A silicon $P-N$ junction is shown in the figure. The doping in the $P$ region is $5 \times 10^{16} \mathrm{~cm}^3$ and doping in the $N$ region is $10 \times 10^{-16} \mathrm{~cm}^{-3}$. The parameters given are

Built-in voltage $\left(\phi_{b i}\right)=0.8 \mathrm{~V}$

Electro charge $(q)=1.6 \times 10^{-19} \mathrm{C}$

Vacuum permittivity of silicon $\left(\varepsilon_{s i}\right)=12$

The magnitude of reverse bias voltage that would completely deplete one of the two regions ( $P$ or $N$ ) prior to the other (rounded off to one decimal place) is $\_\_\_\_$ V.

A bar of silicon is doped with boron concentration of $10^{16} \mathrm{cm}^{-3}$ and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of $10^{20} \mathrm{~cm}^{-2} \mathrm{~s}^{-1}$. If the recombination lifetime is $100 \mu \mathrm{~s}$, intrinsic carrier concentration of silicon is $10^{10} \mathrm{~cm}^{-3}$ and assuming $100 \%$ ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is

The energy band diagram of a $p$-type semiconductor bar of length $L$ under equilibrium condition (i.e., the Fermi energy level $E_F$ is constant) is shown in the figure. The valance band $E_V$ is sloped since doping is non-uniform along the bar. The different between the energy levels of the valence band at the two edges of the bar is $\Delta$.

If the charge of an electron is $q$, then the magnitude of the electric field developed inside the semiconductor bar is