A pure semiconductor crystal has 5 times 10^{ | vedclass

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(A) The concentration of atoms is $N = 5 	imes 10^{22} cm^{-3}$.Since it is doped with $1$ ppm (part per million) of a pentavalent element,the concen

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$4.5 	imes 10^3 cm^{-3}$

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(A) The concentration of atoms is $N = 5 	imes 10^{22} cm^{-3}$.Since it is doped with $1$ ppm (part per million) of a pentavalent element,the concentration of donor electrons $(n_e)$ is:$n_e \approx N_D = 10^{-6} 	imes 5 	imes 10^{22} = 5 	imes 10^{16} cm^{-3}$.Using the law of mass action for semiconductors:$n_e 	imes n_h = n_i^2$where $n_i = 1.5 	imes 10^{10} cm^{-3}$.$n_h = \frac{n_i^2}{n_e} = \frac{(1.5 	imes 10^{10})^2}{5 	imes 10^{16}}$$n_h = \frac{2.25 	imes 10^{20}}{5 	imes 10^{16}} = 0.45 	imes 10^4 = 4.5 	imes 10^3 cm^{-3}$.

$A$ pure semiconductor crystal has $5 \times 10^{22}$ atoms per $cm^3$. It is doped by $1$ ppm concentration of pentavalent element. The number of holes in the doped semiconductor is (given that $n_i = 1.5 \times 10^{10} cm^{-3}$):

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