\(N.2\pi a_{\psi}=\lambda_\beta\)
was formulated before \(a_{\psi\,c}\) was discovered. We might have,
\(\sqrt[3]{N}.2\pi a_{\psi\,c}=\lambda_\beta\)
from,
\(\sqrt[3]{\cfrac{1}{N}}=\cfrac{a_{\psi\,c}}{a_{\psi\,\small{N}}}\)
and so,
\(\cfrac{ave.\,E\alpha}{E\beta}=\sqrt[3]{\cfrac{N+1}{N}}\)
From which we may find \(a_{\psi\,c}\), but for the resonance of the electron cloud that we seek, we need the electron cloud \(a_{\psi\,cloud}\),
not \(\sqrt[3]{N}.a_{\psi\,c}\) nor \(a_{\psi\,\,c}\)
The rationale for making the estimate \(N\) and \(N+1\) is that given,
\(N.2\pi a_{\psi}=\lambda_\beta\)
and
\((N+1)2\pi a_{\psi}=\lambda_{ave\,\,\alpha}\)
no matter what \(a_{\psi}\) is, the addition of one more particle pushes \(a_{\psi}\) outwards and the allowable energy states splits into two narrowly spaced energy states. When the resonance frequency is applied using an appropriate current, the particles injected resonate. X-ray radiations at the \(E\alpha\) values will dominate and is adjusted as the current is reduced by an integer divisor.
This way, there is no need to assume that \(a_{\psi}=a_{\psi\,c}\). The only assumption is that the \(E\alpha\)'s and \(E\beta\) are consecutive, due to the injection and emission of one particle \(a_{\psi}\).
Furthermore,
\(N.2\pi a_{\psi}=\lambda_\beta\)
\((N+1)2\pi a_{\psi}=\lambda_{ave\,\,\alpha}\)
\((\cfrac{\lambda_\beta}{2\pi a_{\psi}}+1)2\pi a_{\psi}=\lambda_{ave\,\,\alpha}\)
\(\lambda_\beta+2\pi a_{\psi}=\lambda_{ave\,\,\alpha}\)
If \(f \propto \cfrac{1}{m}\) (the approximation part) such that,
\(m.f=\,\,constant\)
\(m_\beta c.f_{\beta}.\lambda_\beta+m_\psi c.f_{\psi}.2\pi a_{\psi}=(m_\beta+m_\psi) c.f_{ave\,\,\alpha}.\lambda_{ave\,\,\alpha}\)
that the total work done in going around \(\lambda_\beta\) and around the additional particle \(a_{\psi} \), is the work done going around \(\lambda_{ave\,\,\alpha}\).
Furthermore,
\(N.2\pi a_{\psi}=\lambda_\beta\)
\((N+1)2\pi a_{\psi}=\lambda_{ave\,\,\alpha}\)
\((\cfrac{\lambda_\beta}{2\pi a_{\psi}}+1)2\pi a_{\psi}=\lambda_{ave\,\,\alpha}\)
\(\lambda_\beta+2\pi a_{\psi}=\lambda_{ave\,\,\alpha}\)
If \(f \propto \cfrac{1}{m}\) (the approximation part) such that,
\(m.f=\,\,constant\)
\(m_\beta c.f_{\beta}.\lambda_\beta+m_\psi c.f_{\psi}.2\pi a_{\psi}=(m_\beta+m_\psi) c.f_{ave\,\,\alpha}.\lambda_{ave\,\,\alpha}\)
that the total work done in going around \(\lambda_\beta\) and around the additional particle \(a_{\psi} \), is the work done going around \(\lambda_{ave\,\,\alpha}\).
Goodnight.
