Q. Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 × 10¹⁶ m⁻³. Doping by indium increases nh to 4.5 × 10²² m⁻³. The doped semiconductor is of
A) n-type with electron concentration ne = 2.5 × 10²³ m⁻³
B) p-type having electron concentration ne = 5 × 10⁹ m⁻³
C) n-type with electron concentration ne = 2.5 × 10²² m⁻³
D) p-type with electron concentration ne = 2.5 × 10¹⁰ m⁻³
Explanation
[Solution from source] Indium is trivalent, so the doped crystal is p-type. By the mass-action law ne = ni²/nh = (1.5 × 10¹⁶)²/(4.5 × 10²²) = 2.25 × 10³²/4.5 × 10²² = 5 × 10⁹ m⁻³. The n-type choices are wrong because an acceptor dopant was added. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.