# Electrons Ejected Due of Laser Irradiation of Deuterons Layer on Metals Surface - A New Source of Instant Electrical Energy

Stefan Mehedinteanu

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Stefan Mehedinteanu. Electrons Ejected Due of Laser Irradiation of Deuterons Layer on Metals Surface -A New Source of Instant Electrical Energy. 2018. hal-01319725v3

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# Electrons Ejected Due to Light Irradiation of Hydrogen/Deuterons Layer on Metals
- A New Source of Instant Electrical Energy

Stefan Mehedinteanu¹

¹ CITON-Romania (retired) E-Mail: [mehedintz@yahoo.com](mailto:mehedintz@yahoo.com);
[s.mehedinteanu@yahoo.com](mailto:s.mehedinteanu@yahoo.com);
[s.mehedinteanu@gmail.com](mailto:s.mehedinteanu@gmail.com)

July 31 2018

PACS numbers: 52.25.-b,31.15.-p,32.70.-n

## Abstract

Based on the previously author works about models on nucleons structure and on the bias current inside valence nucleons during β-decay stimulation by a laser, in the present one is analyzed the feasibility of these experiments. Thus, by using QM&MD programme: fhi98md is confirmed the apparition of high Hydrogen (Deuterium) coverage (at ~pm distances) of the surface of Pd lattice, mainly due of the Hydrogen or Deuterium screening by Pd valence electrons, the same for the graphene case, just near nucleons' Einstein Ring. Also is proved the author's model of vortex assisted photon beta decay, when a laser photon makes this process much more probable by creating a spot (melt) in nucleon with suppressed order parameter that lowering the energy barrier for vortex crossing together with an heavy electron (bias current $e^±$) as resulting from the decay of the permanent rate of bosons pairs as produced inside nucleons by a Schwinger effect of the nucleon's inner Electro-Magnetic (EM) field. Then, if is used lasers of modest power it can appears a net gain of ~20. At this high coverage fraction, the H(D) ions at a deep screening near ground state is followed by an electron expulsion, and of proton charge neutralized (like a neutron). In the following, is definitively elucidated the physical basis of the Einstein ring, as due of a QCD operation in case of bulk nucleons when is itself infused a quantum energy teleported (QET) that corresponds to stress-energy tensor as discretized into n-sheeted Riemann surfaces glued, through quantum vacuum bosonic fields alias the *remnant gravitons* of the Quantum existing primordial micro-black (white) halls (μB(W)Hs) decayed to $W^±$ pairs during Spacetime fabric along Universe Evolution Epochs, to the boundaries (CFT) giving the first geodesic (object dimension). This necessary energy-QET in the nucleon to create $W^±$ pairs is always here as explain the free nucleons decay.

This advanced ions screening, that permit the neighborhood ions to fall-in the Einstein Ring due of lost of the kinematic equilibrium of the electron on the orbit, that enhancement of the packing into a nucleus (nucleosynthesis).

## 1. The state of art
Based on the previously author works about models on nucleons structure and on the bias current inside valence nucleons during β-decay stimulation by a laser, in the present one is analyzed the feasibility of some experiments to prove this theoretical finding [1].

The adsorption and absorption of hydrogen on palladium are described in [2a], when the binding energies of about 22, 25, and 35 kcal/mole were observed, and the saturation coverage at 300 °K was 0.39 H/Pd and at 200 °K this value was increased to 0.95 H/Pd.

In [2b], the neutron diffraction studies have shown that hydrogen atoms randomly occupy the octahedral interstices in the metal lattice (in a fcc lattice there is one octahedral hole per metal atom). The limit of absorption at normal pressures is PdH₀.₇, indicating that approximately 70% of the octahedral holes are occupied. The absorption of hydrogen is reversible, and hydrogen rapidly diffuses through the metal lattice.

The process of surface absorption of hydrogen has been shown by scanning tunneling to require aggregates of at least three vacancies on the surface of the crystal to promote the dissociation of the hydrogen molecule [3].

Since the structure of the atoms is:

Pd: [Kr] 4d¹⁰ 2, 8, 18, 18;

Kr: [Ar] 3d¹⁰ 4s² 4p⁶ 2, 8, 18, 8

Ar: [Ne] 3s² 3p⁶ 2, 8, 8

Ne: [He] 2s² 2p⁶ He: 1s² 2

, in order to obtain the inputs fort.11 and fort.12 together with start.inp for fhi98start and fhi98md we use for fhiPP the input:

fort_22.inp for Pd

46.00 8 3 8 0.00 : z nc nv iexc rnlc
1 0 2.00 : n l f
2 0 2.00
2 1 6.00
3 0 2.00
3 1 6.00
3 2 10.00
4 0 2.00
4 1 6.00
4 2 9.00
5 0 1.00
5 1 0.00
2 h : lmax type

Xenon
[Kr] 4d¹⁰ 5s² 5p⁶

per shell

2, 8, 18, 18, 8

¹³⁵Xe

syn

9.14 h

β⁻

1.16MeV

¹³⁵Cs

Fort_22.inp for Xe

54.00 9 2 8 0.00 : z nc nv iexc rnlc

1 0 2.00

: n 1 f

2 0 2.00

2 1 6.00

3 0 2.00

3 1 6.00

3 2 10.00

4 0 2.00

4 1 6.00

4 2 10.00

5 0 2.00

5 1 6.00

2 h

: lmax s_pp_def

An atom's *n*th electron shell can accommodate
*2*n*² electrons,

Core charge can also be calculated as 'atomic number' minus 'all electrons except those in the outer shell'. For example: Chlorine (element 17), with electron configuration 1s² 2s² 2p⁶ 3s² 3p⁵, has 17 protons and 10 inner shell electrons (2 in the first shell, and 8 in the second) so:

$$
\text{Core charge} = 17 - 10 = +7
$$

A subshell is the set of states defined by a common [azimuthal quantum number](<https://en.wikipedia.org/wiki/Azimuthal_quantum_number), *ℓ*, within a shell. The values *ℓ* = 0, 1, 2, 3 correspond to the *s*, *p*, *d*, and *f* labels, respectively. The maximum number of electrons that can be placed in a subshell is given by 2(2*ℓ* + 1). This gives two electrons in an *s* subshell, six electrons in a *p* subshell, ten electrons in a *d* subshell and fourteen electrons in an *f* subshell.

In [4a], about the interaction of hydrogen with palladium surfaces, although bulk palladium can absorb large amounts of hydrogen, the most favorable position for hydrogen is on the surface, not in the bulk. Here, the adsorption energy of atomic H on Pd(100) in the fourfold hollow ~ 0.45 eV and in the bridge position ~0.15 eV that are near constant as a function of the coverage (0.25÷1), that were determined by density functional theory calculations within the local density approximation (LDA) using the FP-LMTO method and by DFT calculations employing the generalized gradient approximation (GGA) for the exchange-correlation functional and using ultrasoft
pseudotentials (US-PP). The coverage is defined as the number of hydrogen atoms per primitive surface unit cell. Adsorption in the fourfold hollow position is much more favorable than at the bridge position. This is a general trend for hydrogen adsorption on low-index Pd surfaces: hydrogen generally prefers to adsorb at the highly coordinated sites.

In [4b], they report on the ability of freestanding single layer graphene to provide tens of electrons for charge neutralization of a slow highly charged ion within a few femtoseconds. The amount of charge transferred to the highly charged ions (HCI) can be estimated by measuring the distribution of exit charge states $q_{out}$ and the energy of highly charged $^{124}_{54}\text{Xe}$ ions ((Z = 54; n = 70) after transmission through single layer graphene (SLG).

Initial ion charge states of $10 < q_{in} < 35$ and velocities below $0.5 \text{ nmfs}^{-1}$ were used. Ions are transmitted through SLG under normal incidence and analyzed with respect to their charge state and kinetic energy by an electrostatic analyzer. $h = 3.4 \text{ Å}$ is the width of the graphene layer. In this case, the elastic collisions (nuclear stopping) may cause direct knockout of carbon atoms, but less than one carbon atom is sputtered on average by a 10–100 keV Xe ions. Even if point defects are produced, they will likely disappear due to dissociation of ubiquitous hydrocarbon molecules. The absence of any traces of large-scale lattice deformations thus confirms the intrinsic ability of suspended SLG to locally sustain exceptionally high current densities, even though it cannot efficiently diffuse heat to a substrate.

The HCI starts capturing electrons from graphene already at 9Å. The strong attractive potential accelerates electrons towards the HCI and, approximately, half of them end up captured by it along the incoming path before penetration into the graphene sheet. The induced electronic charge density as the HCI approaches the graphene layer has two components: one is formed by the convoy electrons around the HCI position, forming an asymmetric wake potential that slows down the ion, and the other one is located at the graphene layer due to the target polarization. Both components merge as the HCI gets closer to the target and forward electron emission starts. Finally, after crossing the layer the projectile is nearly neutralized and the corresponding induced electronic charge is centred around the HCI along its outgoing path. The actual HCI is probably not fully relaxed at the instant of crossing the graphene layer and, therefore, it still suffers a number of autoionization processes and subsequent de-excitation without a significant energy loss.

