91Ó°ÊÓ

The given data can be listed below as:

• The length of the one mole of atom is $I=5\mathrm{cm}$.
• The breadth of the one mole of atom is $b=4\mathrm{cm}$.
• The height of the one mole of atom is $h=0.5\mathrm{cm}$.

The interatomic distance is the distance of the atoms within a crystal system. Moreover, the interatomic distance is referred to as the cube root of the division of the product of structure value and mass and density and Avogadro’s number.

The equation of the interatomic distance is expressed as:

$a=\sqrt[3]{\frac{ZM}{dN}}$ …(¾±)

Here, a is the interatomic distance, Z is the structure value, N is the Avogadro’s number $\left(6.023×{10}^{23}\right)$, Mis the mass and d is the density.

The equation of the density is expressed as:

$d=\frac{M}{V}$

Here, $M$is the mass and $V$is the volume.

Substitute the given values in the equation (i).

$a=\sqrt[3]{\frac{ZV}{N}}$

Substitute all the values in the above equation.

$$\begin{gathered} a=\sqrt[3]{\frac{5\mathrm{cm}×4\mathrm{cm}×0.5\mathrm{cm}×1}{6.023×{10}^{23}}} \\ =\sqrt[3]{\frac{10{\mathrm{cm}}^3}{6.023×{10}^{23}}} \\ =2.55×{10}^{-8}\mathrm{cm} \end{gathered}$$

Thus, the interatomic distance will be $2.55×{10}^{-8}\mathrm{cm}$.