\(f_{res}=0.061\cfrac { c }{ a_{\psi} }\)
\(a_{\psi}\) 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 \(n-1\).
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, \(n-1\).
We estimate the value of \(n\) and the next lower energy state \(n-1\) by taking the ratio of,
\(\cfrac{ave.\,E\alpha}{E\beta}=\cfrac{n}{n-1}\)
since,
\(n.2\pi a_{\psi}=\cfrac{c}{f_n}=\lambda_n\)
\(\cfrac{\lambda_n}{\lambda_{n-1}}=\cfrac{n}{n-1}\)
Using the average value of \(E\alpha1\) and \(E\alpha2\) and \(E\beta\) to find \(a_{\psi1}\) and \(a_{\psi2}\) respectively.
\(a_{\psi}=\cfrac{\lambda_n}{n.2\pi }\)
For \(Cu\), 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.