## The Einstein Ring around nucleons

In quantum gravity, a **virtual black hole** is a black hole that exists temporarily as a result of a quantum fluctuation of spacetime [3] cited in [12]. Theoretical arguments suggest that virtual black holes should have mass and the lifetime on the order of the Planck
particles. Therefore, we consider the Micro-black-holes pairs, that occur with a number
density of approximately one per Quantum bubble[3a]cited in [12], that means one
per quantum bubble volume ≅ a³ at the end dumping of quantum oscillations (when
k = aH; k = H, a = 1), that means nₚ = a³/dₕ⁻¹ ≅ 10⁶⁰; where Hubble constant
dₕ⁻¹ ≅ H₋₁ = 10⁻²⁰ m a more precise approach with the same result is given in [12].

$R_U = 1.6 \times 10^{26} [m]$, the radius of Universe; $t = 5 \times 10^{17} s$, the age of the Universe.

**Considerations on The Fabric of Spacetime Around Objects**

These are obtained by using well-known Inflation models [16a-d] cited in [12], where the
scale leaving the horizon at a given epoch is directly related to the number N(φ) of e-
folds of slow-roll inflation that occur after the epoch of horizon exit. Indeed, since H -the
Hubble length is slowly varying, we have d ln k = d (ln(aH)) ≅ d ln a = &dtd/a = Hdt . From
the definition this gives d ln k = -dN(φ) , and therefore ln(k<sub>end</sub>/k) = N(φ) , or,
k<sub>end</sub> = ke<sup>N</sup>[m], where k<sub>end</sub> is the scale leaving the horizon at the end of slow-roll
inflation, or usually k<sup>-1</sup> << k<sub>end</sub><sup>-1</sup>[m], the correct equation being k = k<sub>end</sub>e<sup>N</sup>[m<sup>-1</sup>].
During Universe evolution, the horizon leave is when a<sub>leave</sub> = k<sub>leave</sub>/H<sub>leave</sub> = 1, and
k<sub>leave</sub><sup>-1</sup> = H<sub>leave</sub><sup>-1</sup> = 10<sup>-27</sup> [m], t<sub>leave</sub> = H<sub>leave</sub><sup>-1</sup>/c = 3.3×10<sup>-36</sup> s at the Electroweak epoch.
In Newtonian interpretation, the Friedmann equations are equivalent to this pair of
equations:

$$
\frac{d\&}{2} = \frac{G \frac{4\pi a^3}{3} \rho}{a}
\left(\frac{d\&}{a}\right)^2 = -\frac{8\pi G}{c^2} p + \Lambda c^2
\rho[\text{kg/m}^3]; \text{energy density } p = -\rho c^2
$$

If we divide with a² we obtain for outside (low density) the object as a pushing GW
wave (BH, planets, stars etc.)

$$
\frac{d\&^2}{2a^2} = H^2 [s^{-2}] = \frac{GM}{a^3} \Rightarrow \frac{GE}{c^2 a^3} \rightarrow \frac{GE}{c^4 a^3} = \frac{G}{c^4 a^3} \cdot n_{\text{at-merging}} \cdot \varepsilon_g [m^{-2}]; \quad (3)
$$

, where: n_g = E/ε_g; n_at-merging = (n_g / (ε_P / ε_g)); M = (4πa³ / 3) ρ; E = Mc²; M = n_μBH m_μBH; the
scalar curvature is R² = 6H².

For inside of the objects as given by equation (v) from [12]:
$$ \frac{8\pi G\rho}{3n} = H^2[s^{-2}] \quad ; \quad n \text{-the number of boundary points} \qquad (v) $$

During Universe evolution at Electroweak epoch (EW) or Reheating due of the quantum fluctuations a huge number of the micro-black holes as Planck particles $n_p$ are generated, $m_{\mu BH} = m_p$; the graviton energy being at horizon leave ;

$$ \varepsilon_g = \eta c / a_{end} = 2 \times 10^{-26} J ; \text{ when } a_{leave} = k_{leave} / H_{leave} = 1, a_{end} \approx 1; M_U = 2.2 \times 10^{53} \text{ kg} ; $$

$$ \lambda_C = \eta c / m_{\mu BH} c^2 ; n_{\mu BH} \approx 1 / H_{-1}^3 \cdot V_{available} \approx n_p \approx 10^{61-62}, $$

First step

$$ n_p = E_U / \varepsilon_p = 10^{70} J / 10^9 J = 10^{61} \text{ with } \varepsilon_{\mu BH\_EW} = 10^{19} / a_{end-EW} = 10^{11} \text{ GeV} \rightarrow 17J, \text{ it} $$

results with $k_{leave}^{-1} = H_{leave}^{-1} = 10^{-27}[m]$ by iteration for $N = 2.5$; in eq. (2);

$$ k_{end} = 9.2 \times 10^{19} [m^{-1}] ; a_{end-EW} = 8.2 \times 10^{7} ; H^{-1} = 10^{-18} [m] ; t = 3.3 \times 10^{-27} s ; $$

$$ R = 6.5 \times 10^{-20} [m] ; l_C = l_p = 1.5 \times 10^{-27} [m] ; $$

Second step

or when immediately, $\varepsilon_{\mu BH\_EW} = 10^{11}/a_{EW} = 10^{11} \text{ GeV} \rightarrow 17J$ at Reheating, that results with $k_{leave}^{-1} = H_{leave}^{-1} = 10^{-27}[m]$ by iteration for $N = 16.2$; in eq. (2); $k_{end} = 9.2 \times 10^{19} [m^{-1}]$; $a_{EW} = 0.92$; $H^{-1} = 10^{-20}[m]$; $t = 3.3 \times 10^{-29} s$; $R = 6.5 \times 10^{-20}[m]$;

$$ l_C = l_p = 1.7 \times 10^{-27} [m] ; a_{EW} = 0.92 ; \text{ where } n_g = \frac{\varepsilon_p}{\varepsilon_g} \cdot n_p \rightarrow 10^{96} \text{ as a "fix" number of} $$

gravitons potentially to escape (an inverse process of a black-hole) from the micro-blacks holes during merging in order to generate particles.

In both steps the number of gravitons that has been released following µBHs merging is

$$ \text{only } n_{at-merging} = \frac{n_g}{(\varepsilon_{\mu BH\_EW} / 10^{-26})} = \frac{10^{96}}{(17/10^{-26})} \approx 10^{68}, \text{ and these generate the curvature} $$

$$ \text{radius of the object } R = d_H = 10^{-20}. $$

In other words the contribution to the space-time deformation in this phase is due of only $n_H^{EW} = 10^{68}/2n_p = 5 \times 10^5$, for two µBHs merging.

**Electroweak symmetry breaking-quarks epoch (QGP)**

The µBH particles decay to QGP; $m_{QGP} = 1.4 \times 10^{-22} \text{ kg} \rightarrow 7.9 \times 10^4 \text{ GeV}$, or

$$ \varepsilon_{\mu BH\_QGP} = \varepsilon_{\mu BH\_EW} / a_{end\_QGP} = 7.9 \times 10^4 \text{ GeV} \rightarrow 1.26 \times 10^{-5} J, $$

$$ \varepsilon_{\mu BH\_EW} = 10^{19} / a_{end-EW} = 10^{11} \text{ GeV} \rightarrow 17J ; a_{end-EW} = 8.2 \times 10^{7} ; k_{end}^{-1} = 7.9 \times 10^{-14} [m], $$

$$ a_{end\_QGP} = 1.27 \times 10^6, \lambda_{C-ee} = 2.3 \times 10^{-21} [m], \text{ with eq. (2) } R = 1.2 \times 10^{-7} m, \text{ and} $$

$$ H_{end}^{-1} = 10^{-7} [m], \quad t_{end} = H_{end}^{-1} / c = 3.3 \times 10^{-16} s, \quad H_{leave}^{-1} = k_{leave}^{-1} = H_{EW}^{-1} = 10^{-20} [m] \text{ we found} \\ N = 15.88 $$
The number of gravitons that has been released following µBHs merging is only

$$n_{at-merging} = \frac{n_g}{(\varepsilon_{\mu BH\_QGP}/10^{-26})} = \frac{10^{96}}{(1.26 \times 10^{-5}/10^{-26})} \approx 7.9 \times 10^{74}, \text{ and these generate the curvature radius of the object R.}$$

In other words the contribution to the space-time deformation is due of

$$n_H^{QGP} = 7.9 \times 10^{74} / 2n_p = 4 \times 10^{12}, \text{ for two } \mu\text{BHs} \text{ merging.}$$

Thus, from the overall balance the number of gravitons which are consumed to push the space time is $:= 5 \times 10^5 \cdot 4 \times 10^{12} = 2 \times 10^{18} < n_{\mu BH-p} = 10^{35}$, and the remnant number is $n_{g-\mu BH-remnant} = 5 \times 10^{16}$.

