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 atom c m- 3 (B) 2.5 × 1035 atom c m- 3 (C) 1 × 1013 atom c m- 3 (D) 1 × 1015 atom c m- 3

Tardigrade Admin · Accepted Answer

Number density of atoms in silicon specimen
=

5 \times 1 0^{28}

atoms-

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

atoms

c m^{- 3}

.
Since, 1 atom of indium is doped in

5 \times 1 0^{7}

silicon atoms, so total number of indium atoms doped per

c m^{3}

of silicon will be

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

atom

c m^{- 3}

.

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

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

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

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

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 m−3 , then the number of acceptor atoms in silicon per cubic centimetre will be

2.5×1030 atom cm−3

2.5×1035 atom cm−3

1×1013 atom cm−3

1×1015 atom cm−3

Number density of atoms in silicon specimen = 5×1028 atoms- m−3=5×1022 atoms cm−3 . Since, 1 atom of indium is doped in 5×107 silicon atoms, so total number of indium atoms doped per cm3 of silicon will be n=5×1075×1022​=1015 atom cm−3 .

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

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

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

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