91Ó°ÊÓ

The given data is listed below as,

• Number of atoms in one mole of tungsten is,$N_A=6.02×{10}^{23}$
• The mass of one tungsten’s mole is, 184 g
• The density of the tungsten atom is,$d=19.3\mathrm{g}/{\mathrm{cm}}^3$

The density of a substance is directly proportional to the mass of the substance and inversely proportional to the volume of the substance.

It can be expressed as follows,

$\mathbf{d}\mathbf{=}\frac{\mathbf{m}}{\mathbf{v}}$

Here, m is the mass of the element andv is the volume of the element.

The mass of one tungsten atom can be expressed as,

$m=\frac{184g}{N_A}$

Here, $N_A$is the Avogadro’s number.

For ,$N_A=6023x{10}^2$

$m=\begin{array}{c}\frac{184g}{60023×{10}^8}\\ =3.05×{10}^{2}9\end{array}$

The expression of the volume of the tungsten atom can be expressed as,

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

Here, is the volume of the tungsten atom, is the mass of one tungsten atom and is the density of the atom.

For $m=305x{10}_9^2$and $d=1930\mathrm{mm}$,

$$\begin{gathered} v=\frac{3.05×{10}^{-22}\mathrm{g}}{19.3\mathrm{g}/{\mathrm{cm}}^3} \\ =1.58×{10}^{-23}{\mathrm{cm}}^3 \end{gathered}$$

Another expression for the volume of the tungsten atom can be expressed as,

$v=a^3$

Here, a is the diameter of the atom

role="math" localid="1656673927324" $$\begin{gathered} \mathrm{For}\mathrm{v}=1.58×{10}^{-23}{\mathrm{cm}}^3 \\ 1.58×{10}^{-23}{\mathrm{cm}}^3={\mathrm{a}}^3 \\ \mathrm{a}={(1.58×{10}^{-23}{\mathrm{cm}}^3)}^{\frac{1}{3}} \\ =2.5×{10}^{-8}\mathrm{cm} \end{gathered}$$

Thus, the approximate diameter of the tungsten atom in a solid block of the material is $2.5×{10}^{-8}\mathrm{cm}$.