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A pure Si crystal has $4 \times 10^{28}$ atoms per $\mathrm{m}^3$. It is doped by $1 \mathrm{ppm}$ concentration of antimony. The number of free electrons available will be
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The correct answer is:
$4 \times 10^{22} \mathrm{~m}^{-3}$
Given: Density of Si atoms $=4 \times 10^{28}$ atoms $/ \mathrm{m}^3$ After doping with $1 \mathrm{ppm}$ of Sb,
$\begin{aligned}
\text { No. of Sb atoms } & =\frac{4 \times 10^{28}}{10^6} \\
& =4 \times 10^{22}
\end{aligned}$
The above number of $\mathrm{Sb}$ atoms donates 1 electron each.
$\therefore \quad$ The total number of free electrons will be $4 \times 10^{22} \mathrm{~m}^{-3}$
$\begin{aligned}
\text { No. of Sb atoms } & =\frac{4 \times 10^{28}}{10^6} \\
& =4 \times 10^{22}
\end{aligned}$
The above number of $\mathrm{Sb}$ atoms donates 1 electron each.
$\therefore \quad$ The total number of free electrons will be $4 \times 10^{22} \mathrm{~m}^{-3}$
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