Question

A silicon specimen is made into a p-type semiconductor by doping on the 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 will be.

• A. $2.5\times {10}^{30}atom/c{m}^3$
• B. $1.0\times {10}^{15}atom/c{m}^3$
• C. $1.0\times {10}^{13}atom/c{m}^3$
• D. $2.5\times {10}^{34}atom/c{m}^3$

Answer 1

Answer: (B)

$1.0\times {10}^{15}atom/c{m}^3$

Number density of atoms in Si
$=5\times {10}^{28}\frac{atoms}{m^3}=5\times {10}^{22}\frac{atoms}{c{m}^3}$
Since $1$ atom of indium is doped in $5\times {10}^7$ silicon atoms,
$\therefore$ Total number of doped indium atoms
$=\frac{5\times {10}^{22}}{5\times {10}^7}=1\times {10}^{15}\frac{atoms}{c{m}^3}$
$\therefore$ Number of acceptor atoms in silicon
$=1\times {10}^{15}\frac{atoms}{c{m}^3}$.