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

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(B) The number density of $Si$ atoms is $N = 5 	imes 10^{28} 	ext{ m}^{-3}$.Since the doping concentration is $1 	ext{ ppm}$ (part per million),the

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

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(B) The number density of $Si$ atoms is $N = 5 	imes 10^{28} 	ext{ m}^{-3}$.Since the doping concentration is $1 	ext{ ppm}$ (part per million),the number density of donor atoms $N_{D}$ is $10^{-6} 	imes 5 	imes 10^{28} = 5 	imes 10^{22} 	ext{ m}^{-3}$.Since $N_{D} \gg n_{i}$,the number of electrons $n_{e} \approx N_{D} = 5 	imes 10^{22} 	ext{ m}^{-3}$.Using the mass action law $n_{e} n_{h} = n_{i}^{2}$,the number of holes $n_{h}$ is given by:$n_{h} = \frac{n_{i}^{2}}{n_{e}} = \frac{(1.5 	imes 10^{16})^{2}}{5 	imes 10^{22}} = \frac{2.25 	imes 10^{32}}{5 	imes 10^{22}} = 0.45 	imes 10^{10} = 4.5 	imes 10^{9} 	ext{ m}^{-3}$.

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

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