A silicon specimen is made into a p-type semiconductor by doping, on average, one indium atom per \( 5\times10^{7} \) silicon atoms. If the number density of atoms in the silicon specimen is \( 5\times10^{28}\,\mathrm{atoms\,m^{-3}} \), the number of acceptor atoms in silicon per cubic centimetre will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.0\times10^{15} \)
Indium is trivalent, so each indium atom in silicon is an acceptor. With one indium atom per \( 5\times10^{7} \) silicon atoms, the number density of acceptors is:
\( \begin{aligned} N_{\mathrm{A}} &= \frac{5\times10^{28}}{5\times10^{7}}\,\mathrm{m^{-3}} \\ &= 1.0\times10^{21}\,\mathrm{m^{-3}} \end{aligned} \)
Since \( 1\,\mathrm{m^{3}}=10^{6}\,\mathrm{cm^{3}} \), per cubic centimetre this is:
\( \begin{aligned} N_{\mathrm{A}} &= \frac{1.0\times10^{21}}{10^{6}}\,\mathrm{cm^{-3}} \\ &= 1.0\times10^{15}\,\mathrm{cm^{-3}} \end{aligned} \)