Given Data:
- Intrinsic carrier concentration, $n_i = 5 \times 10^8 \text{ m}^{-3}$
- Hole concentration in extrinsic semiconductor, $n_h = 8 \times 10^{12} \text{ m}^{-3}$
Step 1 (Identification and Governing Law): [0.5 Mark]
Since the hole concentration ($n_h$) is significantly greater than the intrinsic carrier concentration ($n_i$), the semiconductor is doped with an acceptor impurity (trivalent atoms like Boron, Aluminium, etc.).
- (i) Type of dopant: Acceptor impurity (Trivalent).
- (ii) Extrinsic semiconductor: p-type semiconductor.
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
According to the Law of Mass Action for semiconductors:
$$n_e \cdot n_h = n_i^2$$Substituting the given values to find the electron concentration ($n_e$):
$$n_e = \frac{n_i^2}{n_h}$$ $$n_e = \frac{(5 \times 10^8)^2}{8 \times 10^{12}}$$ $$n_e = \frac{25 \times 10^{16}}{8 \times 10^{12}}$$ $$n_e = 3.125 \times 10^4 \text{ m}^{-3}$$Step 3 (Final Concluded Answer): [0.5 Mark]
The electron concentration in the extrinsic semiconductor is $3.125 \times 10^4 \text{ m}^{-3}$.
💡 EL10 Concept Insight: Think of the semiconductor like a crowded room where the product of "boys" and "girls" always stays the same; if you add a lot of "girls" (holes) by doping, the number of "boys" (electrons) must drop drastically to keep that balance constant!