Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
For an intrinsic semiconductor, the law of mass action states that the product of electron concentration ($n_e$) and hole concentration ($n_h$) remains constant at a given temperature, equal to the square of the intrinsic carrier concentration ($n_i^2$):
$$n_e \cdot n_h = n_i^2$$Given data:
- Intrinsic hole concentration ($n_i$) = $5 \times 10^8 \text{ m}^{-3}$
- New electron concentration ($n_e$) = $4 \times 10^{12} \text{ m}^{-3}$
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
Since the semiconductor is intrinsic, the initial electron concentration is equal to the initial hole concentration ($n_e = n_h = n_i$). Therefore, $n_i^2 = (5 \times 10^8)^2$.
Using the law of mass action to find the new hole concentration ($n_h'$):
$$n_h' = \frac{n_i^2}{n_e}$$Substituting the values:
$$n_h' = \frac{(5 \times 10^8)^2}{4 \times 10^{12}}$$ $$n_h' = \frac{25 \times 10^{16}}{4 \times 10^{12}}$$ $$n_h' = 6.25 \times 10^4 \text{ m}^{-3}$$Step 3 (Final Concluded Answer): [0.5 Mark]
The new hole concentration is $6.25 \times 10^4 \text{ m}^{-3}$.
Since the electron concentration ($n_e = 4 \times 10^{12} \text{ m}^{-3}$) is much greater than the hole concentration ($n_h' = 6.25 \times 10^4 \text{ m}^{-3}$), the semiconductor is an n-type semiconductor.
💡 EL10 Concept Insight: Think of the semiconductor like a balanced scale; when you add extra electrons (doping), the number of holes must drop to keep the total "product" of charges balanced, just like how adding more of one ingredient in a recipe forces you to adjust the others to keep the flavor consistent.