Question

In an intrinsic semiconductor, the number of conduction electrons is $9\times {10}^{19}$ per cubic meter. Find the total number of current carriers $\left(\text{electrons and hole}\right)$ in a semiconductor of size $1\text{cm}\times 1\text{cm}\times 1\text{mm}$

• A. $9\times {10}^{12}$
• B. $18\times {10}^{12}$
• C. $9\times {10}^7$
• D. $8\times {10}^7$

Answer 1

The correct option is B $18\times {10}^{12}$

Let $n_e$ be the number of electrons and $n_h$ be the number of holes.

In an intrinsic semiconductor, $n_e=n_h$

Total charge carriers per unit volume, $n=n_e+n_h$

Given that, the Volume of semiconductor $\left(V\right)$,

$V=1\text{cm}\times 1\text{cm}\times 1\text{mm}=1\times {10}^{-2}\times 1\times {10}^{-2}\times 1\times {10}^{-3}$

$\Rightarrow V=1\times {10}^{-7}{\text{m}}^3$

Total number of charge carrier in volume $V$,

$N=(n_e+n_h)\times V$

$\Rightarrow N=(9\times {10}^{19}+9\times {10}^{19})\times {10}^{-7}$

$\Rightarrow N=18\times {10}^{19}\times {10}^{-7}=18\times {10}^{12}$

Hence, option $\left(B\right)$ is correct.