Wednesday, June 4, 2014

Why two Ejection Rates? Fido

If Fodo Effects is actually the probabilistic interaction of 2 lengths, \(r\) photon circular radius and \({r}_{e}\) electrons orbital radius, then we have an analogy between this and Buffon's Needle Problem.  Where a needle of length \(l\) is toss into a lined paper of line spacing \(d\),  the probability of the needle hitting one of the lines is,

\(P(x)=\)
                  \(=\frac{2x}{\pi}\)      for \(x\lt1 \)    where    \(x=\frac{l}{d}\)
                  \(=\frac{2}{\pi}(x-\sqrt{x^2-1}+arcsec(x))\)            for \(x\ge1\)

A plot of the probability curve is shown below,


From the plot we see that as \(l\) increases the probability of a hit increases,  in the case of photon-electron interaction as we understand it, as \(r\) decreases or \(r_e\) increases relatively, electrons are ejected after a certain frequency threshold.  That means \(x=\frac{l}{d}=\frac{r_e}{r}\), by analogy.  Intuitively, when spacing \(d\) increases the probability of a hit decreases to zero, just as increasing \(r\) drive the Fodo process falls below the frequency \((r=\frac{2\pi.f}{c})\) threshold and no electrons are emitted.

The straight gradient  for values of  \(x\lt 1\) is \(\frac{2}{\pi}\).  If the interaction between a photon in circular motion of radius \(r\) and an electron in orbit of radius \({r}_{e}\) is analogous to the Buffon's Needle Problem, then every hit corresponds to an electron of some KE being ejected.

When the photons of a frequency are interacting with two electron orbits, one lower and one higher, then from the graph, the probability of a hit is different for each orbit radius.  And so the rate of ejection of electrons are different for these two orbits.   It is expected that the photon circular path radius be between the two orbital radii and so we are concerned with \(P(x)\) about \(x\) = 1, that is 0.6 < \(P(x)\) < 0.7.  From the graph, the larger orbital radius (\(\frac{r_e}{r}\gt 1\)) has a higher collision probability and so a correspondingly higher ejection rate.  It also has a lower work function with a correspondingly higher ejected electron \(KE_{max}\).

It is possible that process with a lower ejection rate, eject electrons with higher \(KE_{max}\).  Ejection rate depends on the probability of a collision between the photon and electron, whereas \(KE_{max}\) depends on the work function.  The photons has constant energy.

This probability model is an oversimplification.  In actual, both orbital radius and circular path radius are in 3D space.  The straight path of the photons however, allows for simplification in the plane containing the photon path. So the photon see not 2D circular projections of sphere but a line with marks, marking the extend of the electron orbits.  A hit on this markings on the line, is then a collision with an electron.

The main simplification is to consider all interactions to be along the diameter of the orbit.  The probability that a collision occur off this axis is even smaller.  Off axis only four points of the circles (electron orbits and the photon circular path) meet; along the diameter axis, large part of the perimeters come close together.