## The Confinement into nucleons

With $\varepsilon_{\bar{q}q\_g} = \varepsilon_{\mu BH\_QGP} / a_{end\_qq\_g} = 8.9 \times 10^4 / 3.7 \times 10^3 = 2.1 \text{GeV} \rightarrow 3.4 \times 10^{-10} \text{J}$ at Confinement, where $\varepsilon_{\mu BH\_QGP} = \varepsilon_{\mu BH\_EW} / a_{end\_QGP} = 7.9 \times 10^4 \text{GeV} \rightarrow 1.26 \times 10^{-5} \text{J}$, and when $k_{end}^{-1} = 0.4[m]$, $a_{end\_\bar{q}q\_g} = 3.7 \times 10^3$, with eq. (2) $H_{end}^{-1} = 150[\text{m}]$, $R = 144[\text{m}]$; $t_{end} = H_{end}^{-1} / c = 5 \times 10^{-6} \text{s}$, and $H_{leave}^{-1} = k_{leave}^{-1} = H_{QGP}^{-1} = 10^{-7}[\text{m}]$ we found we found $N = 15.2$.

The number of gravitons that has been released following µBHs merging is only

$$n_{at-merging} = \frac{n_g}{(\varepsilon_{\bar{q}q\_g}/10^{-26})} = \frac{10^{96}}{(3.4 \times 10^{-10}/10^{-26})} \approx 2.9 \times 10^{79}, \text{ and these generate the curvature radius of the object R.}$$

In other words the contribution to the space-time deformation is due of

$$n_H^{qq\_g} = 2.9 \times 10^{79} / 2n_p \approx 10^{17}, \text{ for two } \mu\text{BHs} \text{ merging.}$$

Thus, from the overall balance of µBHs the number of gravitons consumed (via EW $\rightarrow$ QGP $\rightarrow$ QCD) to push the space time is $= 1/8 \cdot 5 \times 10^5 \cdot 4 \times 10^{12} \cdot 10^{17} \approx 0.25 \times 10^{35} \approx n_p = 10^{35}$.

In other words, at Confinement it remains at least few ($\approx 4$) gravitons, so

$$\varepsilon_{\text{nucleon}} = 4 \times 10^{-26} / \lambda_C \approx 10^{-10} \text{J} \approx 0.6 \text{GeV}.$$

Now, from [12] when, with the gravitons released from the nucleon's horizon as a quantized mass, that of an energy $\varepsilon_g = \eta c / a_{EW} = 2 \times 10^{-26} \text{J}$ (when it born !); and when

$$a_{leave} = k_{leave} / H_{leave} = 1, \text{ of } W^{\pm} \text{ mass } M_{U\_W^{\pm}} = 10^{-25} \text{kg inside the nucleon;}$$

$$\lambda_C = \eta c / m_{\mu BH\_W} c^2;$$

In the following, by a QCD operation in case of bulk nucleons is self infused a quantum energy teleported (QET) that corresponds to stress-energy tensor as discretized into n-sheeted Riemann surfaces glued, through quantum vacuum bosonic fields alias the remnant gravitons of the Quantum existing µB(W)Hs decayed to $W^{\pm}$ pairs during Spacetime fabric along Universe Evolution Epochs, to the boundaries (CFT) giving the first geodesic (object dimension) [12]. This necessary energy-QET in the nucleon to
create $W^±$ pairs is always here (see free nucleons decay, below) as being
of $\epsilon_{\mu BH \_W} = \epsilon_{H-g}/a_g = 85 \text{GeV}/a_g = 84 \text{GeV} \rightarrow 1.3 \times 10^{-8} J$, it results with

$k_{leave}^{-1} = H_{leave}^{-1} = 1 \times 10^{-16} [m]$ by iteration for $N = 13.8$; in eq. (2) below;

$k_{end} = 9.8 \times 10^{-11} [m]$; the deformation of space-time around the nucleon results

$H^{-1} = 1 \times 10^{-10} [m]$; $t = 3.3 \times 10^{-19} s$; or $R = 1.3 \times 10^{-10} [m]$; $l_c = 2.2 \times 10^{-18} [m]$; $a_g = 1$;

$M_U = 1.4 \times 10^{-25} \text{Kg}$

In other words, in our case of continuously interaction, for the primordially
$\mu B(W)H_s$ which were $W^\pm$ particles in the neutralized protons we have after
Confinement $n_g \approx 4$ gravitons as above, and

$$n_{at-merging} = \frac{n_g}{(\epsilon_{\mu BH \_W} / 10^{-26})} = \frac{1}{(1.3 \times 10^{-8} / 10^{-26})} \approx 7.4 \times 10^{-19}, \text{ which is a very small}$$

interaction gravitons fraction of $W^±$ transferred as QET (see below) to the horizon where is generated as gravitational wave (GW) that deforms the space it creating the Einstein Ring, or the curvature radius given by eq. (2) from [12], $R = 1.3 \times 10^{-10} [m]$. If the electrons screening of nearer nucleons is sufficiently to attain this dimension, the break-up of theirs kinematic equilibrium (centrifugal force and Spining) it means theirs fall in this curvature radius (Einstein Ring) that producing a packing like in nucleonsynthesis.

But what happen in case of solid compact objects (with nucleonic structure)?

We know, that the penetration of gluons magnetic $B$ flux inside of an nucleon [20],[21], see figure 4., is

$$\lambda = \left( \frac{\epsilon_0 \cdot m \cdot c^2}{n_s \cdot e^2} \right)^{1/2} [m], \quad m\text{-gluons mass, } n_s\text{-number of gluons per } [m]^3,$$

$$\Phi_0 = \pi\eta c/e \rightarrow \text{usually } \frac{\pi\eta}{e} = 2.07e-15[Tm^2] \rightarrow Js/C$$

, where from [9] we have

$$B(x) = \frac{2\Phi_0}{2\pi\lambda^2} \log\left(\frac{2\lambda}{\gamma x}\right), \quad x \ll \lambda$$

, where $\gamma = e^C \approx 1.78(C$ is Euler's constant), and with $x \approx \lambda/10 \rightarrow \log(\ldots) \Rightarrow \approx 1$ on the boundary, or the dual gauge component of the total electrical field

$$B = \frac{2\Phi_0}{2\pi\lambda^2} = \frac{\eta}{e\lambda^2} \left[ \frac{J}{Am^2} \right]$$

In the case of a homogeneous potential directed along the z-axis [18], the Einstein stress-energy tensor is:
$T^{00} = T^{11} = T^{22} = -T^{33} = \rho_B = \frac{\epsilon_0 c^2 B^2}{8\pi}$; $T^{0i} = 0$, where $\rho_B[J/m^3]$-the magnetic energy density.

The equivalence between the Lorenz force energy which squeezes the electrical field $E_{e^\pm}$ is $\epsilon_L = ec\lambda_C B$, and at the interface between normal and superconducting phase we have $B \cong E/c$, with $e^\pm$ pair giving E as: $k_B T = \eta\nu = \epsilon_L = ec\lambda_C \frac{\eta}{e\lambda_C^2} = c\frac{\eta}{\eta}mc = mc^2$, and accounting that the inverse of the penetration length $\lambda \cong \lambda_C$.

Here, the Hubble constant is defined as

$$H^2 = \frac{8\pi G V^4}{3(\eta c)^3 c^4} \quad \text{in} \ [m^{-2}], \qquad (\text{v}')$$

at Compton length $l_C = \eta/mc$ that result the electric field:

$$E^2 = \frac{(V)}{\epsilon_0 (\lambda_c^{e*})^3},$$

The energy is $\epsilon = \frac{V_{vol} \epsilon_0 c^2 B^2}{8\pi} = \rho_B V_{vol} = V[J]$; $V_{vol} = 2\pi\lambda_C\lambda_c(4\lambda_C) \cong 8\pi\lambda_C^3$,

We can rewrite eq. (v') as: with $\lambda_C = \eta/M_U c$

$$H^2 = \frac{8\pi G V^4}{3(\eta c)^3 c^4} \rightarrow \frac{8\pi G V^4}{3\lambda_C^3 (M_U c^2)^3 c^4} \rightarrow \frac{8\pi G}{3} \frac{nV}{n\lambda_C^3 c^4} [m^{-2}]$$

Or, $\frac{d^2}{a^2} = H^2 = \frac{8\pi G M_U}{3n\lambda_C^3 c^2} = \frac{8\pi G \rho}{3nc^2}$; where the total mass is $M_U = n\frac{V}{c^2}$; $M_U = \frac{U}{c^2}$;

$U = \frac{GM^2}{R}$ -the gravitational potential, or $\rho = M_U / nl_C^3$; $M_U = M_{object}$

Thus, we obtain the expression

$$\frac{d^2}{2a^2} = H^2[s^{-2}] = \frac{G \frac{4\pi a^3}{3} \rho}{a^3} [s^{-2}] \rightarrow \frac{4\pi G \rho}{3c^2} = \frac{4\pi G \frac{M_U}{n_{at-merging} l_C^3}}{3c^2} [m^{-2}] \quad (2)$$

$$R = H^{-1}$$

Also, the deformation could be done as (author' approach based on the eq. (12) from [19]):

$$\theta = \sqrt{\kappa^{-1} T_{ab} \cdot k_{end}};$$

where $\kappa = \frac{c^2}{8\pi G}$ as Newton constant, and the stress-energy tensor arrived in CFT layer (horizon) is
$$T_{ab} = \rho = \frac{n_{g\_horizon} \cdot \epsilon_g}{c^2 \cdot V_{GW-layer}} = 1.43 \times 10^{28} [kg/m^3]; \text{ where } V_{GW-layer} = n_{g\_at\_merging} \cdot \lambda_C^3,$$

$n_{g\_horizon} = 1$ the number of gravitons just arrived at horizon (CFT) by QET, and the wave number $k_{end} = 2\pi/\lambda$. Thus, with the above data on the boundary $k_{end} = 9.8 \times 10^{-11} [m]$, it results $\theta = 1.6 \times 10^{-9}$. To note that this deformation is the same as in case of the lensing of Earth [12]

A such GW can deform the space around the nucleons, that any incoming outwards nucleon fall-in, that facilitates the eventually theirs fusion if is screened the Coulomb repulsion, like above, when the Xe lost theirs electrons.

