 Solution

The illuminated area and incident power areA $412\mathrm{nm}$ photon has energy $hc/\lambda =4.821\times 10^{-19}\mathrm{J}=3.01\mathrm{eV}$, above the silicon band gap. With one collected electron per photon,Use the ideal-diode approximation with ideality factor $n=1$, negligible shunt leakage and the same dark diode parameters under illumination. At $300\mathrm{K}$, $V_T=k_{B}T/q=0.025852\mathrm{V}$. The measured dark point then fixesAt open circuit the forward diode electric current balances the photocurrent:This is below the $1.1\mathrm{eV}$ band gap expressed per elementary charge. Excess photon energy does not translate directly into terminal voltage; much of it thermalizes.

The paper does not explicitly specify an ideality factor. If it is not taken as unity, one dark operating point cannot determine both $I_0$ and $n$. The corresponding family isThus $0.625\mathrm{V}$ is a stated-model estimate, not a parameter-free inference from that one measurement.