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A silicon specimen is made into a P-type semiconductor by doping, on an average, one Indium atom per $5 \times 10^{7}$ silicon atoms. If the number density of atoms in the silicon specimen is $5 \times 10^{28}$ atoms /m³ then the number of acceptor atoms in silicon per cubic centimetre will be

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$2.5 \times 10^{36} {atoms/cm}^{3}$

$1.0 \times 10^{15} {atoms/cm}^{3}$

$1.0 \times 10^{13} {atoms/cm}^{3}$

$2.5 \times 10^{30} {atoms/cm}^{3}$

answer is C.

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Detailed Solution

Number density of atoms in silicon specimen $= 5 \times 10^{28} {atoms/cm}^{3}$ $= 5 \times 10^{22} {atoms/cm}^{3}$ Since one atom of indium is doped in $= 5 \times 10^{7}$ Si atom. So number of indium atoms doped per cm3 of silicon

$n = \frac{5 \times 10^{22}}{5 \times 10^{7}} = 1 \times 10^{15} {atoms/cm}^{3} .$

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