Question

A semiconductor has equal electron and hole concentration of $6 \times 10^{8}$ per $m^{3}$ on doping with certain impurity, electron concentration increases to $9 \times 10^{12}perm^{3}$ The new hole concentration is

• A. $2 \times 10^{4}$per $m^{3}$
• B. $2 \times 10^{2}$ per $m^{2}$
• C. $4 \times 10^{4}$ per $m^{3}$
• D. $4 \times 10^{2}$per $m^{3}$

Answer 1

Answer: (C)

$4 \times 10^{4}$ per $m^{3}$

Answer 2

: As $n_{e}n_{h} = n_{i}^{2}$

Here, $n_{i} = 6 \times 10^{8}perm^{3}$and $n_{e} = 9 \times 10^{12}perm^{3}$

$\therefore n_{h} = \frac{n_{i}}{n_{e}} = \frac{(6 \times 10^{8})^{2}}{9 \times 10^{12}} = 4 \times 10^{4}perm^{3}$