If ne and nh represent the number of free electrons and holes respectively in a semiconducting material, then for n-type semiconducting material

Question

If ne and nh represent the number of free electrons and holes respectively in a semiconducting material, then for n-type semiconducting material (A) nE << nH (B) nE >> nH (C) nE = nH (D) nE = nH=0

Tardigrade Admin · Accepted Answer

In an n-type semiconductor, majority charge carriers are electrons and minority charge carriers are holes. Therefore, ne >> nH

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

$n_{E} << n_{H}$

$n_{E} >> n_{H}$

$n_{E} = n_{H}$

$n_{E} = n_{H} = 0$

In an $n$ -type semiconductor, majority charge carriers are electrons and minority charge carriers are holes.
Therefore, $n_{e} >> n_{H}$

Q. If ne​ and nh​ represent the number of free electrons and holes respectively in a semiconducting material, then for n−type semiconducting material

nE​<<nH​

nE​>>nH​

nE​=nH​

nE​=nH​=0