 Solution

Taking room temperature as $300\mathrm{K}$, the thermal voltage is $V_T=k_{B}T/q=0.025852\mathrm{V}$. For the stated ideal diode, $n=1$, soThe $-1$ term is negligible here. A different precise interpretation of room temperature changes the result noticeably because of the exponential; the calculation above states its temperature explicitly.

For a lower-bandgap material, the intrinsic carrier concentration generally rises, increasing the reverse saturation current density, approximately through $J_0\propto n_i^2$ with $n_i^2\propto \exp (-E_g/k_{B}T)$ when other parameters are held comparable. Thus the forward electric current at the same voltage would normally increase. If $J_0$ were artificially held fixed at the supplied numerical value, however, the ideal-diode equation would give the same electric current: the band gap dependence enters through the material parameters rather than as a separate term in that equation.