In a p-n junction diode, the thickness of the | vedclass

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(C) Given: Barrier potential, $V = 0.3 \,V$. Thickness of depletion layer, $d = 2 	imes 10^{-6} \,m$. The electric field $E$ is given by the relation

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$1.5 	imes 10^5 \,Vm^{-1}$ from $n$ to $p$ side

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(C) Given: Barrier potential, $V = 0.3 \,V$. Thickness of depletion layer, $d = 2 	imes 10^{-6} \,m$. The electric field $E$ is given by the relation $E = \frac{V}{d}$. Substituting the values, we get: $E = \frac{0.3}{2 	imes 10^{-6}} = 0.15 	imes 10^6 = 1.5 	imes 10^5 \,V/m$. In a $p-n$ junction, the electric field is directed from the $n$-region to the $p$-region due to the accumulation of positive ions on the $n$-side and negative ions on the $p$-side of the depletion layer. Therefore, the correct option is $C$.

In a $p-n$ junction diode, the thickness of the depletion layer is $2 \times 10^{-6} \,m$ and the barrier potential is $0.3 \,V$. The intensity of the electric field at the junction is

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