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A silicon specimen is made into a P-type semiconductor by doping, on an average, one Indium atom per 5 x $10^{7}$ silicon atoms. If the number density of atoms in the silicon specimen is 5 x $10^{28}$ atoms/ $m^{3}$ , then the number of acceptor atoms in silicon per cubic centimetre will be

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

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

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

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

answer is C.

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Detailed Solution

Number density of atoms in silicon specimen = $5 x 10^{28} atom / m^{3} = 5 x 10^{22} / {cm}^{3}$

Since one atom of Indium is doped in $5 x 10^{7}$ Si atom, so number of indium atoms doped per ${cm}^{3}$ of silicon.

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

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