Concept:Because electron mobility is roughly 2.5 to 3 times higher than hole mobility (\( \mu_n > \mu_p \)), N-channel devices achieve lower on-resistance and faster carrier transit times (\( \tau = L / v = L / (\mu E) \)).
Formula:$$\tau_{\text{transit}} = \frac{L^2}{\mu V} \implies \text{Higher } \mu_n \implies \text{Shorter Transit Time } \tau \implies \text{Faster Switching}$$
Solution:- Higher electron mobility (\( \mu_n \)) allows carriers to drift faster under an applied electric field.
- This reduces transit times through device channels, enabling higher switching speeds and better high-frequency performance.
Why other options are incorrect:- Option B: All semiconductor materials contain atomic nuclei.
- Option C: P-type silicon is widely manufactured and used across the semiconductor industry.
- Option D: Electrons always carry a negative charge of \( -1.6 \times 10^{-19}\text{ C} \).
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