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The number of silicon atoms per $\mathrm{m}^3$ is $5 \times 10^{28}$. This is doped with $4.5 \times 10^{21}$ atoms $/ \mathrm{m}^3$ of Arsenic. The ratio of number of electrons to number of holes after doping is (Take $n_i=$ Number of thermally-generated electrons $\left.=1.5 \times 10^{16} / \mathrm{m}^3\right)$
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Verified Answer
The correct answer is:
$9 \times 10^{12}$
We have
$$
\mathrm{n}_{\mathrm{e}}=4.5 \times 10^{21}
$$
As $\mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{e}}}=\frac{\left(1.5 \times 10^{16}\right)^2}{4.5 \times 10^{21}}$
$$
\Rightarrow \mathrm{n}_{\mathrm{h}}-5 \times 10^{10}
$$
So, $\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{h}}}=\frac{4.5 \times 10^{21}}{5 \times 10^{10}}=9 \times 10^{10}$
$$
\mathrm{n}_{\mathrm{e}}=4.5 \times 10^{21}
$$
As $\mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{e}}}=\frac{\left(1.5 \times 10^{16}\right)^2}{4.5 \times 10^{21}}$
$$
\Rightarrow \mathrm{n}_{\mathrm{h}}-5 \times 10^{10}
$$
So, $\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{h}}}=\frac{4.5 \times 10^{21}}{5 \times 10^{10}}=9 \times 10^{10}$
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