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A silicon specimen is made into a P-type semi-conductor 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 $m^{- 3},$ then the number of acceptor atoms in silicon per cubic centimetre will be

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

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

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

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

answer is C.

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Detailed Solution

No of acceptor atoms doped per cubic meter= $\frac{5 \times 10^{28}}{5 \times 10^{7}} = 10^{21}$

No of acceptor atoms per cubic cm= $\frac{10^{21}}{10^{6}} = 10^{15}$

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