If the potential barrier across a p-n junction is 0.6 V. Then the electric field intensity, in the depletion region having the width of $$6 \times 10^{-6}$$ m, will be _____ $$\times 10^5$$ N C$$^{-1}$$

We are given a p-n junction with a potential barrier $$V = 0.6$$ V and the width of the depletion region is $$d = 6 \times 10^{-6}$$ m. We need to find the electric field intensity in the depletion region.

The electric field in the depletion region can be approximated as $$E = \dfrac{V}{d}$$, assuming a uniform field across the depletion width.

Substituting the values: $$E = \dfrac{0.6}{6 \times 10^{-6}} = \dfrac{0.6}{6} \times 10^{6} = 0.1 \times 10^{6} = 1 \times 10^{5}$$ N C$$^{-1}$$.

Hence, the correct answer is 1.