Suppose a pure Si crystal has 5 times 10^{28} | vedclass

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(D) The number of $As$ atoms per $m^3$ is given by the doping concentration of $1$ ppm (part per million).$n_e \approx N_D = 1 	imes 10^{-6} 	imes (

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$4.5 	imes 10^{9} m^{-3}$

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(D) The number of $As$ atoms per $m^3$ is given by the doping concentration of $1$ ppm (part per million).$n_e \approx N_D = 1 	imes 10^{-6} 	imes (5 	imes 10^{28} m^{-3}) = 5 	imes 10^{22} m^{-3}$.Using the law of mass action,$n_e n_h = n_i^2$.$n_h = \frac{n_i^2}{n_e} = \frac{(1.5 	imes 10^{16})^2}{5 	imes 10^{22}}$.$n_h = \frac{2.25 	imes 10^{32}}{5 	imes 10^{22}} = 0.45 	imes 10^{10} m^{-3} = 4.5 	imes 10^9 m^{-3}$.

Suppose a pure $Si$ crystal has $5 \times 10^{28}$ atoms $m^{-3}$. It is doped by $1$ ppm concentration of pentavalent $As$. Calculate the number of holes. Given that $n_i = 1.5 \times 10^{16} m^{-3}$.

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