Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]

The relationship between the concentration of electrons ($n_e$), holes ($n_h$), and intrinsic charge carriers ($n_i$) in a semiconductor is governed by the Law of Mass Action:

$$n_e \cdot n_h = n_i^2$$

Given Data:

• Density of Si atoms ($N$) = $5 \times 10^{28} \text{ m}^{-3}$
• Doping concentration = $1 \text{ ppm} = 10^{-6}$
• Concentration of holes ($n_h$) = $4.5 \times 10^9 \text{ m}^{-3}$

Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]

First, calculate the concentration of donor atoms ($N_D$), which represents the concentration of electrons ($n_e$) in an n-type semiconductor:

$$n_e \approx N_D = 10^{-6} \times (5 \times 10^{28} \text{ m}^{-3}) = 5 \times 10^{22} \text{ m}^{-3}$$

Now, substitute the values into the Law of Mass Action formula to find $n_i$:

$$n_i^2 = n_e \cdot n_h$$ $$n_i^2 = (5 \times 10^{22}) \times (4.5 \times 10^9)$$ $$n_i^2 = 22.5 \times 10^{31} = 2.25 \times 10^{32}$$ $$n_i = \sqrt{2.25 \times 10^{32}}$$ $$n_i = 1.5 \times 10^{16} \text{ m}^{-3}$$

Step 3 (Final Concluded Answer): [0.25 Mark]

The concentration of intrinsic charge carriers is $1.5 \times 10^{16} \text{ m}^{-3}$.

Correct Option: (B)

💡 EL10 Concept Insight: Think of the semiconductor like a crowded room where the total number of "partnerships" between electrons and holes stays constant; if you add a bunch of extra electrons by doping, the number of holes must drop significantly to keep that balance, allowing us to calculate the original "natural" state of the material.