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