### Casimir effect a proof of gravitons existence

The Casimir energy is $\frac{E}{A} = \frac{\eta c \pi^2}{720 a^3}$, for $A = a^2$ it results $E \approx \frac{\eta c}{a}$, where the $a \le 10^{-9} [m]$ is the distance between the two attractive metallic plates (fall-in), that corresponds with the results of our model when only one graviton release of $W^\pm$ as GW, we have GW energy of $E_{GW} \approx \frac{\eta c}{a}$.

### How the stress-energy tensor is effectively teleported to CFT-boundary?

The ground-state energy is called the zero-point energy. This simple example exposes that zero-point fluctuation is capable of carrying nonzero energy. And it is not only the harmonic oscillator but also other general interacting many-body systems that have zero-point fluctuation with nonvanishing energy in the ground state, like bosonic field etc. Each subsystem is fluctuating with nonzero energy density in the ground state.

In the ground state of an ordinary many-body system, like for a quantum field, there exists a quantum correlation called entanglement [18] among zero-point fluctuations of the subsystems. Zero-point fluctuations of the vacuum in regions A and B are correlated due to the kinetic term of its Hamiltonian ref. [22] cited in [15]. By virtue of the existence of entanglement, when local zero-point fluctuation is measured at a position, the measurement result includes information about quantum fluctuation at a distant position. This vacuum-state entanglement is at the heart of the quantum energy teleportation (QET) protocol with quantum fields, see [15].

This QET mechanism can be summarized using an analogy as follows. The zero-point energy, which will become the output energy $+E_B$ of QET, is analogous to the energy $E_{zero-point}$ saved in the locked safe underground. In QET, we get information about a key to the safe, allowing us to extract the zero-point energy by a remote measurement at A via the vacuum-state entanglement. However, we must pay the quantum fluctuation at A for this extraction. The cost is energy $E_A$, which is larger than the extracted zero-point energy $E_{zero-point} (= E_B)$ taken from the safe at B.

Therefore, a such infused energy (the cost pay) is due of QCD and is necessarily to extract **from quantum vacuum W$^\pm$ bosons pairs of the** bulk of the object, and the energy extracted $E_{zero-point} (= E_B)$ at CFT (object boundary-horizon) it results from the W$^\pm$ **decay** as gravitons release (GW).
These results are in range of the experimentally [4b] determined energy loss for ions with exit charge state $q_{out} = 2$ and $q_{out} = 4$ as a function of the incident charge state $q_{in}$, around 3.2keV for $q_{in} = 32$. The energy of the projectiles was kept constant for all $q_{in}$ at $E = 40keV$. Here is calculated that HCIs with initial charge states $q_{in} = 10, 20$ and $40$ capture ~9, ~17 and ~34 electrons during the passage through the graphene layer, respectively.

In the following, we can say that these experimental results legitimate more our previously works [7a] and the present one, when we substitute the ruly pseudo-potential (QM-code input) for hydrogenlike atoms (including alpha) with the *screened Coulomb potential* as deduced from the well-known Gamow alpha decay theory [7a].

Potential energy (chemisorptions) of a H atom over a fcc hollow site of a Pd(111) surface as a function of the distance from the surface for a frozen substrate, have been determined by DFT-GGA calculations as of ~0.23÷0.3÷0.2eV for distances ~1÷0÷1 Å, for a coverage of $\theta=1$, and of $\theta=1/3$.
In present work we done a similar calculation with the program **fhi98md** [5], [6].
In the following being calculated the electrical current as resulting from laser irradiation of these H(D) Pd coverage.

## The gravitational charge quantization

Now, to remember from author's work [15], $\sqrt{G}m_p$ is the gravitational charge defined by L. Motz as resulting from:

$$F_G = \frac{\sqrt{G}m_1 \sqrt{G}m_2}{r^2}, \text{ where it can say that the gravitational charge } q = \sqrt{G}m_1 \text{ is the source of gravitational field } \sqrt{G}m_1/r^2 \text{ at the distance } r, \text{ and that this field is coupled to the gravitational charge } \sqrt{G}m_2 \text{ at the position } r \text{ (relative to the source of the field) via the product of the field strength and the charge. L. Motz derived the quantization condition on gravitational charge in a similar manner by noting that moving particle with velocity } \mathbf{v} \text{ and with gravitational charge } \sqrt{G}m_p \text{ is coupled not only to the Newtonian gravitational fields of all other particles (the gravielectric field) in the usual way, but also to the Coriolis field } m\omega \times \mathbf{v} \text{ divided by gravitational charge } \sqrt{G}m_p, \text{ due of the inertial Coriolis force, see figure 2 (defined as } F_C = -2m_p\omega \times \mathbf{v} = -mv^2/r \text{), by means of the cross product } (\sqrt{G}m_p \mathbf{v}/c) \times (2\frac{\partial c}{\partial \mathbf{G}}/\sqrt{G}), \text{ see below.}$$

The gravimagnetic field, in GEM equations, we have: $F_e = q(E + \mathbf{v} \times B)$, or equivalently $F = m(E_g + 4\mathbf{v} \times B_g)$; $L = rmv$; the electrical field $E = \frac{q}{4\pi\epsilon_0 r^2}$; the gravitelectric field
$$E_g \cong g = \frac{F}{m} \Leftrightarrow 4\nu \times B_g[ms^{-2}] \rightarrow B_g \equiv \omega[s^{-1}]; \text{ the mass with the charge and the mass}$$

density with the charge density, so, the gravimagnetic field is $E_g = \frac{F_c}{q} = \frac{F_c}{\sqrt{Gm_p}} = \frac{\omega c}{\sqrt{G}}$.

This cross product term give rice to an angular momentum is proportional to moment of inertia $I$ and angular speed $\omega$, $L = I\omega$. Because $I = r^2m$ for a single particle and $\omega = v/r$ for circular motion, angular momentum can be expanded as $L = rmv$, or,

$L = rm_p v$ the component in the motion of the particle which is of the order of $L = 2m_p r^2 \omega$ and is parallel to the field $\omega$.

Therefore, it is obtained the quantization condition as: $L = 2m_p r^2 \omega = 2\eta$.

Since it was interested in the quantization of fundamental gravitational charge, it is found a value for $r$ and $\omega$ that must be associated with such charge. To do this, L.Motz considered the Universe to two such charges in gravitational equilibrium and resolving about each other in the first Bohr orbit. The radius of this orbit is just $r = \eta^2/Gm_p^3$, which it was taken as to be $r$. In this case (unlike the standard Coriolis force), the force is not fictional, but is due to frame dragging induced by the rotating body when $\omega \leftrightarrow \frac{2GL}{c^2r^3}$.

Since, for a spherical body the Coriolis field is the gravitomagnetic field $B_g$ near a rotating body can be derived from the GEM equations. It is exactly half of the Lense-

Thiirring precession rate is $B_g = \omega = \frac{GL}{2c^2r^3}$, and introducing into Coriolis force

expression $F_c = 2m\omega v = \frac{2mGL}{2c^2r^3}v = \frac{m^2G}{r^2} = F_G$; for $v=c$. He also noted that the resulting Coriolis field $2\omega c/\sqrt{G}$ must, according to Mach's principle, produces the centrifugal force $m\omega^2 c$. "A very general statement of Mach's principle is "*Local physical laws are determined by the large-scale structure of the universe*".

