A silicon specimen is made into a P -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 atoms/ m 3, then the number of acceptor atoms in silicon per cubic centimetre will be

Question

A silicon specimen is made into a P -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 atoms/ m 3, then the number of acceptor atoms in silicon per cubic centimetre will be (A) 2.5 × 1030 atoms / cm 3 (B) 1.0 × 1013 atoms / cm 3 (C) 1.0 × 1015 atoms / cm 3 (D) 2.5 × 1036 atoms / cm 3

Tardigrade Admin · Accepted Answer

Number density of atoms in silicon specimen

= 5 \times 1 0^{28}

atom

/ m^{3} = 5 \times 1 0^{22} / c m^{3}

Since one atom of indium is doped in

5 \times 1 0^{7}

Si atom, so number of indium atoms doped per

c m^{3}

of silicon.

n = \frac{5 \times 1 0 ^{22}}{5 \times 1 0 ^{7}} = 1 \times 1 0^{15}

atom

/ c m^{3}

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

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

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

$2.5 \times 1 0^{36}$ atoms $/ c m^{3}$

Q. A silicon specimen is made into a P -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 atoms/ m3, then the number of acceptor atoms in silicon per cubic centimetre will be

2.5×1030 atoms /cm3

1.0×1013 atoms /cm3

1.0×1015 atoms /cm3

2.5×1036 atoms /cm3

Number density of atoms in silicon specimen =5×1028 atom /m3=5×1022/cm3 Since one atom of indium is doped in 5×107 Si atom, so number of indium atoms doped per cm3 of silicon. n=5×1075×1022​=1×1015 atom /cm3

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

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

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

$2.5 \times 1 0^{36}$ atoms $/ c m^{3}$