Step 1: Concept:
The question asks for the approximate position of the Fermi level ($E_F$) in the energy band diagram of an extrinsic semiconductor after doping.
Step 2: Key Formula or Approach:
- A semiconductor doped with pentavalent impurities (Group 15 elements like P, As, Sb) creates an n-type semiconductor.
- These pentavalent atoms have one extra electron that doesn't fit into the primary covalent bonding network. This creates a highly localized, easily ionized energy level called the donor level ($E_d$), positioned just slightly below the bottom edge of the conduction band ($E_c$).
- In an n-type semiconductor at moderate temperatures, the Fermi level shifts upward from the intrinsic mid-gap position toward the conduction band to reflect the massive increase in electron concentration.
Step 3: Step-by-step Explanation:
• Option (A): "Exactly between conduction and valence band." This describes an undoped, intrinsic semiconductor.
• Option (C): "Between acceptor level and top of valence band." This describes a p-type semiconductor (doped with trivalent impurities like Boron).
• Option (D): "Between donor level and top of valence band." The donor level is near the top of the gap, and the valence band is at the bottom. This statement encompasses almost the entire bandgap and is too vague/incorrect for a moderately doped n-type material.
• Option (B): "Between donor level and bottom of conduction band." At absolute zero (0 K), the Fermi level sits exactly halfway between $E_d$ and $E_c$. At moderate (room) temperatures, as donor electrons are excited into the conduction band, the Fermi level drops slightly but generally remains situated in the narrow energy window between the donor level ($E_d$) and the conduction band edge ($E_c$).
Step 4: Final Answer:
The Fermi level lies between the donor level and the bottom of the conduction band, matching option (B).