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(B) Given, the intrinsic carrier concentration $n_{i} = 1.6 	imes 10^{16} \,m^{-3}$.The hole density in the doped semiconductor is $n_{h} = 4 	imes

Vedclass · Accepted Answer

$6.4 	imes 10^9 \,m^{-3}$

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(B) Given, the intrinsic carrier concentration $n_{i} = 1.6 	imes 10^{16} \,m^{-3}$.The hole density in the doped semiconductor is $n_{h} = 4 	imes 10^{22} \,m^{-3}$.According to the law of mass action for semiconductors, $n_{e} n_{h} = n_{i}^2$.Therefore, the electron density $n_{e}$ is given by:$n_{e} = \frac{n_{i}^2}{n_{h}} = \frac{(1.6 	imes 10^{16})^2}{4 	imes 10^{22}}$$n_{e} = \frac{2.56 	imes 10^{32}}{4 	imes 10^{22}}$$n_{e} = 0.64 	imes 10^{10} \,m^{-3} = 6.4 	imes 10^9 \,m^{-3}$.

In a pure silicon, the number of electrons and holes per unit volume is $1.6 \times 10^{16} \,m^{-3}$. If silicon is doped with Boron in a way that the hole density increases to $4 \times 10^{22} \,m^{-3}$, then the electron density in the doped semiconductor will be:

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