= Solution
The illuminated area and incident <power> are
$$
A=(0.04)^2=1.6\times10^{-3}\,\mathrm{m^2},\qquad
P_{\rm in}=87A=0.1392\,\mathrm W.
$$
A $412\,\mathrm{nm}$ <photon> has <energy> $hc/\lambda=4.821\times10^{-19}\,\mathrm J=3.01\,\mathrm{eV}$, above the <silicon> <band gap>. With one collected <electron> per <photon>,
$$
I_{\rm ph}=q\frac{P_{\rm in}\lambda}{hc}=0.0462562\,\mathrm A.
$$
Use the ideal-diode approximation with \b[ideality factor $n=1$], negligible shunt leakage and the same dark <diode> parameters under illumination. At $300\,\mathrm K$, $V_T=k_BT/q=0.025852\,\mathrm V$. The measured dark point then fixes
$$
I_0=\frac{10^{-4}}{\exp(0.466/V_T)-1}=1.4844\times10^{-12}\,\mathrm A.
$$
At open circuit the forward <diode> <electric current> balances the photocurrent:
$$
\boxed{V_{\rm oc}=V_T\ln\!\left(1+\frac{I_{\rm ph}}{I_0}\right)=0.62465\,\mathrm V\simeq0.625\,\mathrm V}.
$$
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 is
$$
V_{\rm oc}(n)=nV_T\ln\!\left[1+\frac{I_{\rm ph}}{10^{-4}}\left(\exp\!\left(\frac{0.466}{nV_T}\right)-1\right)\right].
$$
Thus $0.625\,\mathrm V$ is a stated-model estimate, not a parameter-free inference from that one measurement.
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