Question

In ${\mu }_e$ and ${\mu }_h$ are electron and hole mobility. E be the applied electric field, the current density $J$ for intristic semiconductor is equal to

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

Answer 1

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

The holes move in the direction of applied electric field and the current density due to them is given as ${\mu }_h{n}_{i}eE$.

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 ${\mu }_e{n}_{i}eE$.

Hence total current density $=J_h+J_e$ $=n_{i}e({\mu }_e+{\mu }_h)E$