This line of best fit has gradient\(\rightarrow\infty\) as \(x\rightarrow\infty\). For any such line fitted closer to the origin, it has gradient \(\lt\infty\). "Threshold", the x-intercept of this line is not unique.
This is a line to fit V-I characteristics of a gas discharge tube at the onset of glow discharge,
where \(x=log(I)\). The drop in voltage here is attributed to negative charges created.
This should not be confused with,
\(K_{max}=hf-\Phi\)
where \(K_{max}\) is the maximum kinetic energy of ejected electrons and \(\Phi=hf_o\) the Work Function defined by \(f_o\), the threshold frequency. This relation answers the question: "What happens to the excess energy of the photon beyond \(\Phi\) after impact?"
Using sodium, \(Na\)
\(f_o=4.39*10^{14}\,Hz\)
If we reverse and find \(a_{\psi}\) using
\(f_{osc}=c\sqrt{\cfrac{2\pi}{a_{\psi}}}\) --- (*)
with a change in notation,
\(f_{hole}=c\sqrt{\cfrac{2\pi}{a_{\psi\,hole}}}\)
\(a_{\psi\,hole}=2\pi\left(\cfrac{c}{f_{hole}}\right)^2=2.93\text{e-}12\,m\)
What is this small hole?
Could it be that \(f_{osc}\) of sodium, opens up this small hole and subsequent photons that falls into it, is split into two parts,
\(K_{max}=hf-\Phi\)
\(\Phi=hf_{hole}\), \(f\) is the frequency of the impacting photons and \(K_{max}\) is the maximum kinetic energy of the ejected particles. \(f_{osc}\) of sodium is given by,
\(f_{osc}=c\sqrt{\cfrac{2\pi}{a_{\psi\,Na}}}\)
where \(a_{\psi\,Na}\) is the size of sodium \(\psi\) cloud.
\(f_{osc}\) opens up a hole denoted by \(a_{\psi\,hole}\), and all photons at greater size than \(a_{\psi\,hole}\) (lower frequency than \(f_{hole}\), as indicated by (*)) are rejected by the hole. If this is the role of \(f_{osc}\), then using a narrow bandwidth of photons at \(f_{osc}\) and illuminating the metal with photons at bigger size than \(a_{hole}\) will cause the metal to appear very bright and without the emission of charged particles. All illuminating photons is rejected by the hole opened by \(f_{osc}\).
Photons of size smaller, \(a_{\psi}\lt a_{\psi\,hole}\) falls into the hole. When do such photons return after falling into the hole? What dictates the size of the hole? Hole of different sizes could explain the seemingly different processes that occurs simultaneously during photoelectric emission where two \(K_{max}\) values were noted. The post related to this is "More Fodo Effects" dated 04 Jun 2014; two related diagrams are shown below,
Impurities could explain the two \(K_{max}s\).
\(f_{hole}\) is the threshold frequency. Given \(f_{\psi}\), \(a_{\psi}\) can be derived given,
\(2\pi a_{\psi}=n\lambda=n\cfrac{c}{f_{\psi}}\)
\(f_{\psi}=\cfrac{c}{2\pi a_{\psi}}\), with \(n=1\)
but an impacting photon interact through,
\(f_{\psi}=c\sqrt{\cfrac{2\pi}{a_{\psi}}}\)
when \(f_{\psi}\gt f_{hole}\), \(a_{\psi}\lt a_{\psi\,hole}\) and the photon goes through the hole, and its energy is split via,
\(K_{max}=hf-\Phi\)
\(\Phi=hf_{hole}\) and \(K_{max}\) is the maximum kinetic energy of a emitted particle.
\(f_{osc}\) that triggers \(f_{hole}\) is lower than \(f_{hole}\). \(f_{hole}\) alone does not cause photoelectric emission. \(f_{osc}\), a frequency below the threshold, has first to create the hole, via a mode of oscillation with resonance at,
\(f_{osc}=c\sqrt{\cfrac{2\pi}{a_{\psi}}}\)
where \(a_{\psi}\) is the size of the \(\psi\) ball.
This hole is in effect, a region of negative energy that minus off energy of subsequent impacting photons of higher frequencies. Photons of lower frequencies are too big to fit into the hole. And only the resonance frequency, \(f_{osc}\) opens the hole in the metal.
Besides being convoluted, this explanation does not hold \(a_{\psi\,hole}\) at all, apart from the exertion that \(f_{osc}\) caused it. It however, allows for many size holes that can account for more than one \(K_{max}\) and allows \(f_{osc}\) to be different from the threshold frequency. But charges are still created within the metal at \(f_{osc}\) and, higher frequency photons interacting with the holes created can be outside the metal, ie emitted, when they return from the time dimension.
How does \(f_{osc}\) create charges? \(f_{osc}\) set the \(\psi\) cloud of the atom into resonance and the atom loses its orbiting valence electrons. And, how a photon of higher energy \(f_{\psi}\gt f_{hole}\) ejects a particle of kinetic energy \(K_{max}\) after filling in the hole? The hole requires \(E=hf_{hole}\), what remains of the impacting photons with higher energy after making good for the hole is like a torus photon (posts "What Donuts? Dipoles?" and "Torus Photons" dated 29 Dec 2017).
\(K_{max}=hf_{\psi}-hf_{hole}\)
When the torus photon collapses however, the resulting particle may not have a full unit charge. It may be a partial charge of size smaller than \(a_{\psi\,c}\).
Is \(f_{osc}\) valid?
Note: The previous version of this post has mistaken \(f_{osc}\gt f_{o}\) or \(f_{hole}\) that was wrong.
