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: Read the situation.
Pure silicon has equal numbers of electrons and holes, here $1.6 \times 10^{16}$ per cubic metre. After adding Boron the hole count shoots up to $4 \times 10^{22}$ per cubic metre. We want the new electron count.

Step 2: The rule that ties them together.
In any sample at a fixed temperature, the product of electron density and hole density stays the same. This is the law of mass action: $n_e \cdot n_h = n_i^2$, where $n_i$ is the pure (intrinsic) value.

Step 3: Rearrange for what we want.
We want $n_e$, so $n_e = \dfrac{n_i^2}{n_h}$.

Step 4: Square the intrinsic value.
$n_i^2 = (1.6 \times 10^{16})^2 = 2.56 \times 10^{32}$.

Step 5: Divide by the new hole density.
$n_e = \dfrac{2.56 \times 10^{32}}{4 \times 10^{22}} = 0.64 \times 10^{10}$.

Step 6: Tidy up the answer.
$0.64 \times 10^{10} = 6.4 \times 10^{9}$ per cubic metre. Adding holes pushes the electron count way down, which makes sense, and this is option (B).
\[ \boxed{n_e = 6.4 \times 10^{9}\ \text{m}^{-3}} \]