This will be so only if $v = \omega r \cong c$ is of the order of $c$ since the centrifugal force (being charge times field) is just $(\sqrt{Gm}) \cdot (2\omega c/\sqrt{G})$. Introducing these two relationships into the above quantization equation, is obtained $L = 2m_p c r = 2\eta$, or, substituting for $r$ as Bohr orbit when the electron is held in a circular orbit by electrostatic attraction, after some manipulations:

$$\frac{m \cdot m_e v^2 \cdot r^2}{r^2 \cdot r} = \frac{Z k_e e^2 \cdot m}{r^2} \quad ; m = m_p, k_e = 1/4\pi\epsilon_0 \quad ; \text{with}$$

$$mvr = n\eta, \quad (mvr)^2 = \eta^2; \quad n=1; \quad r = \frac{\eta^2}{k_e e^2 m}, \quad r = \frac{\eta^2}{k_e e^2 m} = \frac{\eta^2}{Gm^2 m}; \quad \text{when } e \rightarrow \sqrt{Gm_p} \text{ as in}$$

GEM. So, introducing in $L$ we have:
$$2m_p c \frac{\eta^2}{Gm_p^3} = \eta, \text{ or } \frac{Gm_p^2}{c} = \eta$$

With $m_p = 1.7 \times 10^{-8} \text{ kg}$; $\eta = 10^{-34}$, it results

$r_p = 3.3 \times 10^{-35} m \approx r_{\mu BH} = \lambda_c = \eta / mc = 2 \times 10^{-35}$, and for the electron

$m_e = 9 \times 10^{-31}; e = 1.6 \times 10^{-19} \text{ C}$ it results $r_e = 5.9 \times 10^{-11} \text{ m}$.

<figure><img src="image_2.png" /><figcaption>Figure 2. Coriolis field H</figcaption></figure>

## 2. The evaluation of Pd-D reactions with QM&MD programme: fhi96md

This it was demonstrated by using a code package **fhi98md** which is an efficient code to perform density-functional theory (DFT) total-energy calculations for materials ranging insulators to transition metals [5], [6], [7]. The package employs first-principles pseudopotentials, and a plane-wave basis-set, and is used to done a special calculus for some metals (Pd) where are depozided on the surface and implanted interstitially 1;2;3 H ions. The package **fhi98md** is an efficient code to perform density-functional theory total-energy calculations for materials ranging insulators to transition metals. The package employs first-principles pseudopotentials, and a plane-wave basis-set. For exchange and correlation both the local density and generalized gradient approximations are implemented.

In Polly-atomic systems as for example molecules, crystals, defects in crystals, surfaces, it is highly desirable to perform accurate electronic structure calculations, without introducing uncontrollable approximations.

The emergency functional the key variable in DFT is the electron density $n(r)$.

Consequently, the complementary tool from the all package FHIPP it was used to obtain the pseudo-potentials of H(D), Pd. Thus, is suficiently to substitute the ruly pseudo-potential (QM-code input) for hydrogenlike atoms (including alpha) with the screened Coulomb potential as deduced from the well-known Gamow alpha decay theory.
All that our suggest to use for hydrogen-like atoms (for other atoms are kept the original formulations) the pseudo-potential form, so neglecting the non-local contributions at the level of atom itself, respectively:

$$ = \frac{1}{[4\pi\epsilon_0]} \frac{Z_a Z_D e^2}{r} \quad \text{for } r \ge R $$

In [1] cited in [7], an explanation of the large screening was suggested by the Debye plasma model applied to the quasi-free metallic electrons. In this approach, they combined the Drude model of metals (with a kinetic energy 0.5 kT for the quasi-free valence electrons) with the Debye model of plasma: the Drude-Debye model, in short Debye model.

The electron Debye radius around the deuterons in the lattice is given by

$$ R_D = (\epsilon_0 kT / e^2 n_{eff} \rho_a)^{1/2} = 69(T / n_{eff} \rho_a)^{1/2} \text{ (m)}, \text{ with the temperature T of the quasi-free electrons in units K, } n_{eff} \text{ the number of thesis electrons per metallic atom and the atomic density } \rho_a \text{ in units of atoms m}^{-3}. \text{ For } T = 293 \text{ K, } \rho_a = 6 \times 10^{28} \text{m}^{-3}, \text{ and } n_{eff} = 1, \text{ is obtained a radius } R_D, \text{ which is about a factor 10 smaller than the Bohr radius of a hydrogen atom, therefore, we have } r = R_D \approx 0.01 a_0; a_0 = 5.29 \times 10^{-11} m. $$

The input is obtained with fhi98start applied to files start.inp and inp.mod, that obtaining inp.ini as an input to fhi98md plus the pseudopontetials fort for each species [7]. In figure 1(a,b,c,d) are presented the *quart-octahedral* holes in Pd lattice where are placed H(D) atoms. Thus, the main results of application are for: a),b),c) : respectively, simple *quart-octahedral* Pd4 we have : (non-eq) total energy = -78. a.u., for 1 D atom in the proximity of 1Pd (at 0.05Bohrs) of an *internal energy at zero temperature* = - a.u., see fig. 1(b)

<table><tr><td>Palladium</td><td>1</td><td>-2.6175</td><td>-2.6175</td><td>0.0000</td></tr><tr><td>Palladium</td><td>2</td><td>2.6175</td><td>2.6175</td><td>0.0000</td></tr><tr><td>Palladium</td><td>3</td><td>2.6175</td><td>-2.6175</td><td>5.2350</td></tr><tr><td>Palladium</td><td>4</td><td>-2.6175</td><td>2.6175</td><td>5.2350</td></tr><tr><td>Hydrogen</td><td>1</td><td>0.0000</td><td>0.0000</td><td>2.6384</td></tr></table>

The significative difference being: (-233 - (-78))x27.2=-4.18keV, for Pd-4 electrons-valence

Near 0.01Bohrs distance between H-H, fig.1c

Atomic positions tau0 :

<table><thead><tr><th>Species</th><th>Nr.</th><th>x</th><th>y</th><th>z</th></tr></thead><tbody><tr><td>Palladium</td><td>1</td><td>-2.6175</td><td>-2.6175</td><td>0.0000</td></tr><tr><td>Palladium</td><td>2</td><td>2.6175</td><td>2.6175</td><td>0.0000</td></tr><tr><td>Palladium</td><td>3</td><td>2.6175</td><td>-2.6175</td><td>5.2350</td></tr><tr><td>Palladium</td><td>4</td><td>-2.6175</td><td>2.6175</td><td>5.2350</td></tr><tr><td>Hydrogen</td><td>1</td><td>0.0000</td><td>0.0000</td><td>2.6175</td></tr><tr><td>Hydrogen</td><td>2</td><td>0.0000</td><td>0.0000</td><td>2.6280</td></tr></tbody></table>

When, the internal energy at zero temperature = -331 a.u. (Pd- 4 valence electrons)

The difference (-331a.u. - (-78.991727))x27.2=-6.8keV

Pd4H4-fig.1d with the "extracted Hs to be visible"

<table><tr><td>Palladium</td><td>1</td><td>0.0000</td><td>0.0000</td><td>0.0000</td></tr></table>
<table><tr><td>Palladium</td><td>2</td><td>5.2350</td><td>5.2350</td><td>0.0000</td></tr><tr><td>Palladium</td><td>3</td><td>5.2350</td><td>0.0000</td><td>5.2350</td></tr><tr><td>Palladium</td><td>4</td><td>0.0000</td><td>5.2350</td><td>5.2350</td></tr><tr><td>Hydrogen</td><td>1</td><td>2.6175</td><td>2.6175</td><td>2.6175</td></tr><tr><td>Hydrogen</td><td>2</td><td>2.6175</td><td>2.6175</td><td>2.6280</td></tr><tr><td>Hydrogen</td><td>3</td><td>2.6489</td><td>2.6175</td><td>2.6280</td></tr><tr><td>Hydrogen</td><td>4</td><td>2.6175</td><td>2.6489</td><td>2.6280</td></tr></table>

internal energy at zero temperature = -630.947097 a.u.

The difference (-630a.u. - (-78.991727))x27.2=-14.9keV

Pd-He-Hf-fig. 1eI and fig. 1eII with alpha particle (He<sup>+2</sup>) "extracted to be visible

Atomic positions tau0 :

<table><thead><tr><th>Species</th><th>Nr.</th><th>x</th><th>y</th><th>z</th></tr></thead><tbody><tr><td>Palladium</td><td>1</td><td>-2.6175</td><td>-2.6175</td><td>0.0000</td></tr><tr><td>Palladium</td><td>2</td><td>2.6175</td><td>2.6175</td><td>0.0000</td></tr><tr><td>Palladium</td><td>3</td><td>2.6175</td><td>-2.6175</td><td>5.2350</td></tr><tr><td>Palladium</td><td>4</td><td>-2.6175</td><td>2.6175</td><td>5.2350</td></tr><tr><td>Helium</td><td>1</td><td>0.0000</td><td>0.0000</td><td>2.6175</td></tr><tr><td>Hafnium</td><td>1</td><td>0.0000</td><td>0.0000</td><td>2.6594</td></tr></tbody></table>

internal energy at zero temperature = -4147.420020 a.u.

The difference (-4147 a.u. - (-78.991727))x27.2=-110keV

PdHe=fig. 1eAtomic positions tau0 :

<table><thead><tr><th>Species</th><th>Nr.</th><th>x</th><th>y</th><th>z</th></tr></thead><tbody><tr><td>Palladium</td><td>1</td><td>-2.6175</td><td>-2.6175</td><td>0.0000</td></tr><tr><td>Palladium</td><td>2</td><td>2.6175</td><td>2.6175</td><td>0.0000</td></tr><tr><td>Palladium</td><td>3</td><td>2.6175</td><td>-2.6175</td><td>5.2350</td></tr><tr><td>Palladium</td><td>4</td><td>-2.6175</td><td>2.6175</td><td>5.2350</td></tr><tr><td>Xenon</td><td>1</td><td>0.0000</td><td>0.0000</td><td>2.6384</td></tr></tbody></table>

internal energy at zero temperature = -475 a.u.

