Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 426 of 494
In an extrinsic P-type semiconductor, how does the Fermi energy level (\( E_F \)) shift relative to its intrinsic position?
A
It shifts upward toward the top of the conduction band
B
It moves outside the crystal lattice boundary
C
It remains fixed at the exact center of the bandgap
D
It shifts downward closer to the top edge of the valence band
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: It shifts downward closer to the top edge of the valence band
Concept:

Acceptor dopants increase the hole concentration in the valence band, shifting the Fermi level closer to the valence band edge.

Formula:

$$E_i - E_F = k_B T \ln\left(\frac{N_A}{n_i}\right) > 0$$

Solution:

  • Doping with acceptor atoms increases the hole concentration (\( p \approx N_A \gg n_i \)).


  • This increases the probability of unoccupied states near the valence band, shifting the Fermi level (\( E_F \)) downward toward \( E_v \).


Why other options are incorrect:

  • Option A: An upward shift toward the conduction band occurs in N-type semiconductors.
  • Option B: The Fermi level is an internal energy parameter of the semiconductor material.
  • Option C: The midpoint position characterizes undoped intrinsic material.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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