Question

A silicon specimen is made into ap-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 $5 \times 10^{28}$ atom/m$^3$ then the number of acceptor atoms in silicon per cubic centimeter will be

• A. $2.5 \times 10^{30}$ atom /$cm^3$
• B. $2.5 \times 10^{35}$ atom/$cm^3$
• C. $1.0 \times 10^{13}$ atom/$cm^3$
• D. $1.0 \times 10^{15}$ atom/$cm^3$

Answer 1

Answer: (D)

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