In μe and μh are electron and hole mobility. E be the applied electric field, the current density τ for intristic semiconductor is equal to

Question

In μe and μh are electron and hole mobility. E be the applied electric field, the current density τ for intristic semiconductor is equal to (A) ni e(μe+μh) E (B) ni e(μe-μh) E (C) (ni e(μe+μh)/E) (D) (E/ni e(μe+μn))

Tardigrade Admin · Accepted Answer

The holes move in the direction of applied electric field and the current density due to them is given as

μ_{h} n_{i} e E

.
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

μ_{2} n_{1} e E

.
Hence total current density

= J_{h} + J_{e} = n_{i} e (μ_{e} + μ_{h}) E

Q. In $μ_{e}$ and $μ_{h}$ are electron and hole mobility. $E$ be the applied electric field, the current density $τ$ for intristic semiconductor is equal to

$n_{i} e (μ_{e} + μ_{h}) E$

$n_{i} e (μ_{e} - μ_{h}) E$

$\frac{n _{i} e ( μ _{e} + μ _{h} )}{E}$

$\frac{E}{n _{i} e ( μ _{e} + μ _{n} )}$

Q. In μe​ and μh​ are electron and hole mobility. E be the applied electric field, the current density τ for intristic semiconductor is equal to

ni​e(μe​+μh​)E

ni​e(μe​−μh​)E

Eni​e(μe​+μh​)​

ni​e(μe​+μn​)E​

Q. In $μ_{e}$ and $μ_{h}$ are electron and hole mobility. $E$ be the applied electric field, the current density $τ$ for intristic semiconductor is equal to

$n_{i} e (μ_{e} + μ_{h}) E$

$n_{i} e (μ_{e} - μ_{h}) E$

$\frac{n _{i} e ( μ _{e} + μ _{h} )}{E}$

$\frac{E}{n _{i} e ( μ _{e} + μ _{n} )}$