A pure semiconductor has equal electron and hole concentration of 1016m-3 . Doping by indium increases nh to 5× 1022m-3 . Then, the value of ne in the doped semiconductor is

Question

A pure semiconductor has equal electron and hole concentration of 1016m-3 . Doping by indium increases nh to 5× 1022m-3 . Then, the value of ne in the doped semiconductor is (A)  106/m3  (B)  1022/m3  (C)  2× 106/m3  (D)  1019/m3

Tardigrade Admin · Accepted Answer

(ni)2=nenh (1016)2=ne× 5× 1022 ∴ ne=(1016× 1016/5× 1022) ne=2× 109/m3

Q. A pure semiconductor has equal electron and hole concentration of $10^{16} m^{- 3}$ . Doping by indium increases $n_{h}$ to $5 \times 10^{22} m^{- 3}$ . Then, the value of $n_{e}$ in the doped semiconductor is

$10^{6} / m^{3}$

$10^{22} / m^{3}$

$2 \times 10^{6} / m^{3}$

$10^{19} / m^{3}$

$2 \times 10^{9} / m^{3}$

$(n_{i})^{2} = n_{e} n_{h}$
$(10^{16})^{2} = n_{e} \times 5 \times 10^{22}$
$∴$ $n_{e} = \frac{10 ^{16} \times 10 ^{16}}{5 \times 10 ^{22}}$
$n_{e} = 2 \times 10^{9} / m^{3}$

Q. A pure semiconductor has equal electron and hole concentration of 1016m−3 . Doping by indium increases nh​ to 5×1022m−3 . Then, the value of ne​ in the doped semiconductor is

106/m3

1022/m3

2×106/m3

1019/m3

2×109/m3

(ni​)2=ne​nh​ (1016)2=ne​×5×1022 ∴ ne​=5×10221016×1016​ ne​=2×109/m3