇ � ∇θ ¼ 0, we have.

2

∂θ ∂t þ 1 2

F ¼ pσ<sup>2</sup> ¼ p0σ<sup>1</sup>

<sup>Δ</sup><sup>p</sup> <sup>¼</sup> <sup>p</sup> � <sup>p</sup><sup>0</sup> <sup>¼</sup> <sup>σ</sup><sup>1</sup>

∂P

In order to verify Eq. (51), <sup>∇</sup>:<sup>u</sup> <sup>¼</sup> 0, we need to obtain <sup>∇</sup><sup>r</sup> <sup>¼</sup> <sup>x</sup>

∇P

r :

∇:u ¼ �2ν∇:

∇ ∂θ ∂t þ 1 2

is6the logistic probability function P xð Þ¼ , <sup>y</sup>, <sup>x</sup>, <sup>t</sup> <sup>1</sup>

incompressibility as follows.

ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup>

Nuclear Materials

conditions (50) and (51).

We can write,

It is equivalent to,

transformation to another species.

putting (52) in (53) we have

<sup>∇</sup>:∇<sup>r</sup> <sup>¼</sup> <sup>y</sup>2þz<sup>2</sup> ð Þþ <sup>x</sup>2þz<sup>2</sup> ð Þþ <sup>x</sup>2þy<sup>2</sup> ð Þ

38

<sup>x</sup>2þy2þz<sup>2</sup> ð Þ3=<sup>2</sup> <sup>¼</sup> <sup>2</sup>

ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup> <sup>¼</sup> ð Þ <sup>∇</sup>θ:<sup>∇</sup> <sup>∇</sup><sup>θ</sup> <sup>¼</sup> <sup>1</sup>

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)

$$\Delta W = \Delta E\_{\text{max}-\text{excess}}(Z, A) = \int\_0^{r\_0} \int\_0^{\pi} \int\_0^{2\pi} p dV = p(\zeta)V(r\_0) \tag{61}$$

The fine-structure constant, <sup>α</sup> <sup>¼</sup> <sup>1</sup>

DOI: http://dx.doi.org/10.5772/intechopen.95405

hc <sup>λ</sup> <sup>¼</sup> <sup>e</sup><sup>2</sup> 4πε0ð Þ 2r

electromagnetic field and the nuclear force.

pp pN

nucleus, which are under the action of the same nuclear force, F.

pn pN

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

¼ pp pN = pn pN

divergence theorem and the Navier Stokes equations. For an incompressible fluid, whose velocity field u

Writing the Divergence theorem. Ð Ð u

term must be equal to zero, that is:

pp pn <sup>¼</sup> ppV pNV <sup>¼</sup> <sup>2</sup>

<sup>¼</sup> <sup>F</sup>=πð Þ <sup>0</sup>:<sup>84184</sup> <sup>2</sup> <sup>F</sup>=<sup>π</sup> <sup>1</sup>:2A<sup>1</sup>=<sup>3</sup> � �<sup>2</sup> <sup>¼</sup>

For the chemical element with maximum nuclear pressure, <sup>62</sup>

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

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

ððð <sup>∇</sup> :u

<sup>α</sup><sup>0</sup> <sup>¼</sup> <sup>e</sup><sup>2</sup>

is identical to <sup>1</sup>

nucleus pp pN ¼ 2 α :

<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:

and vice versa

volume and vice versa.

fulfilled.

41

<sup>4</sup>πε0ð Þ <sup>2</sup><sup>r</sup> <sup>=</sup>

angular wavelength r; according to Planck relation).

<sup>137</sup>:035999174 35 ð Þ is the ratio of two energies: the

e2 4πε0ℏc

<sup>α</sup> <sup>¼</sup> <sup>274</sup>:<sup>07</sup> (66)

ð Þ <sup>0</sup>:<sup>84184</sup> <sup>2</sup> (67)

ð Þ x, y, z is given, ∇

 :u ¼ 0 is

dxdydz <sup>¼</sup> 0, the first

<sup>28</sup>Ni we have pn

pN ¼

¼ 1 2

α (65)

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

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

Thus, we already know the pressure relation between the proton and the atomic

Now we find the relation of pressures between the neutron and the atomic

4πε0ð Þ 2r

<sup>α</sup><sup>0</sup> , since the atomic nucleus interacts with the proton through the

r ℏc ¼ 1 2

<sup>1</sup>:2A<sup>1</sup>=<sup>3</sup> � �<sup>2</sup>

¼ 274:07=31:830 ¼ 8:6104

4.2 An action on the nuclear surface produces a reaction in the nuclear volume

Theorem 13 An action on the nuclear surface produces a reaction in the nuclear

Proof. The volume and the nuclear surface are connected through the Gaussian

dxdydz ¼ 0

dS <sup>¼</sup> ÐÐÐ <sup>∇</sup>

 :u

:n

2πr hc <sup>¼</sup> <sup>e</sup><sup>2</sup>

Exergy: Mechanical Nuclear Physics Measures Pressure, Viscosity and X-Ray Resonance…

that according to Quantum Mechanics, pð Þζ ,V rð Þ<sup>0</sup> are the observable values. By 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 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> r2 0 <sup>μ</sup> with <sup>r</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup>:2A1=<sup>2</sup> fm, from where:

$$p(\zeta) = \frac{\Delta E\_{\text{mass\\_excess}}(Z, A)}{\frac{4}{3}\pi r\_0^3} = \frac{3p\_0}{\mu r\_0}.\tag{62}$$

For Yukawa's potential, it is often assumed μ≈ <sup>1</sup> r0 :

### 3.6 Model of nuclear viscosity

Using the fundamental expression <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup> <sup>2</sup><sup>η</sup> <sup>¼</sup> <sup>1</sup> <sup>T</sup>1=<sup>2</sup> obtained from the resolution of 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 nuclear decay constant as follows: <sup>k</sup> <sup>¼</sup> <sup>p</sup> 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:

$$
\overline{\eta} = \frac{\overline{p}}{2\overline{k}}\tag{63}
$$

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, replacing the dynamic viscosity <sup>ν</sup> <sup>¼</sup> <sup>η</sup> ρ0 m2 s h i, we obtain: j j <sup>u</sup> <sup>¼</sup> <sup>2</sup> <sup>η</sup> ρ0 � �μð Þ <sup>1</sup> � <sup>P</sup> :

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

$$\eta = \frac{|\mathbf{u}| \rho\_0}{2\mu(1 - P)} = \frac{|\mathbf{u}| \rho\_0 rP}{4(1 - P)}\tag{64}$$

Eq. (64) has a more complicated form and depends on the speed of the fluid particles and the radius from which these particles leave.

#### 3.7 Calculation of nuclear pressure and viscosity

The nuclear decay constant k is determined by the nuclear pressure p0, and the dynamic nuclear viscosity <sup>η</sup>, as follows: <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup> 2η :
