From the post "A Shield" and "A \(\Psi\) Gun" both dated 27 May 2016,
\(f_{res}=0.061\cfrac { c }{ a_{\psi} }\)
\(a_{\psi}\) is driven to oscillate about \(x=\cfrac{\pi \sqrt { 2{ mc^{ 2 } } } }{G}\). The work done in moving \(\psi\) forward is given by,
\(\int_{\pi-A_w}^{\pi+A_p}{tanh(x)}dx\)
On the return, at position \(x\), \(\psi\) radiates this energy gained. The energy radiated is,
\(X=\int_{\pi-A_w}^{\pi+A_p}{tanh(x)}dx-\int_{\pi-A_w}^{x}{tanh(x)}dx\)
\(X=\left[ ln(cosh(x)) \right] _{ \pi -A_{ w } }^{ \pi +A_{ p } }-\left[ ln(cosh(x)) \right] _{ \pi -A_{ w } }^{ x }\)
\( X=X_{ o }-ln(cosh(x))\)
with, \( X_{ o }=ln(cosh(\pi +A_{ p }))\)
A plot of log(cosh(pi+2))-log(cosh(x)) shows an almost linear decrease in energy radiated given the position \(x=a_{\psi}\),
\(U_{1/2}\) marks the value of \(a_{\psi}\) where work done travelling on the left and right side are equal.
We would also expect an almost linear increase in \(X\) as the amplitude \(A_p\) is increased at any given position, for example \(x=\pi\). ie
\( X_{\pi}=ln(cosh(x_{1/2} +x))-ln(cosh(x_{1/2}))\)
is almost increasing linearly. In addition, when \(a_{\psi}\) drops below
\(n.2\pi f_n a_{\psi}=c\) --- (*)
where the particle(s) is in resonance along \(2\pi a_{\psi}\) with \(n\) wavelengths along the perimeter of a circle of radius \(a_{\psi}\), a radiation peak occurs as \(a_{\psi}\) move to a lower energy state \(n\rightarrow n-1\).
In particular, the driving field is affecting the inner electron clouds that surrounds the nucleus below the orbiting \(T^{+}\) particles as described in the post "The Rest Are In The Clouds" dated 16 Apr 2016.
The following table is the X ray emission data using high energy electron bombardment,
| Element | ave \(E\,\alpha\) | \(E\,\alpha1\) | \(E\,\alpha2\) | \(E\,\beta\) | n | \(a_{\psi}1\) | \(a_{\psi}2\) |
| Cr2,8,13,1 | 2.29100 | 2.28970 | 2.29361 | 2.08487 | 11 | 0.0331517775 | 0.0331858844 |
| Fe2,8,14,2 | 1.93736 | 1.93604 | 1.93998 | 1.75661 | 10 | 0.0308378963 | 0.0310675679 |
| Co2,8,15,2 | 1.79026 | 1.78897 | 1.79285 | 1.62079 | 10 | 0.0284964345 | 0.0286654428 |
| Cu2.8.18.1 | 1.54184 | 1.54056 | 1.54439 | 1.39222 | 9 | 0.0272691257 | 0.0277007991 |
| Mo2,8,18,13,1 | 0.71073 | 0.70930 | 0.71359 | 0.63229 | 8 | 0.0141412915 | 0.0143778083 |
\(a_{\psi}1\) and \(a_{\psi}2\) are calculated data.
The split into \(E\,\alpha1\) and \(E\alpha2\) is consistent with the split in the solution for \(a_{\psi}\) in the post "Two Quantum Wells, Quantum Tunneling, \(v_{min}\)" dated 19 Jul 2015. \(E \beta\) is due to the energy state available at the next lower level, \(N\), where the electron cloud contains \(N\) particles.
We estimate the value of \(N\), the energy state of the electron cloud and \(N+1\) the next higher state after the electron accepts a particle by taking the ratio of,
\(\cfrac{ave.\,E\alpha}{E\beta}=\cfrac{N+1}{N}\)
since,
\(N.2\pi a_{\psi}=\cfrac{c}{f_n}=\lambda_\beta\)
\(\cfrac{\lambda_{\alpha\,\,ave}}{\lambda_{\beta}}=\cfrac{ave.\,E\alpha}{E\beta}=\cfrac{N+1}{N}\)
Using the average value of \(E\alpha1\) and \(E\alpha2\) and \(E\beta\) to find \(a_{\psi1}\) and \(a_{\psi2}\) respectively. And for the inner electron cloud,
\(a_{\psi}=\cfrac{\lambda_{\beta}}{N.2\pi }\)
For \(Cu\),
\(N=9\)
The inner electron cloud has a radius of \(a_{\psi}=0.02447*10^{-10}\)
which is one fifth the size of the atomic radius at 1.45 A. This lead us to the resonance frequency,
\(f_{res}=0.061\cfrac { 299792458 }{ 0.02447*10^{-10}}=7.473*10^{18} Hz=7.473 EHz\)
needed to drive \(a_{\psi}\) to resonance. Given an elementary electron charge of 1.602176565*10^{-19} C, a bombardment at \(f_{res}\) is
\(I_{res}=q_e.f_{res}=1.602176565*10^{-19}*7.473*10^{18}=1.1974 A\)
This radiation due to the excitation of the inner electron clouds is in the \(X\) ray region. If this is true, with \(f_{res}\), \(X\) ray production is safer and cheaper... Hurrah! Just before you electrocute yourself, that's 1.1974 A per electron cloud. Hurrah!
The good news is any integer division of \(f_{res}\) will still set the system into resonance but slowly. For example,
\(f=\cfrac{f_{res}}{1000}\)
will still resonate but has a slow buildup.
Note: As the atomic size increases the inner electron cloud is compressed to a smaller radius.
