Question

The structure of aluminum is fcc and its density is$\mathit{\rho }{\mathbf{\text{= 2.70g cm}}}^{\mathbf{-}\mathbf{3}}$

(a) How many Al atoms belong to a unit cell?

(b) Calculate a, the lattice parameter, and d, the nearest neighbor distance.

Short answer

a) The number of Al atoms that belong to a unit cellis 4.

b) The lattice parameter and the nearest neighbor distance are $3.22\stackrel{°}{A}$and$2.8\stackrel{°}{A}$ respectively.

Step-by-step solution

Subpart (a) The numbers of Al atoms that belong to a unit cell.

As the structure is face-centered cubic lattice;

There are 8 corners in one cube and the contribution of one atom to the unit cell is$\frac{1}{8}$ $\frac{1}{8}$; as each atom contributes to 4 unit cells up and 4 unit cells down.

Therefore, the contribution of Al atoms in the corners is

$\frac{1}{8}\times 8=1$

There are 6 faces in one cube and the contribution of one atom to the unit cell is$\frac{1}{2}$ ; as each atom contributes to 2 unit cells.

Therefore, the contribution of Al atoms in the face center is

$\frac{1}{2}\times 6=3$

The total numbers of Al atoms that belong to a unit cell is:

$1+3=4$

Subpart (b) The lattice parameter and the nearest neighbor distance.

As density;

$\rho =\frac{\text{zM}}{{\text{a}}^{\text{3}}{\text{N}}_{\text{A}}}$

Here;

Given the density as$2.70{\text{g cm}}^{\text{- 3}}$$2.70{\text{g cm}}^{\text{- 3}}$.

For fcc; z = 4

$\text{M(Al)}=27$

Putting the values in the equation;

$2.70{\text{g cm}}^{\text{- 3}}=\frac{\text{2×}27}{{\text{a}}^{\text{3}}{\text{×6.022×10}}^{23}}$

Therefore, the lattice parameter is:

$$\begin{gathered} {\text{a}}^{\text{3}}=\frac{\text{2×}27\text{g}}{2.70{\text{g cm}}^{\text{- 3}}\text{6.022}\times {\text{10}}^{23}} \\ {\text{a}}^{\text{3}}=\frac{\text{54g}}{{16.26\times 10}^{23}{\text{g cm}}^{\text{- 3}}} \\ {\text{a}}^{\text{3}}=33.2\times {10}^{-24}{\text{cm}}^{\text{3}} \end{gathered}$$

And so;

$\text{a}=3.22\times {10}^{-8}\text{cm}$

Changing the unit;

$1\stackrel{°}{A}={10}^{-8}\text{cm}$

$\text{1cm =}{10}^8\stackrel{°}{A}$

$$\begin{gathered} 3.22\times {10}^{-8}\text{cm}=3.22\times {10}^{-8}\times {\text{10}}^8\stackrel{°}{A} \\ 3.22\times {10}^{-7}\text{cm}=3.22\stackrel{°}{A} \end{gathered}$$

The atomic radius of aluminium.

As in fcc;

$\text{r}=\frac{a\sqrt{2}}{4}$

$\text{r}=\frac{(3.22){\displaystyle \stackrel{°}{A}}\sqrt{2}}{4}$

$\text{r}=1.4\stackrel{°}{A}$

width="58">r=1.4A°

Therefore, the nearest neighbor distance;

$$\begin{gathered} \text{d = 2}\times \text{r} \\ =2\times 1.4\stackrel{°}{A} \\ =2.8\stackrel{°}{A} \end{gathered}$$