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.

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

To understand why the current in the n-type semiconductor is greater than the current in the p-type semiconductor under the same applied electric field, let us analyze the factors that determine electrical conduction in semiconductors.

The electrical current $I$ in a semiconductor is directly proportional to its electrical conductivity $\sigma$ under a given applied electric field $E$ and cross-sectional area $A$, as described by Ohm's law:
$I=\sigma AE$

For an n-type semiconductor, the majority charge carriers are electrons. Therefore, its conductivity ${\sigma }_n$ is primarily due to electrons and is given by:
${\sigma }_n\approx e{n}_e{\mu }_e$
where:
- $e$ is the elementary charge,
- $n_e$ is the electron concentration, and
- ${\mu }_e$ is the electron mobility.

For a p-type semiconductor, the majority charge carriers are holes. Its conductivity ${\sigma }_p$ is primarily due to holes and is given by:
${\sigma }_p\approx e{n}_h{\mu }_h$
where:
- $n_h$ is the hole concentration, and
- ${\mu }_h$ is the hole mobility.

According to the problem statement, the electron concentration in the n-type semiconductor is equal to the hole concentration in the p-type semiconductor:
$n_e=n_h$

Since both semiconductors are subjected to the same external electric field ($E$) and assuming identical physical dimensions ($A$), the comparison between the currents depends entirely on the mobilities of the respective charge carriers.

Electrons conduct in the conduction band, where they are free from atomic bonds and have a smaller effective mass. Holes, on the other hand, conduct in the valence band, representing the movement of empty states (bound electron vacancies) which requires bound valence electrons to repeatedly break and form covalent bonds. Consequently, the mobility of electrons is always greater than the mobility of holes:
${\mu }_e>{\mu }_h$

Since ${\mu }_e>{\mu }_h$ and $n_e=n_h$, the conductivity of the n-type semiconductor is greater than that of the p-type semiconductor:
${\sigma }_n>{\sigma }_p$

This directly results in a larger current flowing through the n-type semiconductor than the p-type semiconductor:
$I_{\text{n-type}}>I_{\text{p-type}}$