Light, Indeed Light.

From the post "Planck Constant not Constant"

$E_{max}(f)=E[KE(f)]=\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }.f$

From the web, a set of  photoelectric data by Lulu Liu (Partner: Pablo Solis)∗
MIT Undergraduate of MIT 2007 "Determination of Planck’s Constant Using the Photoelectric Effect".  The target metal was potassium, K e(2,8,8,1) of work function 2.29 eV, and atomic radius 235 pm.

wavelength      frequency Hz        Stoppping Volatge
365.0 nm        8.2135e14              0.76 V
404.7 nm        7.4078e14              0.60 V
546.1 nm        5.4897e14              0.44 V
577.0 nm        5.1957e14              0.44 V

The last entry in the table seem to suggest saturation where $\cfrac{r_e}{r}>1$.  We will not use this point.   Further more the data seem to be for the region beyond the first plateau for the Electron E_max vs Photon Frequency graph in the post "Fodo Electric Effect".

A plot of  stopping voltage (V) vs frequency (*e14 Hz) is reproduced below,

The regression line is y=0.11x-0.18  error in slop = +0.0371, error in y intercept = +0.2647

Therefore,  from the gradient of the regression line,

$\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }=0.11e-14$

$KE_{max}=\cfrac {(0.11e-14)*c}{ 4{ r }_{ e } }$

where ${r}_{e}$ is the atomic radius of K.

${KE}_{max}$=(0.11e-14)*299792458/(4*(235e-12))

Since,

$KE_{max}={m}_{p}c^2-\Phi$

${m}_{p}=\cfrac{KE_{max}+\Phi}{c^2}$

So,

${m}_{p}$ = ((0.11e-14)*299792458/(4*(235e-12))+2.29)/299792458^2 *(1.60217657e-19)

the last factor converts eV to Joules (J).  And so my second attempt at mass of a photon is,

${m}_{p}$ = 6.2948e-34 kg

compared to that of an electron, 9.10938291e-31 kg and the mass of a proton 1.67262178e-27 kg.

Light, indeed light.