A silicon p-n junction has a barrier potential of 0.60 V across a depletion layer that is 0.50 µm…

Step 1. The field. $E = \dfrac{V_0}{W} = \dfrac{0.60\ \text{V}}{0.50 \times 10^{-6}\ \text{m}} = 1.2 \times 10^6\ \text{V/m}$.

Step 2. The energy. $eV_0 = 1.602 \times 10^{-19}\ \text{C} \times 0.60\ \text{V} = 9.61 \times 10^{-20}\ \text{J}$, which is $0.60$ eV.

Step 3. Forward bias. The barrier falls to $V_0 - V = 0.60\ \text{V} - 0.40\ \text{V} = 0.20\ \text{V}$.

Step 4. Reverse bias. The barrier rises to $V_0 + V = 0.60\ \text{V} + 3.0\ \text{V} = 3.6\ \text{V}$.