MCQ
A silicon speciman is made into a $P-$type semi-conductor by dopping, on an average, one Indium atom per $5 \times {10^7}$silicon atoms. If the number density of atoms in the silicon specimen is $5 \times {10^{28}}{\rm{atoms}}/{m^3}$ then the number of acceptor atoms in silicon per cubic centimetre will be
  • A
    $2.5 \times {10^{30}}atoms/c{m^3}$
  • B
    $1.0 \times {10^{13}}atoms/c{m^3}$
  • $1.0 \times {10^{15}}atoms/c{m^3}$
  • D
    $2.5 \times {10^{36}}atoms/c{m^3}$

Answer

Correct option: C.
$1.0 \times {10^{15}}atoms/c{m^3}$
c
(c)Number density of atoms in silicon specimen $= 5 × 10^{28} \,\,atom/m^{3} = 5 × 10^{22}\,\, atom/cm^{3}$
Since one atom of indium is doped in $5 × 10^{7}$ Si atom. So number of indium atoms doped per $cm^3 $ of silicon.$n = \frac{{5 \times {{10}^{22}}}}{{5 \times {{10}^7}}} = 1 \times {10^{15}}\,atom/c{m^3}$

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