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.

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

Explanation:

To compare the electric current flowing through an n-type semiconductor and a p-type semiconductor, let us look at the fundamental formula for electric current in a semiconductor:

$I=n·q·A·v{d}_{}$

Here, the drift velocity $v{d}_{}$ is related to the applied electric field $E$ by the mobility $\mu$:

$v{d}_{}=\mu ·E$

Therefore, the current density $J$ or current $I$ can be expressed as:

$I=n·q·A·\mu ·E$

1. For the n-type semiconductor:

The current is primarily conducted by majority charge carriers, which are electrons. Thus, the current $I$_n is:

$I{n}_{}=n{e}_{}·e·A·\mu e_{}·E$

where $n{e}_{}$ is the electron concentration, and $\mu e_{}$ is the electron mobility.

2. For the p-type semiconductor:

The current is primarily conducted by majority charge carriers, which are holes. Thus, the current $I$_p is:

$I{p}_{}=n{h}_{}·e·A·\mu h_{}·E$

where $n{h}_{}$ is the hole concentration, and $\mu h_{}$ is the hole mobility.

Given Conditions:

- Electron concentration in the n-type equals hole concentration in the p-type: $n{e}_{}=n{h}_{}$.
- The applied electric field $E$ and cross-sectional area $A$ are the same for both.

Key Physical Concept:

Electrons move in the conduction band where they are free from atomic bounds, while holes represent vacancies in the valence band moving via continuous breaking and reforming of covalent bonds. As a result, the mobility of electrons is significantly greater than the mobility of holes:

$\mu e_{}>\mu h_{}$

Since $\mu e_{}>\mu h_{}$, it directly follows that:

$I{n}_{}>I{p}_{}$

Hence, the current flowing in the n-type semiconductor is greater than the current in the p-type semiconductor (current in n-type > current in p-type).