Question

The effective radius of an iron atom is $\sqrt{2}A$. It has FCC structure. Calculate its density ($\text{Fe = 56 amu}$, ${{N}_{A}}=6\times {{10}^{23}}$)

Answer 1

Find the number of atoms per unit cell in FCC structure. Then use the radius and lattice constant relation for FCC to find the value of lattice constant $\left( a \right)$ for a given radius and then find out the density.

Formula Used:
$r=\dfrac{a}{2\sqrt{2}}$,
density = Mass of unit cell/ Volume of unit cell $=\dfrac{m}{\text{V}}=\dfrac{z\times m}{{{a}^{3}}}=\dfrac{z\times M}{{{a}^{3}}\times {{N}_{A}}}$

Complete step by step answer:
Given radius =$\sqrt{2}A$
For FCC structure there are 4 atoms per unit cell, so $\text{z = 4}$. Atoms are arranged at the corners and center of each cube face of the cell.
There are eight corners and contribution at each corner is $\dfrac{1}{8}$.
So, the no. of atoms at corners $=8\times \dfrac{1}{8}=1$
There are six faces of a cube and the contribution at each face is $\dfrac{1}{2}$. So, the no. of atoms at cube faces $=6\times \dfrac{1}{2}=3$
So, the total no. of atoms per unit cell $=1+3=4$
For radius of FCC structure, we have, $r=\dfrac{a}{2\sqrt{2}}$ so $a=2\sqrt{2}r$
Therefore, $a=2\sqrt{2}\times \sqrt{2}\text{A}$
$\begin{aligned} & =2\times 2\times {{10}^{-8}}\text{ cm} \\\ & \text{= 4}\times \text{1}{{\text{0}}^{-8}}\text{ cm} \\\ \end{aligned}$
To calculate density we have, $\rho =\dfrac{zM}{{{a}^{3}}{{N}_{A}}}$. Given $\text{M = 56 amu}$ and ${{\text{N}}_{A}}=6\times {{10}^{23}}$
So,

& \rho =\dfrac{4\times 56gm}{{{(4\times {{10}^{-8}}cm)}^{3}}\times 6\times {{10}^{23}}} \\\ & =0.5833\times {{10}^{-23+24}} \\\ & =0.5833\times {{10}^{+1}} \\\ & =5.833\text{ g c}{{\text{m}}^{-3}} \end{aligned}$$ **Note:** Face centered cubic lattice like all lattices, has lattice points at the eight corners of the unit cell plus additional points at the centers of each face of the unit cell. It has unit cell vectors $a=b=c$ and interaxial angles $$\alpha =\beta =\gamma ={{90}^{0}}$$. In FCC structures the spheres fill 74% of the volume. There are 26 metals that have the FCC lattice. Density of unit cell: It is given as the ratio of mass and volume of unit cell. The mass of the unit cell is equal to the product of no. of atoms in a unit cell and the mass of each atom in the unit cell. $\begin{aligned} & \text{Mass of the unit cell = Number of atoms in unit cell }\times \text{ Mass of each atom} \\\ & \text{= z }\times \text{ m } \end{aligned}$ Mass of an atom can be given with the help of Avogadro no. and Molar mass as : $$\dfrac{M}{{{N}_{A}}}$$ Volume of unit cell, $\text{V = }{{a}^{3}}$ So, density = Mass of unit cell/ Volume of unit cell $$=\dfrac{m}{\text{V}}=\dfrac{z\times m}{{{a}^{3}}}=\dfrac{z\times M}{{{a}^{3}}\times {{N}_{A}}}$$