Concept:The minority carrier concentration in a doped semiconductor at equilibrium is calculated using the Law of Mass Action.
Formula:$$p_n = \frac{n_i^2}{n_n} \approx \frac{n_i^2}{N_D}$$
Solution:- Calculate \( n_i^2 \):
- $$n_i^2 = (1.5 \times 10^{10})^2 = 2.25 \times 10^{20}\text{ cm}^{-6}$$
- Using \( n_n \approx N_D = 1.0 \times 10^{16}\text{ cm}^{-3} \), calculate minority hole concentration:
- $$p_n = \frac{2.25 \times 10^{20}}{1.0 \times 10^{16}} = 2.25 \times 10^4\text{ cm}^{-3}$$
Why other options are incorrect:- Option A: \( 1.5 \times 10^6\text{ cm}^{-3} \) results from dividing \( n_i \) by \( N_D \) without squaring \( n_i \).
- Option C: \( 2.25 \times 10^{10}\text{ cm}^{-3} \) results from an arithmetic exponent error.
- Option D: \( 1.0 \times 10^{16}\text{ cm}^{-3} \) is the majority electron concentration \( n_n \).
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