A pure semiconductor crystal has 8 times 10^{ | vedclass

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(B) The number of atoms per $	ext{m}^3$ in the semiconductor crystal is $N = 8 	imes 10^{28} 	ext{ atoms/m}^3$.The doping concentration is $2 	ext

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$6.25 	imes 10^8 	ext{ m}^{-3}$

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(B) The number of atoms per $	ext{m}^3$ in the semiconductor crystal is $N = 8 	imes 10^{28} 	ext{ atoms/m}^3$.The doping concentration is $2 	ext{ ppm}$,which means $2$ atoms per $10^6$ atoms.The number of donor atoms $(n_d)$ per $	ext{m}^3$ is given by $n_d = 2 	imes 10^{-6} 	imes 8 	imes 10^{28} = 16 	imes 10^{22} 	ext{ atoms/m}^3$.Since each pentavalent impurity atom provides one free electron,the electron concentration is $n_e \approx n_d = 16 	imes 10^{22} 	ext{ m}^{-3}$.Using the law of mass action,$n_e \cdot n_h = n_i^2$,where $n_h$ is the hole concentration.$n_h = \frac{n_i^2}{n_e} = \frac{(1 	imes 10^{16})^2}{16 	imes 10^{22}} = \frac{10^{32}}{16 	imes 10^{22}} = 0.0625 	imes 10^{10} = 6.25 	imes 10^8 	ext{ m}^{-3}$.

$A$ pure semiconductor crystal has $8 \times 10^{28} \text{ atoms/m}^3$. It is doped with a $2 \text{ ppm}$ concentration of pentavalent atoms. The number of holes formed in the semiconductor crystal is (Intrinsic carrier concentration,$n_i = 1 \times 10^{16} \text{ m}^{-3}$).

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