 Photoelectric contact-potential correction

Define $V_c$ as positive when the contact field adds to the retarding barrier, and let $V_{\mathrm{app}}$ be the signed externally imposed retarding voltage. In this chosen sign convention the barrier is $e(V_{\mathrm{app}}+V_c)$, henceThis convention specifies a model of the gap; an instrument's polarity labels must be translated into it. With a fixed $V_c$, subtraction at two frequencies removes both the work function and the offset:For an original example with the same $\Phi =2.30\mathrm{eV}$ and an additional $V_c=+0.200\mathrm{V}$, choose ${\nu }_1=6.00\times 10^{14}\mathrm{Hz}$ and ${\nu }_2=1.00\times 10^{15}\mathrm{Hz}$. The predicted terminal endpoints are $-0.018599$ and $1.635668\mathrm{V}$. The negative first endpoint calls for a small externally accelerating voltage because the contact field already over-retards the fastest Electrons. Their difference is $1.654267\mathrm{V}$, giving $h/e\simeq 4.135668\times 10^{-15}\mathrm{V}\mathrm{s}$. Ignoring $V_c$ in the intercept would falsely infer $2.50\mathrm{eV}$ for the work function. The terminal line's zero would then be mistaken for a material threshold. If $V_c$ drifts between the two readings, even the slope is biased. These are calculated examples, not Millikan's measurements.