If one indium atom is doped per 5 times 10^7 | vedclass

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(C) The number density of silicon atoms is given as $N_{Si} = 5 	imes 10^{28} \, 	ext{atoms}/	ext{m}^3$. Since $1 \, 	ext{m}^3 = 10^6 \, 	ext{cm}

Vedclass · Accepted Answer

$1.0 	imes 10^{15} \, 	ext{atoms}/	ext{cm}^3$

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(C) The number density of silicon atoms is given as $N_{Si} = 5 	imes 10^{28} \, 	ext{atoms}/	ext{m}^3$. Since $1 \, 	ext{m}^3 = 10^6 \, 	ext{cm}^3$,we convert the density to $	ext{cm}^3$: $N_{Si} = \frac{5 	imes 10^{28}}{10^6} \, 	ext{atoms}/	ext{cm}^3 = 5 	imes 10^{22} \, 	ext{atoms}/	ext{cm}^3$. It is given that $1$ indium atom is doped per $5 	imes 10^7$ silicon atoms. The number density of acceptor (indium) atoms $N_A$ is calculated as: $N_A = \frac{N_{Si}}{5 	imes 10^7} = \frac{5 	imes 10^{22}}{5 	imes 10^7} = 1 	imes 10^{15} \, 	ext{atoms}/	ext{cm}^3$.

If one indium atom is doped per $5 \times 10^7$ silicon atoms,and the number density of silicon atoms is $5 \times 10^{28} \, \text{atoms}/\text{m}^3$,find the number density of acceptor atoms in $\text{atoms}/\text{cm}^3$.

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