A particular semiconductor in equilibrium has 1× 1016 cm- 3 donor atoms, 1.1 × 1017 cm- 3 acceptor atoms. If the intrinsic carrier density (ni) of the semiconductor is 1012 cm- 3 , then the electron density in it will be

Question

A particular semiconductor in equilibrium has 1× 1016 cm- 3 donor atoms, 1.1 × 1017 cm- 3 acceptor atoms. If the intrinsic carrier density (ni) of the semiconductor is 1012 cm- 3 , then the electron density in it will be (A) 1016cm- 3 (B) 1012cm- 3 (C) 1.1× 1017cm- 3 (D) 107cm- 3

Tardigrade Admin · Accepted Answer

From mass action law. ne nh=ni2 and nh=(1.1 × 1017-1016) cm -3=1017 cm -3 ∴ ne=((1012)2/1017)=107 cm -3

Q. A particular semiconductor in equilibrium has $1 \times 1 0^{16} c m^{- 3}$ donor atoms, $1.1 \times 1 0^{17} c m^{- 3}$ acceptor atoms. If the intrinsic carrier density $(n_{i})$ of the semiconductor is $1 0^{12} c m^{- 3}$ , then the electron density in it will be

$1 0^{16} c m^{- 3}$

$1 0^{12} c m^{- 3}$

$1.1 \times 1 0^{17} c m^{- 3}$

$1 0^{7} c m^{- 3}$

From mass action law.
$n_{e} n_{h} = n_{i}^{2}$
and $n_{h} = (1.1 \times 1 0^{17} - 1 0^{16}) c m^{- 3} = 1 0^{17} c m^{- 3}$
$∴ n_{e} = \frac{( 1 0 ^{12} ) ^{2}}{1 0 ^{17}} = 1 0^{7} c m^{- 3}$

Q. A particular semiconductor in equilibrium has 1×1016cm−3 donor atoms, 1.1×1017cm−3 acceptor atoms. If the intrinsic carrier density (ni​) of the semiconductor is 1012cm−3 , then the electron density in it will be

1016cm−3

1012cm−3

1.1×1017cm−3

107cm−3