The difference (-475 a.u. - (-78.991727))x27.2=-10.7 keV

<figure><img src="image_3.png" /><figcaption>a)</figcaption></figure>
<figure><img src="image_4.png" /><figcaption>c)</figcaption></figure>

<figure><img src="image_5.png" /></figure>

<figure><img src="image_6.png" /></figure>
<figure><img src="image_7.png" /><figcaption>1e-I)</figcaption></figure>

<figure><img src="image_8.png" /><figcaption>1e-II)</figcaption></figure>
<figure><img src="image_9.png" /><figcaption>Fig.1. The quart-octahedral hole of Pd-H coordinates.</figcaption></figure>

<figure><img src="image_10.png" /></figure>

<figure><img src="image_11.png" /><figcaption>b) Fig.2 Paladium structure</figcaption></figure>

3. How can obtain an electrical current on metallic support of H/D LAYER or grapheme as induced by laser pulses
### 3.1 The bias current model of β-decay stimulation by a thermal spike of a photon

In order to accelerate the β-decay by a single photon reaction, a new model it was proposed in [7], [8], [10] to calculate a direct reaction of single photon with one of nucleon of the valence n-n; p-p; n-p pairs (see IBM model [30,31]cited in [7]) of the nucleus, that being in the unstable state (a β-decaynuclide), they are the most susceptible to react with the photon, see some of model's results from [7], figures 1÷4, respectively.

The interaction between a photon of high energy and of low band width $\Delta E/E \le 10^{-3}$ and of nucleon into state of excitation has been characterized by the beta decay energy $Q_β$ from the nuclei, that is viewed as a direct reaction, without the formation of a compound nucleus.

Essentially, the general picture of this model described in details in [8], [9], [10a], [10b] is that the vortex (boson $W$, figure 3.) crossing may trigger the $s \to n$ transition. A photon makes this process much more probable by creating a spot (melt) with suppressed order parameter and thus with lower energy barrier for vortex crossing. A sketch of the strip and of the belt across are shown in Figure 4, the induced vortex crossing together with an electron (travel current $e^\pm$), which turns superconducting hot belt into the normal state resulting in a vortex assisted photon beta decay.

As a consequence of the Lorentz force acting on a vortex crossing a thin and narrow current-biased strip the energy $\Phi_0 I/c$ is released, which for currents $I > 0.6$Ic suffices to create a normal belt across the entire width $w$ of the strip (extending to a few correlation lengths $\xi$ along the strip).

Therefore, by using the same nucleon model we can account for a vortex (W boson) assisted photon count rate, as in [8], [9a], [9b]:

$$R_{pc} = R_h [1 - \exp(-\Re)]$$

where:

$$R_v(I, v_h) = \frac{4k_B T R_{eff}}{\pi \Phi_0^2} \frac{\xi}{w} \left(\frac{v_h}{2\pi}\right)^{1/2} \left(\frac{I}{I_{ch}}\right)^{v_h+1} \\ \Re \approx \frac{\tau_{GL} R_v}{\nu_0 - \nu_h} \frac{1 - (I_{ch}/I)^{v_h - \nu_h - 1}}{\ln(I_{ch}/I)}$$

The current being

$$I_{ch} = I_{c0} (v_h / \nu_0)^{3/2} \\ I_{c0} = \frac{\pi c^2 \Phi_0 \epsilon_0}{8 (2\pi^3 x \lambda^5)^{1/2}} e^{-x/\lambda} \times 10^{15} [A/fm^2]$$

The effective ohmic resistance is $R_{eff} = R_s R/(R_s + 2\pi(\xi/w)^2 R)$.
We can suppose than along the hot belt induced by the incident photon, the charge $e^{-} + W^{-}$ creates a bias current ($I > 2/3e[(1/(v_{h}/v_{0})^{3/2}) \approx 3] \Rightarrow 2e$, see below, who circulates due of the potential difference between the vortex and the rest of isotope.

At the first sight, the ohmic resistance of this ad-hoc electrical circuit created by the bias current is given as:

$$R^{-1} = \frac{U_{\beta}}{\tau_{GL}} \frac{1}{V_{vortex}^{2}}$$

, where the vortex potential is $V_{vortex} = H_{0}\xi$,

$H_{0}$-an “external” electro-magnetic field of a dipole created by the pair $u\bar{u}$ (the chromoelectrical field)

$$H_{0} = E_{0} = \frac{de}{4\pi\varepsilon_{0}r^{3}} = 8.33e24\left[\frac{N}{C}\right]$$

, where, $r \approx 0.05[fm]$-is the electrical flux tube radius, $d = 0.7[fm]$-the distance between the two quarks charges, usually $H[A/m]$, but here is used as $B = \mu_{0}H\left[\frac{J}{Am^{2}}\right]$,
, and the characteristic distance $\xi \le \lambda$, the coherence length,

and the power is $U_{\beta}/\tau \approx \varepsilon_{vortex(W^{\pm})}/\tau$, with $\tau_{GL} = \pi\eta/(8k_{B}T_{c}) = 1.5e-24[s]$-the Ginzburg-Landau life time of $W^{\pm}$ bosons.

Numerically, with $T_{c-11} = 5 \times 10^{11} K$ (ELI laser); and $T_{c-9} = 10^{9} K$ (Nd:YAG laser), $E_{prag-11} = k_{B}T_{c-11} \approx 43MeV$; $E_{prag-9} = 0.09MeV$ , result $v_{0} = \varepsilon_{W}/k_{B}T_{c} = 1.0e-09/(1.38e-23*2.12) = 36.2$ , where $T_{c} = 10^{12} K$ at confinement, and where $\varepsilon_{W}$ results from eq. (2) from [8] as $\varepsilon_{W} = \varepsilon_{int}(d = x - \lambda; x = 0.14) * 0.11[fm] \approx 1.0e-09[J]$; $R^{-1} = 143\Omega$; $R_{s} = 1000\Omega$; $R_{eff} = 141\Omega$, $w \approx 1fm$.

With $\Phi_{0} = 2 \times 10^{-15} [Tm^{2}]$; $\lambda = 0.117 fm$; $x \approx \lambda$ , it results $I_{c0} \approx 3.6 \times 10^{7} [A/fm^{2}]$, $I_{cow} = I_{c0} \cdot W_{rate} = 3.6 \times 10^{7} \cdot 10^{-8} = 0.36 [A/fm^{2}]$

$$v_{h-11} = \varepsilon_{0h} / E_{prag-11} = 5 \times 10^{-11} / E_{prag-11} = 6.9,$$

$$v_{h-9} = \varepsilon_{0h-9} / E_{prag-9} = 9.6 \times 10^{-14} / E_{prag-9} = 6.9; \text{ that results } R_{v} = 3.7 \times 10^{18} s^{-1}$$

The total power per pulse of Nd:YAG laser is $P_{tot} = 9.6 \times 10^{-14} \cdot 5 \times 10^{13} = 4.6[J]$

$I_{ch} = I/I_{c0} \cdot 3.6 \times 10^{7} \times \text{area} \times \text{rate}_{W^{\pm}} = 5 \times 10^{6} \cdot 10^{-8} = 50[\text{mA}], \text{area} = 1\text{fm}^{2}$; the production rate of $W^{\pm} = 10^{-8}$, $U = R_{eff}I \approx 7[V]$; $I/I_{c0} = 0.138$ as obtained by trials (figure 5), and where $\varepsilon_{0h}$ is obtained by using the lower critical field

$$B_{0} = H_{c1} = \frac{2\Phi_{0}}{2\pi\lambda^{2}} \log\left(\frac{\lambda}{\xi}\right) = \frac{\pi\eta c}{\pi\lambda^{2}c} \log(\kappa) = 1.15\left[\frac{J}{Am^{2}}\right]$$

, and with $x \approx \xi = 0.10[fm]$;
respectively: $\epsilon_{0h} = Vc^2\epsilon_0(2H_{c1})^2/8\pi = 5.3e-1 \text{ [J]}$

In our first case of muons born due of hot spot $v_h = 330\,\text{MeV}/E_{\text{prag}} = 7.68$ which is obtained by trials. Thus we obtain the average (dc) voltage $V_{dc} = \Phi_0/c \cdot R_v \rightarrow 7400\,\text{V}$.
We obtain

$\tau_0 \approx d^2 \Phi_0/(2\pi\xi^2 2cR_{\text{eff}} I)$, $I=50\text{mA}$, $d=1\text{pm}$, $\tau_0 = 9\text{ns}$, the time-of-flight. In fact, this estimate coincides with the time it takes a vortex to cross the strip being pushed solely by the Lorentz force. With these it results $R_v=10^{18}/\text{s}$.

The value of $E_{\text{prag}}$ is determined by trials in order to have $R_{\text{pc}}/R_h \cong 1$, see figure 5.

