Answer

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**Hint:**First we have to convert the data into cubic meters.

Then we use the concept of intrinsic semiconductor for finding the required solution.

On some simplification, we get the answer.

**Formula used:**

$1cm = {10^{ - 2}}meter$

$1mm = {10^{ - 3}}meter$

The number of the conduction electron $(N)$ in a sample of the volume $(V)$ is,

$N = $ volume of the sample \[ \times \] number of conduction electrons per cubic meter.

The number of electron density is $n$and the number of hole density \[p\] is equal i.e. \[n = p\].

**Complete step by step answer:**

In an intrinsic semiconductor how many electrons become free, the same number of holes are created.

Here the number of electron density is $n$ and the number of hole density \[p\] is equal.

We can write it as, \[n = p\]

It is given that the number of the conduction electrons in a pure semiconductor is $6 \times {10^{19}}$ per cubic meter.

We have to calculate the number of holes in a sample of size $(1cm \times 1cm \times 2mm)$.

First, we convert the given size in the cubic meter,

So we can write it as, $1cm \times 1cm \times 2mm = ({10^{ - 2}} \times {10^{ - 2}} \times 2 \times {10^{ - 3}}){m^3} = (2 \times {10^{ - 7}}){m^3}$

$\therefore $ The number of the conduction electron $(n)$ in the sample of the volume $(2 \times {10^{ - 7}}){m^3}$ is,

$n = $ volume of the sample \[ \times \] number of conduction electrons per cubic meter.

\[\therefore n = 2 \times {10^{ - 7}} \times 6 \times {10^{19}}\]

Let us multiply the terms an we get,

\[ \Rightarrow n = 12 \times {10^{12}}\]

Since the number of the hole is equal to the number of the conduction electron,

\[n = p\]

Hence, the number of the hole(\[p\]) in the sample of the volume $(2 \times {10^{ - 7}}){m^3}$ is equal to $12 \times {10^{12}}$ .

$\therefore $ The number of holes in the sample of the volume is $12 \times {10^{12}}$

**Note:**We take the number of electron density is $n$ and the number of hole density \[p\] is equal.

Let in a pure Silicon crystal the electron situated in the position of ‘A’ in the bond (fig-1) breaks the bond and goes to the position ‘X’.

That means a valence electron becomes a conduction electron.

In the meantime, in the position, ‘A’ there occurs a crisis of an electron, and then in the position ‘A’ there generates a positive charge according to the corresponding electron clouds.

This crisis of electrons in the bond is named Hole.

That’s why in pure semiconductor the number of electron density is $n$ and the number of hole density \[p\] is equal.

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