The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.

To compare the currents in the n-type and p-type semiconductors, let us examine the expression for electrical current in a semiconductor.

The current $I$ flowing through a semiconductor under the influence of an applied electric field $E$ is given by the relation:
$I=nqvA$
where:
- $n$ is the concentration of charge carriers,
- $q$ is the charge of each carrier (magnitude $e$),
- $v$ is the drift velocity of the carriers, and
- $A$ is the cross-sectional area of the semiconductor.

The drift velocity $v$ of a charge carrier is directly proportional to the applied electric field $E$ via its mobility $\mu$:
$v=\mu E$

Substituting the drift velocity into the current equation, we obtain:
$I=nq\mu EA$

Let the electron concentration in the n-type semiconductor be $n_e$ and the hole concentration in the p-type semiconductor be $n_h$. According to the question, we are given that:
$n_e=n_h$

Since identical external electric fields ($E$) are applied across identical geometries ($A$), the current in the n-type semiconductor ($I_n$) and the p-type semiconductor ($I_p$) will depend on the mobilities of their respective majority carriers:
$I_n\propto {\mu }_e$
$I_p\propto {\mu }_h$
where ${\mu }_e$ is the mobility of electrons and ${\mu }_h$ is the mobility of holes.

Electrons in the conduction band have a higher mobility compared to holes in the valence band (${\mu }_e>{\mu }_h$). This is because holes move via the vacant sites in the bound covalent bonds, which requires higher activation energy and faces greater resistance compared to the movement of free electrons in the conduction band.

Since ${\mu }_e>{\mu }_h$, it follows that:
$I_n>I_p$

Therefore, the current in the n-type semiconductor is greater than the current in the p-type semiconductor.

The correct option is: current in n-type > current in p-type.