If ne and nh are electron and hole concentrations in an extrinsic semiconductor and ni is electron concentrations in an intrinsic semiconductor, then

Question

If ne and nh are electron and hole concentrations in an extrinsic semiconductor and ni is electron concentrations in an intrinsic semiconductor, then (A) ((ne/nh))=ni (B) (ne+nh)=ni (C) (ne-nh)=ni2 (D) ne nh=ni2

Tardigrade Admin · Accepted Answer

ne and nh are electron and hole concentrations in an extrinsic semiconductor and ni is electron concentration in an intrinsic semiconductor, then ne nh=ni2

Q. If $n_{e}$ and $n_{h}$ are electron and hole concentrations in an extrinsic semiconductor and $n_{i}$ is electron concentrations in an intrinsic semiconductor, then

$(\frac{n _{e}}{n _{h}}) = n_{i}$

$(n_{e} + n_{h}) = n_{i}$

$(n_{e} - n_{h}) = n_{i}^{2}$

$n_{e} n_{h} = n_{i}^{2}$

$n_{e}$ and $n_{h}$ are electron and hole concentrations in
an extrinsic semiconductor and $n_{i}$ is electron concentration in an intrinsic semiconductor, then
$n_{e} n_{h} = n_{i}^{2}$

Q. If ne​ and nh​ are electron and hole concentrations in an extrinsic semiconductor and ni​ is electron concentrations in an intrinsic semiconductor, then

(nh​ne​​)=ni​

(ne​+nh​)=ni​

(ne​−nh​)=ni2​

ne​nh​=ni2​