Question

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 $m^{- 3}$ , then the number of acceptor atoms in silicon per cubic centimetre will be

• A. $2.5 \times 10^{30}$ atom $c m^{- 3}$
• B. $2.5 \times 10^{35}$ atom $c m^{- 3}$
• C. $1 \times 10^{13}$ atom $c m^{- 3}$
• D. $1 \times 10^{15}$ atom $c m^{- 3}$

Answer 1

Answer: (D)

Number density of atoms in silicon specimen
= $5 \times 10^{28}$ atoms- $m^{- 3}=5 \times 10^{22}$ atoms $c m^{- 3}$ .
Since, 1 atom of indium is doped in $5 \times 10^{7}$ silicon atoms, so total number of indium atoms doped per $c m^{3}$ of silicon will be
$n=\frac{5 \times 10^{22}}{5 \times 10^{7}}=10^{15}$ atom $c m^{- 3}$ .