Since most have found \(h\) the Planck's Constant to be accurate at,
\(h\) = 4.135667516e-15 eV
If we assume that photon-electron interaction is the same in metals with a threshold frequency, as of photoelectric effect. That the mechanism of photon and electron collision is the same across such metals and so share the same probability function. Then,
\(E_{max}(f)=E[KE(f)]=\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }.f=h.f\)
\(h=\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }\)
\(KE_{max}=\cfrac{h.c}{4{r}_{e}}\)
and from
\(KE_{max}={m}_{p}c^2-\Phi\)
\({m}_{p}= \cfrac{{KE}_{max}+\Phi}{c^2}\)
\({m}_{p}= \cfrac{1}{c^2}.(\cfrac{h.c}{4{r}_{e}}+\Phi)\)
and so, using potassium as an example, \({r}_{e}\) = 235 pm and \(\Phi\) = 2.29 V,
\({m}_{p}\) = (4.135667516e-15*299792458/(4*235e-12)+2.29)/(299792458)^2*(1.60217657e-19)
\({m}_{p}\) = 2.3554e-33 kg
This is my third attempt at \({m}_{p}\). From this value we obtain,
Energy of a photon is
\({E}_{p}={m}_{p}c^2\) = (2.3554e-33)*299792458^2 = 2.1169e-16 J
Momentum of a photon is
\({P}_{p}={m}_{p}c\) = (2.3554e-33)*299792458 = 7.0613e-25 kgm s-1
Try once, try twice and maybe third time lucky.
