Q153 / 160 · TS EAMCET 2015 · Physics
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
- A$\left(\frac{n_e}{n_h}\right)=n_i$
- B$\left(n_e+n_h\right)=n_i$
- C$\left(n_e-n_h\right)=n_i^2$
- Correct answerD$n_e n_h=n_i^2$
Solution
$n_e$ and $n_h$ are electron and hole concentrations in an extrinsic semiconductor and $n_i$ is electron concentration in an intrinsic semiconductor, then
$$
n_e n_h=n_i^2
$$
Hence, the correct answer is option: 'D'.