Question

${\mu }_e$ and ${\mu }_h$ are the electron and hole mobilities of a semiconducting crystal respectively. $E$ is the applied electric field. Then the current density $J$ for the intrinsic semiconductor is (Take $n_i$ the intrinsic concentration of the semiconductor)-

• A. $\frac{n_{i}e({\mu }_e+{\mu }_h)}{E}$
• B. $\frac{E}{n_{i}e({\mu }_e+{\mu }_h)}$
• C. $n_{i}e({\mu }_e+{\mu }_h)E$
• D. $n_{i}e({\mu }_e-{\mu }_h)E$

Answer 1

The correct option is C $n_{i}e({\mu }_e+{\mu }_h)E$

The holes move in the direction of applied electric field and the current density due to them is given as ${\mu }_h{n}_i\text{eE}$.
The electrons move in the direction opposite to that of the electric field, and current due to that is in the direction opposite to that of the movement of electrons, that is in the direction of the electric field.
Current density due to movement of electrons is${\mu }_e{n}_i\text{eE}$
Hence, total current density $=J_n+J_e$
$=n_{i}e({\mu }_e+{\mu }_h)E$