The concentration of hole electron pairs in pure silicon at T=300 K is 7×1015 per cubic meter. Antimony is doped into silicon in a proportion of 1 atom in 107Si atoms. Assuming that half of the impurity atoms contribute electron in the conduction band, calculate the factor by which the number of charge carriers increases due to doping. The number of silicon atoms per cubic meter is 5×1028

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The concentration of hole electron pairs in pure silicon at $T = 300 K$ is $7 \times 10^{15}$ per cubic meter. Antimony is doped into silicon in a proportion of 1 atom in $10^{7} Si$ atoms. Assuming that half of the impurity atoms contribute electron in the conduction band, calculate the factor by which the number of charge carriers increases due to doping. The number of silicon atoms per cubic meter is $5 \times 10^{28}$

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$1.8 \times 10^{5}$

$2.8 \times 10^{5}$

$4.2 \times 10^{5}$

$3.1 \times 10^{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} In pure semiconductor electron-hole pair = 7 \times 10^{15} / m^{3} \\ n_{initial} = n_{h} + n_{e} = 14 \times 10^{15} \\ After doping donor Impurity, \\ N_{D} = \frac{5 \times 10^{28}}{10^{7}} = 5 \times 10^{21} a n d n_{e} = \frac{N_{D}}{2} = 2.5 \times 10^{21} \\ S o, n_{final} = n_{h} + n_{e} \\ \Rightarrow n_{final} \approx n_{e} \approx 2.5 \times 10^{21} (∵ n_{e} > > n_{h}) \\ F a c t o r = \frac{n_{final} - n_{initial}}{n_{initial}} = \frac{2.5 \times 10^{21} - 14 \times 10^{15}}{14 \times 10^{15}} \approx \frac{2.5 \times 10^{21}}{14 \times 10^{15}} = 1.8 \times 10^{5} \end{array}$