Question

In a pure silicon, number of electrons and holes per unit volume are $1.6 \times 10^{16}\ \text{m}^{-3}$. If silicon is doped with Boron in a way that on doping hole density increases to $4 \times 10^{22}\ \text{m}^{-3}$. Then electron density in doped semiconductor will be

• A. $6.4 \times 10^{-9}\ \text{m}^{-3}$
• B. $6.4 \times 10^{9}\ \text{m}^{-3}$
• C. $6.4 \times 10^{-10}\ \text{m}^{-3}$
• D. $6.4 \times 10^{10}\ \text{m}^{-3}$

Hint:

The Law of Mass Action states that the product of carrier concentrations in a semiconductor remains constant at a given temperature, even after doping.

Answer 1

The Correct Option is B

Step 1: Use the mass action law.
In any semiconductor at equilibrium, $n_e \cdot n_h = n_i^2$, where $n_i$ is the intrinsic carrier value.

Step 2: Rearrange for electron density.
\[ n_e = \frac{n_i^2}{n_h}. \] Here $n_i = 1.6\times10^{16}$ and the doped hole density $n_h = 4\times10^{22}$.

Step 3: Plug in.
\[ n_e = \frac{(1.6\times10^{16})^2}{4\times10^{22}} = \frac{2.56\times10^{32}}{4\times10^{22}}. \]

Step 4: Simplify.
$\tfrac{2.56}{4}\times10^{10} = 0.64\times10^{10} = 6.4\times10^{9}$. \[ \boxed{6.4\times10^{9}\ \text{m}^{-3}} \]