Official Correct Choice:
Option C: Near middle of bandgap (intrinsic), close to conduction band (N-type), close to valence band (P-type)
Concept:The Fermi level \( E_F \) represents the chemical potential for electrons. Doping shifts \( E_F \) toward the conduction band (N-type) or toward the valence band (P-type).
Formula:$$E_F(\text{intrinsic}) \approx E_i = \frac{E_c + E_v}{2}$$
$$E_F(\text{N-type}) = E_c - k T \ln\left(\frac{N_c}{N_D}\right), \quad E_F(\text{P-type}) = E_v + k T \ln\left(\frac{N_v}{N_A}\right)$$
Solution:- In an intrinsic semiconductor (\( n = p \)), \( E_F \) lies near the middle of the forbidden bandgap.
- In an N-type semiconductor (\( n \gg p \)), \( E_F \) shifts upward close to the conduction band edge \( E_c \).
- In a P-type semiconductor (\( p \gg n \)), \( E_F \) shifts downward close to the valence band edge \( E_v \).
Why other options are incorrect:- Option A: This reverses the Fermi level positions for intrinsic, N-type, and P-type materials.
- Option B: The Fermi level only enters the conduction band in degenerate, heavily doped semiconductors (\( N_D > 10^{19}\text{ cm}^{-3} \)).
- Option D: Fermi levels describe electronic states within the bandgap, not metallic cores.
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