Q. If $n_{E\ }\ and\ n_{H}$ represent the number of free electrons and holes respectively in a semiconducting material, then for $n$-type semiconducting material
A) $n_{E\ }$ < < $n_{H}$
B) $n_{E\ }$ > > $n_{H}$
C) $n_{E\ }$ = $n_{H}$
D) $n_{E\ }$ = $n_{H}$ = 0
Explanation
(b) In an $n$-type semiconductor, majority charge carriers are electrons and minority charge carriers are holes.
Therefore, $n_{E\ }$ > > $n_{H}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.