What fraction of space is actually occupied by ironnuclei in a "solid" piece of iron? (The density of ironis$7.87\times {10}^3\text{kg}/{\text{m}}^3$ ).

Density of iron, $\rho =7.87\times {10}^3\text{kg}/{\text{m}}^3$.

Mass of iron, $m=55.934\text{u}$.

Conversion: $1\text{u}=1.66\times {10}^{-27}\text{kg}$.

The radius of the nucleusis$r=A^{1/3}\times R_o$ .

Where, is the mass number androle="math" localid="1658403049741" $R_0$ is role="math" localid="1658403057498" $1.2\times {10}^{-15}\text{m}$.

Volume of sphere, $r=A^{1/3}\times R_o$.

$V_{\text{total}}=\frac{4}{3}\pi r^3$

Where, $r$ is the radius

Volume of $1$ iron nucleus is given as shown below.

$\begin{array}{l}V=\frac{4}{3}\pi r^3\\ V=A\times \frac{4}{3}\pi R_o^3\\ V=56\left(\frac{4}{3}\pi \right){(1.2\times {10}^{-15}\text{m})}^3\\ V=4.05\times {10}^{-43}{\text{m}}^3\end{array}$

The number of iron atoms in a piece of solid iron of 1 cubic meter is given as:

$\begin{array}{l}N=\frac{themassof1cubicmetcrofiron}{massofoneironatom}\\ N=\frac{\rho \times (1m^3)}{m}\\ N=\frac{(7.87\times {10}^{3}kg/m^3)\times (1m^3)}{(55.934u)\left(\frac{1.66\times {10}^{-2k{e}_e}}{1u}\right)}\\ N=8.48\times {10}^{28}ironnucleus\end{array}$

Total volume occupied by nucleus present in $1{\text{m}}^3$.

$\begin{array}{l}{V}_{\text{total}}=N\times V\\ V_{\text{total}}=(4.05\times {10}^{-43}\frac{{\text{m}}^3}{1\text{nucleus}})(8.48\times {10}^{28}\text{nucleus})\\ V_{\text{total}}=3.43\times {10}^{-14}{\text{m}}^3\end{array}$

So, the fraction of space occupied by the iron nucleus in $1{\text{m}}^3$ is given as:

$\begin{array}{l}=\frac{V_{\text{kat1}}}{1{\text{m}}^3}\\ =\frac{3.43\times {10}^{-14}{\text{m}}^3}{1{\text{m}}^3}\\ =3.43\times {10}^{-14}\end{array}$

Therefore, the fraction in a piece of iron that is occupied by the iron nuclei is approximate $3\times {10}^{-14}$