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

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(A) Given: Number density of $Si$ atoms,$N = 5 	imes 10^{28} 	ext{ m}^{-3}$.Concentration of $As$ doping = $1 	ext{ ppm} = 10^{-6}$.The number dens

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

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(A) Given: Number density of $Si$ atoms,$N = 5 	imes 10^{28} 	ext{ m}^{-3}$.Concentration of $As$ doping = $1 	ext{ ppm} = 10^{-6}$.The number density of donor atoms $(N_D)$ is $N_D = 10^{-6} 	imes 5 	imes 10^{28} = 5 	imes 10^{22} 	ext{ m}^{-3}$.Since $N_D \gg n_i$,the number density of electrons $n_e \approx N_D = 5 	imes 10^{22} 	ext{ m}^{-3}$.Using the law of mass action,$n_e n_h = n_i^2$.$n_h = \frac{n_i^2}{n_e} = \frac{(1.5 	imes 10^{16})^2}{5 	imes 10^{22}} = \frac{2.25 	imes 10^{32}}{5 	imes 10^{22}} = 0.45 	imes 10^{10} = 4.5 	imes 10^9 	ext{ m}^{-3}$.

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

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