The model results show that in order to have instant rates(100% decay), or a beta decay rate of $R_{\text{pc}} = 5 \times 10^{13} \text{ counts/s}$, with the incident of single photons rate of $R_{\text{ph}} = 5 \times 10^{13} \text{ counts/s}$, $R_{\text{pc}}/R_h \cong 1$, for all beta-decay isotopes, i.e. these rates are not dependent of the nuclide type, the photons energy needs to be above a threshold energy value of very precise value $5 \times 10^{11} K \rightarrow 43\,\text{MeV}$, but in this case we don't have a net energy gain due of the small current $I_{\text{ch}} = 5 \times 10^{-2} \text{[A]}$ due of permanent rate of bosons as the source for further electrons (muons) production by Schwinger effect $W^{\pm} \cong 10^{-8}$ inside the nucleon (see section 3.2 below), which can decay into heavy-electrons (muons) but more sure into electrons, and also is released the remnant graviton of the inside nucleon primordial $\mu\text{BHs}$ pair as a QET, as to be later extracted as GW that generates the curvature radius (Einstein Ring).

In the second case of Nd:YAG laser when $T \cong 10^9 K$, the power released by the electrical current is $P_w = R I_{ch}^2 = 0.35[w]$ for each deuteron spike, but for an entirely laser wave (10$^{13}$photons) when pass along a $30\,\mu\text{m}$ (Nd:YAG) it means a total power $P_{\text{tot}} = 0.35 \times (30\,\mu\text{m}/8\,\text{Bohrs})^2 \cong 2 \times 10^9[w]$, for an absorption coverage area of Pd/H~1; 1 Bohr=5.29x10$^{-11}$[m], or about $\sim 2 \times 10^9/10^8 \cong 20$ than the power used to produce a such laser spot. These values are in case of ELI-laser when: the power is $\sim 2\text{Pw}$, the flux $\sim 10^{13}\text{ph/s}$; flux$\sim 10^{24}\text{w/m}^2$, or the extracted power is lower$\sim 1.4 \times 10^8/2 \times 10^{15} \cong 0.7 \times 10^{-7}$ times that used by the laser, or without any energy gain. But we can obtain a net energy gain if we use, for example, a smaller laser Nd:YAG when the flux $\sim 10^{13}\text{ph/s}$, or $T = 10^9 K$, but not smaller than this lower limit value, since the current through the "hot belt" is the same i.e. $\sim I_{ch}$.

A Nd:YAG laser with pulse energy of $< 0.2 \text{ J}$ is recommended to be used, with 5 ns pulses (or $P_w = 0.1 \text{ J}/5 \text{ ns} = 10^8[w]$) at 532 nm and normally 10 Hz repetition rate, of 0.2 J pulses with 5 ns pulse length ejects ions with energies in the MeV range. The ns-resolved signal to a collector can be observed directly on an oscilloscope, showing ions arriving with energies in the range 2-14 MeV/u at flight times 12-100 ns, mainly protons from $n \rightarrow p$ transformation, and deuterons ejected by proton collisions. An expected signal at several mA peak current corresponds to $1 \times 10^{13}$ particles released per laser shot and to an energy release $> 1 \text{ J}$ assuming isotropic formation and average particle energy of 3MeV as observed, or $\cong 1 \text{ J}/10^{13} \cong 10^{-14} \cong \epsilon_{0h,9} = 9.6 \times 10^{-14} \text{ [J]} \rightarrow 0.6 \text{ MeV}$, and
$\tau \approx \eta/10^{-14} J \approx 10^{-20} s$, or in term of pulse duration

$\tau = 5ns/10^{13} \approx 5 \times 10^{-22} [s] \rightarrow \tau_{W^{\pm}} = 3 \times 10^{-25} s$, see the next section.

The results from our calculation, in fact the proton it could be due of the transformation a neutron of deuteron into proton ($n \rightarrow p+e^-$) by the laser stimulated beta decay, so there is “not any fusion”, see below. If an experiment it will be done, that it will be for the first time when the findings confirm author’s models [7], [8], [10], the most important are, the particles energy in MeV range, and the electrical current of ~few mA, remaining to be confirmed the number of particles, respectively of ~$10^{10}$ per laser shot.

This vortex-assisted mechanism may be verified by application of magnetic fields, which effectively enhance $I_{ch}$ along with the vortex crossing rates but do not affect the creation of hot spots by photons.

<figure><img src="image_12.png" /><figcaption>Fig.3. Abrikosov's triangular lattice for a nucleon (author's proposal [8], [9])</figcaption></figure>

<figure><img src="image_13.png" /></figure>
Fig.4. The photonuclear mechanism. From left to right, illustration of incident photon creating superconducting hot spot (hot belt) across nucleon, followed by a thermally induced vortex crossing together with an electron (*bias current*), which turns superconducting hot belt into the normal state resulting in a vortex assisted photon beta decay.

<figure><img src="image_14.png" /><figcaption>FIG.5. The vortex-assisted photon count rate $R_{pc}/R_h$ vs. *bias current*.</figcaption></figure>

## 3.2. The free neutron decay calculation

In the following, we will use some results of section 4.1a from [8] when the Compton length is $\lambda_c = \eta/m_*c = 2.3e-18 [m]$, the effective mass is

$$m_* = 1.44 \times 10^{-25} \text{ kg} \rightarrow V = 81 \text{ GeV} \rightarrow E = 1.1 \times 10^{28}, \text{ the critical field}$$

$$\text{being } E_c = \frac{m_*^2 c^3}{e\eta} \approx 3.5 \times 10^{28} > E = 1.1 \times 10^{28} [N/C]; \quad B = E/c = 3.7 \times 10^{19} [T].$$

From the section (4.1b) of [8], are used the bosons $W^\pm$ pairs generated inside the nucleons as due of one quark $u\bar{u} \leftrightarrow u$ as a resultant of $\times 3$ flux tubes vortex potential, see figure (1.b) of [8], respectively $\varepsilon = mc^2 \approx 81 \text{ GeV}$- which after the release of an electron that it getting the final beta energy as been equally to the out of barrier turning
point after the tunneling, and accounting for the valence nucleons interactions (shell-energy levels). The number of assaults of the barrier, like in Gamow theory [20, 21] cited in [6] is $n_a = v_b/R_{inner}$; where the velocity is $v_b \approx (2\varepsilon/m)^{1/2} = 2.9 \times 10^8$ m/s, where, the inner radius of the barrier is $R_{inner} \approx b = 3.5 \times 10^{-17}$ [m], see below. For only one of the three vortex-flux tubes ($q\bar{q}g$) we have: $\varepsilon = \eta B/m \approx 4 \times 10^{-9}$ [J] $\rightarrow \approx 25$ GeV, with the above (B) which is obtained from eq.(1.a) from [8] with the resultant potential $V = 81$ GeV, that corresponds to $m_{q\bar{q}} \approx 29$ GeV from 4.1a of [8], the energy of the particle for the first Landau level (as above), and we can see that it results to be equally with $\approx 1/3$ rest mass of the $W^\pm$, that resulting $n_a \approx 7.5 \times 10^{24}$ s$^{-1}$.

In case of WKB [20] cited in [6], the transmission coefficient is $T = 2 \frac{\sqrt{2m|V-Q|}}{\eta} \Delta r$, and the decay constant $\Gamma = n_a e^{-T}$.

For the thick barrier the transmission coefficient is $T = 2\pi \frac{Qb}{\eta\nu} = 2\pi \frac{\sqrt{2mQ}}{\eta} b$; where, the kinetic energy of the particle after the barrier at b

is $Q \equiv \frac{1}{2}mv^2$, $b = d_b/2\pi = 3.5 \times 10^{-17}$ [m], see below, that results $T = 63$; and the decay constant $\Gamma = 3 \times 10^{-3}$ s$^{-1} \rightarrow \approx 324$ s

To “materialize” a virtual $e^+ - e^-$ pair in a constant electric field E the separation d must be sufficiently large $eEd = 2mc^2$

Probability for separation d as a quantum fluctuation

$$P \propto \exp\left(-\frac{d}{\lambda_{Compton}}\right) = \exp\left(-\frac{2m^2c^3}{e\eta E}\right) = \exp\left(-\frac{2E_{cr}}{E}\right)$$
The emission (transmission through barrier) is sufficient for observation when $E \approx E_{cr}$, with $Q = 1/2mc^2$, results $T = 2\pi \frac{mcb}{\eta} \approx \frac{2\pi b}{\lambda_c}$, or $b \approx d_b/2\pi$.

