f3=c516πh∗densityZ∗3.4354
prove to give to high a value for f ≈1025. Maybe in the expression,
Einput/particle=8πc3f2∗hf=8πhc3f3
we are wrong to consider photon energy, E=hf as a wave only,
Einput/particle=12c2∗densityZ∗3.4354=8πhc3f3
Instead, if we consider each photon as a particle at light speed, and equate the total energy input to the system as the sum of kinetic energy and potential energy,
Einput/particle=KE+PE
12c2∗densityZ∗3.4354=12c2∗8πc3f2+8πhc3f3
f3+c22hf2−c516πh∗densityZ∗3.4354=0
Since the solution to the cubic equation ax3+bx2+c′x+d=0 is
x=3√(−b327a3+bc′6a2−d2a)+√(−b327a3+bc′6a2−d2a)2+(c′3a−b29a2)3+3√(−b327a3+bc′6a2−d2a)−√(−b327a3+bc′6a2−d2a)2+(c′3a−b29a2)3−b3a
where x=f, a=1, b=c22h, c′=0 and d=−c516πh∗densityZ∗3.4354
b327a3=c6216h3
d2a=−c532πh∗densityZ∗3.4354
b29a2=c436h2
b3a=c26h
So,
f=3√(−c6216h3+c5∗density∗3.435432πhZ)+√(−c6216h3+c5∗density∗3.435432πhZ)2+(−c436h2)3+3√(−c6216h3+c5∗density∗3.435432πhZ)−√(−c6216h3+c5∗density∗3.435432πhZ)2+(−c436h2)3−c26h
Or,
f=3√A+√A2−B3+3√A−√A2−B3−C
where,
B=c436h2
C=c26h
Consider,
(−c6216h3+c5∗density∗3.435432πhZ)2+(−c436h2)3=−2∗c6216h3∗c5∗density∗3.435432πhZ+(c5∗density∗3.435432πhZ)2
=c5∗density∗3.435432πhZ(−2∗c6216h3+c5∗density∗3.435432πhZ)
=c10∗density∗3.435432πh2Z(−2∗c216h2+density∗3.435432πZ)
and,
A=c5h(−c216h2+density∗3.435432πZ)
Since,
c216h2>>density∗3.435432πZ
f is complex and negative! Maybe we should have use,
L=hf−KE
that the photon is at light speed before and after collision and work is done only by the energy in excess of its kinetic energy. E=hf being the total energy of the photon and is not considered its PE. We have instead,
f3−c22hf2−c516πh∗densityZ∗3.4354=0
f=3√A+√A2−B3+3√A−√A2−B3−C
where,
A=c6216h3+c5∗density∗3.435432πhZ
B=c436h2
C=c26h
and
A2−B3=c10∗density∗3.435432πh2Z(2∗c216h2+density∗3.435432πZ)
Still,
c216h2>>density∗3.435432πZ
A2−B3≈c10∗density∗3.435432πh2Z(2∗c216h2)
A>>√A2−B3
f≈c26h which is way too big!
If we formulated f as,
Einput/particle=KE
where the photons are just particles with light speed c,
12c2∗densityZ∗3.4354=12c2∗8πc3f2
f=√c38π∗densityZ∗3.4354
With iron, Z=26 and density 7.874gcm−3
f=√c38π∗787426∗3.4354=3.340e13Hz
Iron has a work function of 4.33eV, so
f=ΨheV=4.334.135667516e−15=1.0470e15Hz
Missed by a mile.
In this attempt to derive threshold frequency, the "charge" ejected is a basic particle of one quarter the normal charge.
It is possible to argue that since it is an atom that is subjected to the bombardment of photons, that we should take the unit atom view instead of a unit particle view. In which case, each atom has Z number of particles of each type that is subjected to collisions with photon,
Einput/atom=12c2∗density∗Z∗3.4354
all Z particles interact as one with a photon, to result in an emission of a basic particle. So,
Einput/atom=12c2∗density∗Z∗3.4354=12c2∗8πc3f2
f=√c38π∗density∗Z∗3.4354
For the case of iron,
f=√c38π∗7874∗26∗3.4354=8.6833e14Hz
Iron has a work function of Ψ=4.33eV, so
f=ΨheV=4.54.135667516e−15=1.0470e15Hz
Iron has a work function of Ψ=4.33eV, so
f=ΨheV=4.54.135667516e−15=1.0470e15Hz
and we are off by 1.21 times.
density∗Z or densityZ ???
Good night...
