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(C) The potential barrier $V$ across the depletion region is related to the electric field $E$ and the width $d$ by the formula $V = E \cdot d$.Given

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$1000$

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(C) The potential barrier $V$ across the depletion region is related to the electric field $E$ and the width $d$ by the formula $V = E \cdot d$.Given the electric field $E = 2 	imes 10^5 \ V/m$.The energy required for a particle with charge $q = 3e$ to overcome the potential barrier is $U = 0.6 \ eV$.The potential difference $V$ is given by $V = \frac{U}{q} = \frac{0.6 \ eV}{3e} = 0.2 \ V$.Now, substituting the values into the formula $V = E \cdot d$:$0.2 = (2 	imes 10^5) \cdot d$$d = \frac{0.2}{2 	imes 10^5} = 0.1 	imes 10^{-5} \ m = 10^{-6} \ m$.Since $1 \ nm = 10^{-9} \ m$, we have $d = 1000 \ nm$.

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