In a semiconductor,the number density of intr | vedclass

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(A) According to the law of mass action for semiconductors,the product of electron density $(n_e)$ and hole density $(n_h)$ is equal to the square of

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$5$

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(A) According to the law of mass action for semiconductors,the product of electron density $(n_e)$ and hole density $(n_h)$ is equal to the square of the intrinsic carrier density $(n_i)$:$n_e \cdot n_h = n_i^2$Given:$n_i = 1.5 	imes 10^{16} \, 	ext{m}^{-3}$$n_h = 4.5 	imes 10^{22} \, 	ext{m}^{-3}$Substituting the values into the formula:$n_e = \frac{n_i^2}{n_h} = \frac{(1.5 	imes 10^{16})^2}{4.5 	imes 10^{22}}$$n_e = \frac{2.25 	imes 10^{32}}{4.5 	imes 10^{22}}$$n_e = 0.5 	imes 10^{10} \, 	ext{m}^{-3} = 5 	imes 10^9 \, 	ext{m}^{-3}$Thus,the electron density is $5 	imes 10^9 \, 	ext{m}^{-3}$.

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