Bonding Energy And $\psi$ Cloud

From photoelectric effect,

$KE=hf_{\small{\Psi}}-\Psi$

under a retarding electric field, in the limit when the ejected electrons have zero velocity,

 $0=hf_{\small{\Psi}}-\Psi$

$f_{\small{\Psi}}=\cfrac{\Psi}{h}$

But according to the scenario of particles acquiring $v_{boom}$ that ejects basic particles of a quarter charge, $v_{boom}\ne0$!  Instead, the ejected particles are always at $v_{boom}$.

And when $f$ is varied, as $f$ is spread among a Boltzmann distribution, $f_{\small{\Psi}}$ is never zero.  At any $f$ close to $f_{\small{\Psi}}$, there will be some photons at $f_{\small{\Psi}}$ ejecting particles at $v_{boom}$, still.  Although a value of $f$ far from$f_{\small{\Psi}}$, photons with $f_{\small{\Psi}}$ can be considered effectively small.  At $f=f_{\small{\Psi}}$, the ejected particles should be significantly higher, as every subsequent collision ejects a particle at $v_{boom}$.

An example of anode current vs. retarding voltage data for a number of wavelengths is shown below. This is from the first edition of Melissinos, where retarding voltage increases to the left.

According the scenario presented here, there will be ejected particles at all $f$, as long as some photons in the spread of $f$ have $f=f_{\small{\Psi}}$, and these ejected particles have a fixed $v_{boom}$.  The number of ejected particles is not affected by the retarding voltage.  The retarding voltage does not stop the particles from being ejected, instead the ejected particles are driven into the reverse direction and form a reverse current.  A horizontal asymptote, suggests that all ejected particles have the same velocity.  If the ejected particles have different velocities then increasing voltage will drive the current higher towards a higher plateau, as slow particles catch up with fast particles slowed to a fixed maximum due to collisions.

$V_s$, the stoppage voltage is at the asymptote when $I_s$, the current is negative (not zero).  The change in current after $V_s$ towards zero, is due to collisions as the number of ejected particles that make their way to the anode increases with voltage.  Collisions knock some ejected particles off this path.

According the scenario presented here, the asymptote occurs naturally without inviting "thermal energy distribution to widen the work function".  By considering a basic particle as being ejected from a particle and not from the cathode, but still within the cathode allows for a negative current.  Otherwise, an ejected particle from the material with zero velocity, is free from the material and should not contribute to current.  A voltage pushing the particle back into the material will have to do the reverse of ejecting the particle from the material.  An incrementally small voltage below $V_s$ cannot do so, this would suggest an immaterial barrier that makes $\Psi$, the associated work function zero.  (Maybe not the same work, but still some work moving the particle into the material.)

In this case, the work function pops a basic particle from a $\psi$ cloud at $v_{boom}$.  $v_{boom}$ and so $KE$ is a constant!  $KE$ does not spread immediately after the particles are ejected.  Only, after the first collision do $KE$ of the ejected particles spreads in value.  Since the ejected particles are basic particles, they will coalesce into big particle.  They increase in momentum as they approach the anode without a driving forward voltage.

Since all $f$ with a significant $f_{\small{\Psi}}$ spread, will pop particles at $v_{boom}$, $\Psi$ must be defined as when $f=f_{\small{\Psi}}$ where the number of ejected particles is maximum.  And the work being done,

$\Psi=h.f_{\small{\Psi}}$

In the case of two $\psi$ cloud sticking together via a basic particle on each side,

$B_{\small{\Psi}}=2\Psi=2h.f_{\small{\Psi}}$

where $B_{\small{\Psi}}$ is the bonding energy of two similar $\psi$ clouds.  This is a covalent bond, in contrast to a electron being share by two paired orbits, this bond is the attachment between two $\psi$ clouds.  Ionization energy does not appear here as defined as the energy needed to eject an electron from orbit around a nucleus; nor does ionization apply when an electron is loss from being shared by two paired orbits.

What else does $\Psi$ implies?