A silicon specimen is made into ap-type semiconductor by doping on an average one indium atom per 5 × 107 silicon atoms. If the number density of atoms in the silicon specimen is 5 × 1028 atom/m3 then the number of acceptor atoms in silicon per cubic centimeter will be

Question

A silicon specimen is made into ap-type semiconductor by doping on an average one indium atom per 5 × 107 silicon atoms. If the number density of atoms in the silicon specimen is 5 × 1028 atom/m3 then the number of acceptor atoms in silicon per cubic centimeter will be (A) 2.5 × 1030 atom /cm3 (B) 2.5 × 1035 atom/cm3 (C) 1.0 × 1013 atom/cm3 (D) 1.0 × 1015 atom/cm3

Tardigrade Admin · Accepted Answer

Number density of atoms in silicon specimen

= 5 \times 1 0^{28} a t o m / m^{3}

= 5 \times 1 0^{22} a t o m / m^{3}

Since, I atom of indium is doped in 5

\times 1 0^{7}

silicon atoms, so total number of indium atom doped per cm

^{3}

of silicon will be

n = \frac{5 \times 1 0 ^{22}}{5 \times 1 0 ^{7}}

= 1 0^{15} a t o m / c m^{3}

Q. A silicon specimen is made into ap-type semiconductor by doping on an average one indium atom per $5 \times 1 0^{7}$ silicon atoms. If the number density of atoms in the silicon specimen is $5 \times 1 0^{28}$ atom/m $^{3}$ then the number of acceptor atoms in silicon per cubic centimeter will be

$2.5 \times 1 0^{30}$ atom / $c m^{3}$

$2.5 \times 1 0^{35}$ atom/ $c m^{3}$

$1.0 \times 1 0^{13}$ atom/ $c m^{3}$

$1.0 \times 1 0^{15}$ atom/ $c m^{3}$

Q. A silicon specimen is made into ap-type semiconductor by doping on an average one indium atom per 5×107 silicon atoms. If the number density of atoms in the silicon specimen is 5×1028 atom/m3 then the number of acceptor atoms in silicon per cubic centimeter will be

2.5×1030 atom /cm3

2.5×1035 atom/cm3

1.0×1013 atom/cm3

1.0×1015 atom/cm3

Number density of atoms in silicon specimen =5×1028atom/m3 =5×1022atom/m3 Since, I atom of indium is doped in 5 ×107 silicon atoms, so total number of indium atom doped per cm3 of silicon will be n=5×1075×1022​ =1015atom/cm3

$2.5 \times 1 0^{30}$ atom / $c m^{3}$

$2.5 \times 1 0^{35}$ atom/ $c m^{3}$

$1.0 \times 1 0^{13}$ atom/ $c m^{3}$

$1.0 \times 1 0^{15}$ atom/ $c m^{3}$