From the post "Sonic Boom" dated 14 Oct 2017,
\(v^{ 2 }_{ { max } }=-c^{ 2 }cos(\theta )*3.4354\)
but
\(cos(\theta )=1\) as \(\theta=0\),
\(v^2_{max}=c^2*3.4354\)
taking amplitude only because we are considering kinetic energy input per particle,
\(\cfrac{1}{2}m_{\rho}v^{ 2 }_{max}=E_{input/particle}=\cfrac{1}{2}c^{ 2 }*\cfrac{density}{Z}*3.4354\)
where division by \(Z\) gives unit per particle, and \(m_{\rho}\) is mass density. If this energy input is to be provided by a heat source (black body radiation), the density of states or the number of photons per unit volume per unit energy at frequency \(f\) is given by,
\(\rho_n=\cfrac{dn_s}{df}=\cfrac{8\pi}{c^3}f^2\) --- (*)
So, given energy per photon,
\(E=hf\)
the energy density input from such a illumination is,
\(E_{input/particle}=\cfrac{8\pi}{c^3}f^2*hf=\cfrac{8\pi h}{c^3}f^3\)
If we dare,
\(\cfrac{1}{2}c^{ 2 }*\cfrac{density}{Z}*3.4354=\cfrac{8\pi h}{c^3}f^3\)
\(f^3=\cfrac{c^{ 5 }}{16\pi h}*\cfrac{density}{Z}*3.4354\) --- (**)
which relates the density and atomic number of a metal to its threshold frequency for photoelectric emission.
Is it necessary to account for Durian constant? If considered, the constant applies to both the receiving and radiating side, the constant cancels out. The expression for \(f\) can be considered as per entanglement volume.
Why black body radiation? The metal is receiving the energy as a black body. Photons, once they leave their source is independent of the source. The photon, however, on reaching the metal, is received in effect the reciprocal of a black body. Given the frequency \(f\), the photon number density received conforms to expression (*). In other word, (*) is a boundary condition, that applies irrespective of the illumination.
Luckily, the expression (**) can be tested...
