In a p-n junction, an electric field of 5 tim | vedclass

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(D) The kinetic energy required for an electron to overcome the potential barrier is given by $K.E. = eV$, where $e$ is the elementary charge $(1.6

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$4 	imes 10^{-5} \ cm$

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(D) The kinetic energy required for an electron to overcome the potential barrier is given by $K.E. = eV$, where $e$ is the elementary charge $(1.6 	imes 10^{-19} \ C)$ and $V$ is the potential barrier.The potential barrier $V$ is calculated as $V = \frac{K.E.}{e} = \frac{3.2 	imes 10^{-20} \ J}{1.6 	imes 10^{-19} \ C} = 0.2 \ V$.The relationship between the electric field $E$, potential $V$, and the width of the depletion region $d$ is $E = \frac{V}{d}$, which implies $d = \frac{V}{E}$.Substituting the values, $d = \frac{0.2 \ V}{5 	imes 10^5 \ V/m} = 0.04 	imes 10^{-5} \ m = 4 	imes 10^{-7} \ m$.Converting this to centimeters: $d = 4 	imes 10^{-7} \ m = 4 	imes 10^{-5} \ cm$.

In a $p-n$ junction, an electric field of $5 \times 10^5 \ V/m$ exists in the depletion region. The minimum kinetic energy of a conduction electron, in order to diffuse from $n$-side to the $p$-side, is found to be $3.2 \times 10^{-20} \ J$. The width of the depletion region is,

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