Concept:The depletion width \( W \) is determined by the doping concentrations and permittivity of the semiconductor, and is typically on the order of one micrometer.
Formula:$$W = \sqrt{\frac{2 \epsilon_s (V_0 + V_R)}{q} \left( \frac{1}{N_A} + \frac{1}{N_D} \right)} \approx 10^{-6}\text{ m} = 1\ \mu\text{m}$$
Solution:- For moderate doping levels (\( 10^{16}\text{ cm}^{-3} \)), the equilibrium depletion thickness spans roughly \( 0.1\ \mu\text{m} \) to \( 1\ \mu\text{m} \) (\( 10^{-6}\text{ m} \)).
Why other options are incorrect:- Option B: \( 10^{-2}\text{ m} \) (1 cm) is macroscopic and far larger than microscopic semiconductor junctions.
- Option C: \( 10^{-12}\text{ m} \) is smaller than the diameter of an atom (\( \approx 10^{-10}\text{ m} \)).
- Option D: \( 10^{-4}\text{ m} \) (100 µm) is roughly the thickness of an entire semiconductor chip wafer.
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