Three Times And I am Light

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.