\(f^3=\cfrac{c^{ 5 }}{16\pi h}*\cfrac{density}{Z}*3.4354\)
prove to give to high a value for \(f\) \(\approx10^{25}\). Maybe in the expression,
\(E_{input/particle}=\cfrac{8\pi}{c^3}f^2*hf=\cfrac{8\pi h}{c^3}f^3\)
we are wrong to consider photon energy, \(E=hf\) as a wave only,
\(E_{input/particle}=\cfrac{1}{2}c^{ 2 }*\cfrac{density}{Z}*3.4354=\cfrac{8\pi h}{c^3}f^3\)
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,
\(E_{input/particle}=KE+PE\)
\(\cfrac{1}{2}c^2*\cfrac{density}{Z}*3.4354=\cfrac{1}{2}c^2*\cfrac{8\pi}{c^3}f^2+\cfrac{8\pi h}{c^3}f^3\)
\(f^{ 3 }+\cfrac { c^{ 2 } }{ 2h } f^{ 2 }-\cfrac { c^{ 5 } }{ 16\pi h } *\cfrac { density }{ Z } *3.4354=0\)
Since the solution to the cubic equation \(ax^{ 3 }+bx^{ 2 }+c^{'}x+d=0\) is
\( x=\sqrt [ 3 ]{ \left( \cfrac { -b^{ 3 } }{ 27a^{ 3 } } +\cfrac { bc^{'} }{ 6a^{ 2 } } -\cfrac { d }{ 2a } \right) +\sqrt { \left( \cfrac { -b^{ 3 } }{ 27a^{ 3 } } +\cfrac { bc^{'} }{ 6a^{ 2 } } -\cfrac { d }{ 2a } \right) ^{ 2 }+\left( \cfrac { c^{'} }{ 3a } -\cfrac { b^{ 2 } }{ 9a^{ 2 } } \right) ^{ 3 } } } \\ +\sqrt [ 3 ]{ \left( \cfrac { -b^{ 3 } }{ 27a^{ 3 } } +\cfrac { bc^{'} }{ 6a^{ 2 } } -\cfrac { d }{ 2a } \right) -\sqrt { \left( \cfrac { -b^{ 3 } }{ 27a^{ 3 } } +\cfrac { bc^{'} }{ 6a^{ 2 } } -\cfrac { d }{ 2a } \right) ^{ 2 }+\left( \cfrac { c^{'} }{ 3a } -\cfrac { b^{ 2 } }{ 9a^{ 2 } } \right) ^{ 3 } } } -\cfrac { b }{ 3a } \)
where \(x=f\), \(a=1\), \(b=\cfrac { c^{ 2 } }{ 2h }\), \(c^{'}=0\) and \(d=-\cfrac { c^{ 5 } }{ 16\pi h } *\cfrac { density }{ Z } *3.4354\)
\(\cfrac { b^{ 3 } }{ 27a^{ 3 } }=\cfrac { c^{ 6 } }{ 216h^3 }\)
\(\cfrac { d }{ 2a }=-\cfrac { c^{ 5 } }{ 32\pi h } *\cfrac { density }{ Z } *3.4354\)
\(\cfrac { b^{ 2 } }{ 9a^{ 2 } }=\cfrac { c^{ 4 } }{ 36h^2 }\)
\(\cfrac { b }{ 3a }=\cfrac { c^{ 2 } }{ 6h }\)
So,
\(f=\sqrt [ 3 ]{ \left( -\cfrac { c^{ 6 } }{ 216h^{ 3 } } +\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ } \right) +\sqrt { \left( -\cfrac { c^{ 6 } }{ 216h^{ 3 } } +\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ } \right) ^{ 2 }+\left( -\cfrac { c^{ 4 } }{ 36h^{ 2 } } \right) ^{ 3 } } } \\ +\sqrt [ 3 ]{ \left( -\cfrac { c^{ 6 } }{ 216h^{ 3 } } +\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ } \right) -\sqrt { \left( -\cfrac { c^{ 6 } }{ 216h^{ 3 } } +\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ } \right) ^{ 2 }+\left( -\cfrac { c^{ 4 } }{ 36h^{ 2 } } \right) ^{ 3 } } } -\cfrac { c^{ 2 } }{ 6h } \)
Or,
\(f=\sqrt [ 3 ]{ A+\sqrt { A^{ 2 }-B^{ 3 } } } +\sqrt [ 3 ]{ A-\sqrt { A^{ 2 }-B^{ 3 } } } -C\)
where,
\(B=\cfrac { c^{ 4 } }{ 36h^{ 2 } } \)
\(C=\cfrac { c^{ 2 } }{ 6h } \)
Consider,
\(\left( -\cfrac { c^{ 6 } }{ 216h^{ 3 } } +\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ } \right) ^{ 2 }+\left( -\cfrac { c^{ 4 } }{ 36h^{ 2 } } \right) ^{ 3 }\\=-2*\cfrac { c^{ 6 } }{ 216h^{ 3 } } *\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ }+\left(\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ }\right)^2 \)