Now, by using the Schwinger effect as in section 2.1 of the companion author's paper [8], the number of $W^\pm$ pairs produced inside the nucleon (more inside of the only one resultant flux tube, see figures 1.a; 1.b from [8]) due to the potential resultant $u\bar{u} \leftrightarrow 3 \times \text{vortex}(q\bar{q}g)$ of $V = 80$ GeV, results as $R/s = R/V \times V_{vol} = 2.3 \times 10^{18}$ s$^{-1} \approx n_a$, where $R/V = 2 \times 10^{71}$ [1/m$^3$s] and the volume is $V_{vol} \approx (\lambda_c)^3 \approx 1.24 \times 10^{-53}$ [m$^3] \ge V_b$, the penetration length being the Compton length $\lambda_c = 2.3 \times 10^{-18}$ [m], and for a four-volume of $V_{Compton} = \lambda_c^4/c \approx 9.5 \times 10^{-80}$ [m$^3$s], that results a *permanently rate* $R \approx R/V \cdot V_{Compton} = 10^{-8} W^\pm$ pairs. Thus, it results a main conclusion of this investigation, namely, that the “interacting” potentials inside the nucleons are that were already established in [8], respectively $\approx 80$ GeV around the valence quarks (u,d) which it
seems to be “locked” at the electroweak symmetry breaking (≈ 100GeV); that of the Giant Vortex (see the insert in fig. 1.a from [8]) at the center of the triangle-the Higgs boson $H = 125\,GeV$; and that resulting from interaction of 2× inter-pairs of flux tubes as been the neutral boson $Z = 90GeV$.

Therefore, in other words is proved that all the time inside the nucleon are available $10^{-8}W^{\pm}$ pairs that seems to correspond to the “weak interaction” coupling constant $10^{-7}$, which is absorbed or emitted by the quarks, resulting an $e^{+}$, or $e^{-}$ which help the quarks transformation like ($u \rightarrow d$), respectively ($d \rightarrow u$) for beta-decay. In our understanding, the created electron takes the energy at the *turning point* out of the barrier equally with the electron itself for unbounded neutrons, or that of the binding energy of nucleon in isotope nucleus, when it *passes* the barrier of gluon condensate characterized by an *quantum tunneling* suppression given as:

$\exp(-\Delta E \tau/\eta) \approx 7.3 \times 10^{-22}$, where, as the lifetime of $W^{\pm}$ being $\tau \approx 3 \times 10^{-25} s$. Here, $\Delta E$ corresponds to the height of gluon condensate barrier, due of the *phase slip* with $2\pi - \varphi$ and of a $\Phi_0$ energy release as: $\Delta E = c^2 \Phi_0^2 \epsilon_0 / d_b$; $d_b \approx k\lambda_c = 1.98 \times 10^{-16}$, $k = 85$, where the Compton length is just the penetration length for $W^{\pm}$ pair $\lambda_c = 2.3 \times 10^{-18} [m]$, or in other words just the barrier size, and $\Delta E = 1.6 \times 10^{-8} [J] \rightarrow 100 \, GeV \approx 3 \times 25 \, GeV$ as for 3 sea quarks color flux tubes, see figures 1.a; 1.b. The value of the resulting flux tube it remains as in (4.2.a) of [8], respectively of 0.4GeV as the string strength.

Thus, the probability (rate) to produce $W^{\pm} \rightarrow e^{\pm}$, into a more simple way- without the external interactions of the neutron (free-not bounded), is given as:

$RV \exp(-\Delta E \tau/\eta) \approx 1.7 \times 10^{-3} s^{-1} \rightarrow \tau_{1/2} \approx 58 [s] \approx 612 s$, that corresponds for *free neutrons decay* ($\beta^{-}$) by emission of an electron and an electron antineutrino to become a proton $n^0 \rightarrow p^+ + e^- + \bar{\nu}_e$, with half-life of 611s, and $Q_{\beta^{-}} = m_e v^2 = 0.5 \, MeV$.

In the classic understanding of $\beta^{-}$ disintegration $n \rightarrow p + e^{-} + \bar{\nu}_{e}$, in ours understanding this occurs when one of the down quarks ($d$) in the neutron (*udd*) transforms into an up quark (*u*) due of interacting with the charge of $W^{+}$ boson of the pair $W^{\pm}$, transforming the neutron into a proton (*uud*). In mean time the other part of this pair $W^{-}$ boson decays into an electron and an electron antineutrino *udd* → *uud* + $e^{-} + \nu_{e}$. Probable the claimed energy of boson $W^{-}$ is the same as to be the necessarily energy to traverse the gluonic barrier, when it decays into $e^{-}$ at the end.

*The free neutron decay*

Consequently, for the $\beta^{-}$ decay process, the energy combines well with the existing one, that releasing an electron which penetrates the barrier:

$$d \rightarrow u + W^{+} + W^{-} \rightarrow u + e^{-} + \bar{\nu}_{e} \\ d(-1/3e) + e^{+} (+3/3e) = u(+2/3e) + e^{-} (-3/3e) \\ \text{, since } W^{-} \rightarrow e^{-}, \text{ and } W^{+} \rightarrow e^{+}$$

In case of $\beta^{+}$ decay, it can only happen inside nuclei when the daughter nucleus has a greater *binding energy* (and therefore a lower total energy) than the mother nucleus. The
difference between these energies goes into the reaction of converting a proton into a neutron, a positron and a neutrino and into the kinetic energy of these particles.

Thus, an opposite process to the above negative beta decay, $\beta^+$ decay of nuclei (only bounded proton) when $p \to n+e^+ + \nu_e$, or energy+$uud+W^++W^- \to udd+e^+ + \nu_e$, or, $u(2/3e)+e^-(3/3e)+\text{energy}=d(-1/3e)+e^+(3/3e)$.

For free proton decay an added energy it seems to be necessarily to reduce the barrier width to $d_b = 9 \times 10^{-17} [m]$, when the production rate is:

$RV \exp(-\Delta E \tau/\eta) \approx 7 \times 10^{-29} s^{-1} \to \tau_{1/2} \approx 10^{28} [s]$, respectively, an increase to $\Delta E = 3.5 \times 10^{-8} [J] \ge 225 \text{GeV}$ from $\Delta E = 1.6 \times 10^{-8} [J] \to 100 \text{GeV}$, as for the free neutron, or near $\ge \nu_{e.v} = 247 \text{GeV}$, like at LHC when the gluonic “cover” of protons it was “melted (at least 2 gluons)”, and the resulted difference ($\approx 225-100=125 \text{GeV}$) being just that of the Higgs boson (a quanta of energy!) which it was, in this spectacular way “released” [10] as $2g \to 2\gamma$.

In the process of electron capture, one of the orbital electrons, usually from K or L electron shell, is captured by a proton in the nucleus, forming a neutron and an electron neutrino.

$$p+e^- \to n+\nu_e$$

About others calculations of beta decay processes of different isotopes, see the author's work [8].

## 4. Conclusions

In the work by using QM&MD programmes: fhi96md is confirmed a high coverage H/Pd~1.

Thus, there are used the prior author models of vortex assisted photon beta decay, when a laser photon makes this process much more probable by creating a spot (melt) in nucleon with suppressed order parameter that lowering the energy barrier for vortex crossing together with a heavy electron (bias current $e^\pm$) as resulting from the decay of the of bosons pairs rate $W^\pm \approx 10^{-8}$ as produced inside nucleons by a Schwinger effect. The electrical power results as $P_w=2 \times 10^9 \text{ w} \ll P_{\text{laser}}=2 \text{Pw}$ for a laser spot of size $30 \mu\text{m}$, that corresponds with ELI laser characteristics, or without a any net gain of energy.

For the prior author's models validation it is necessarily to do laboratory experiments in the spirit of this work, namely based on the use of a Nd:YAG laser with pulse energy of

< 0.2 J, with 5 ns pulses (or $P_w = 0.1 \text{J}/5 \text{ns} = 10^8 [\text{w}]$) at 532 nm and normally 10 Hz repetition rate. Thus in this case of much smaller power lasers when per photons is obtained ~$10^{-5} \text{w} \times 1 \text{ns}(T=10^9 \text{K}) \sim 10^{-14} \text{J}$ of duration ~$1 \text{ns}$ and, respectively ~$10^{-14} \text{J} \cdot 10^{13}$ ~$0.1 \text{J} \sim 10^8 [\text{w}]$ for a pulse composed of ~$10^{13} \text{ph/s}$. The net energy is much higher of
$2 \times 10^8/10^8 = 20$. In such experiments is expected to obtain a particles energy in MeV range, a current of ~few mA, a voltage on the shunt 7[V], and the time-of-flight $10^{-8}$ s.
If these electrons are collected either in the metallic plate in serried into an electrical circuit, or into a spherical conductive cover, we can constitute a reliable source of direct electricity with the period equally that of laser pulse frequency. It is possible that the neutron of D do not transforms into proton, since the *open hot belt* created due of the laser photon incidence to *close* after electron passage, therefore it is not a consume of D.
To obtain a D LAYER the author calculated that is necessary a thermal energy of 2-3eV to be deposited on the Pd plate in the vacuum chamber containing the D gas, this being in serried into an electrical circuit. If is used the co-deposition Pd/D technique, are obtained also cold fusions together with an electrical current, that due of electrons deep screening of deuterons in lattice, as explained by the author early.

It is also found that as in case of other cosmic objects, the nucleons have an Einstein Ring around 1Bohr radius, that if we realize a screening till this radius and further, after that the neighborhood ion fall-in the Einstein Ring that enhances the packing into a atom's nucleus (nucleosynthesis).

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