\(=\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ }\left(-2*\cfrac { c^{ 6 } }{ 216h^{ 3 } }+\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ }\right)\)
\(=\cfrac { c^{ 10}*density*3.4354 }{ 32\pi h^2Z }\left(-2*\cfrac { c }{ 216h^{ 2 } }+\cfrac { density*3.4354 }{ 32\pi Z }\right)\)
and,
\(A= \cfrac{c^5}{h}\left(-\cfrac { c }{ 216h^{ 2 } } +\cfrac { density*3.4354 }{ 32\pi Z } \right) \)
Since,
\(\cfrac { c }{ 216h^{ 2 } }\gt\gt\cfrac { density*3.4354 }{ 32\pi Z }\)
\(f\) 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,
\(f^{ 3 }-\cfrac { c^{ 2 } }{ 2h } f^{ 2 }-\cfrac { c^{ 5 } }{ 16\pi h } *\cfrac { density }{ Z } *3.4354=0\)
\(f=\sqrt [ 3 ]{ A+\sqrt { A^{ 2 }-B^{ 3 } } } +\sqrt [ 3 ]{ A-\sqrt { A^{ 2 }-B^{ 3 } } } -C\)
where,
\(A= \cfrac { c^{ 6 } }{ 216h^{ 3 } } +\cfrac { c^{ 5 }*density*3.4354 }{ 32\pi hZ } \)
\(B=\cfrac { c^{ 4 } }{ 36h^{ 2 } } \)
\(C=\cfrac { c^{ 2 } }{ 6h } \)
and
\( A^{ 2 }-B^{ 3 }=\cfrac { c^{ 10}*density*3.4354 }{ 32\pi h^2Z }\left(2*\cfrac { c }{ 216h^{ 2 } }+\cfrac { density*3.4354 }{ 32\pi Z }\right)\)
Still,
\(\cfrac { c }{ 216h^{ 2 } }\gt\gt\cfrac { density*3.4354 }{ 32\pi Z }\)
\( A^{ 2 }-B^{ 3 }\approx\cfrac { c^{ 10}*density*3.4354 }{ 32\pi h^2Z }\left(2*\cfrac { c }{ 216h^{ 2 } }\right)\)
\(A\gt\gt\sqrt { A^{ 2 }-B^{ 3 } } \)
\(f\approx\cfrac{c^2}{6h}\) which is way too big!
If we formulated \(f\) as,
\(E_{input/particle}=KE\)
where the photons are just particles with light speed \(c\),
\(\cfrac{1}{2}c^2*\cfrac{density}{Z}*3.4354=\cfrac{1}{2}c^2*\cfrac{8\pi}{c^3}f^2\)
\(f=\sqrt{\cfrac{c^3}{8\pi}*\cfrac{density}{Z}*3.4354}\)
With iron, \(Z=26\) and density \(7.874\,gcm^{-3}\)
\(f=\sqrt{\cfrac{c^3}{8\pi}*\cfrac{7874}{26}*3.4354}=3.340e13\,Hz\)
Iron has a work function of \(4.33\,eV\), so
\(f=\cfrac{\Psi}{h_{\small{eV}}}=\cfrac{4.33}{4.135667516e-15}=1.0470e15\,Hz\)
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,
\(E_{input/atom}=\cfrac{1}{2}c^2*{density}*{Z}*3.4354\)
all \(Z\) particles interact as one with a photon, to result in an emission of a basic particle. So,
\(E_{input/atom}=\cfrac{1}{2}c^2*{density}*{Z}*3.4354=\cfrac{1}{2}c^2*\cfrac{8\pi}{c^3}f^2\)
\(f=\sqrt{\cfrac{c^3}{8\pi}*{density}*{Z}*3.4354}\)
For the case of iron,
\(f=\sqrt{\cfrac{c^3}{8\pi}*7874*26*3.4354}=8.6833e14\,Hz\)
Iron has a work function of \(\Psi=4.33\,eV\), so
\(f=\cfrac{\Psi}{h_{\small{eV}}}=\cfrac{4.5}{4.135667516e-15}=1.0470e15\,Hz\)
Iron has a work function of \(\Psi=4.33\,eV\), so
\(f=\cfrac{\Psi}{h_{\small{eV}}}=\cfrac{4.5}{4.135667516e-15}=1.0470e15\,Hz\)
and we are off by \(1.21\) times.
\({density}*{Z}\) or \(\cfrac{density}{Z}\) ???
Good night...
