Question

The atomic mass number of copper is $A=64$. Assume that atoms in solid copper form a cubic crystal lattice. To envision this, imagine that you place atoms at the centers of tiny sugar cubes, then stack the little sugar cubes to form a big cube. If you dissolve the sugar, the atoms left behind are in a cubic crystal lattice. What is the smallest distance between two copper atoms?

Short answer

The smallest distance between two copper atoms is$\mathbf{0}\mathbf{.}\mathbf{228}\mathbf{}\mathit{n}\mathit{m}$

Step-by-step solution

Defining basic concepts for Counting atoms and moles 

A macroscopic sample of matter consists of N atoms (or molecules), each of mass m (the atomic or molecular mass): N = M m

$N_A=6.022\times {10}^{23}mo{l}^{-1}$is Avogadro’s number.

The molar mass$M_{mol},\mathrm{in}kg/mol$, is the numerical value of the atomic or molecular mass in u divided by $1000$.

The atomic mass unit is $1u=1.66\times {10}^{-27}kg$

The number density of the sample is $\frac{N}{V}$ .

Finding expression for distance a

Imagine a large cube made up of smaller cubes, each containing an atom at its center. It is clear that we are looking for side $a$of the smaller cubes, because the smallest distance between two atoms lies along the line connecting the centers of two atoms that are touching.

Since the atoms are at the centers of their respective touching cubes, the total distance is twice half the side, which makes the distance of side $a$
If we consider our larger cube, which consists of $x$cubes along each side, i.e. a total ofrole="math" localid="1648315496762" $x³$ smaller cubes, we can see that its volume is
$V=\left(xa\right)³=x³a³$,
since $ax$is the distance of one side.

We can now express the mass of the larger cube by the molar mass of copper and by its density.
The mass $m\mathrm{of}n\mathrm{moles}\mathrm{or}N$atoms of copper is then
$$\begin{gathered} m=n{M}_{Cu}=\frac{N}{N_A}{M}_{Cu}m \\ \Rightarrow m=\frac{x^3}{N_A}{M}_{Cu} \end{gathered}$$
Since we know the definition of density, the mass of our copper atoms is
$$\begin{gathered} \rho =\frac{m}{V}\Rightarrow m=\rho V \\ \Rightarrow m=\rho x³a³ \end{gathered}$$
If we equate the two equations, we can now find an expression for the distance $a$:
role="math" localid="1648315955629" $$\begin{gathered} \frac{x^3}{N_A}{M}_{Cu}=m=\rho x³a³ \\ \Rightarrow \frac{M_{Cu}}{N_A}=\rho a³ \\ \Rightarrow \mathit{a}\mathbf{}\mathbf{=}\mathbf{}\sqrt[\mathbf{3}]{\frac{{\mathbf{M}}_{\mathbf{C}\mathbf{u}}}{{\mathbf{N}}_{\mathbf{A}\mathbf{}}\mathbf{\rho }}} \end{gathered}$$

;

Recall that the molar mass $M$in grams is approximately equal to the atomic mass number $A$for all elements.

Put $M_{Cu}=64g/mol=0.064kg/mol$,$N_A=6.02\times {10}^{23}\mathrm{and}\rho =8960kg/m³$, we get

$$\begin{gathered} a=\sqrt[3]{\frac{0.064}{6.02\times {10}^{23}\times 8960}}=2.28\times {10}^{-10}m \\ =\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\stackrel{\mathbf{°}}{\mathbf{A}} \end{gathered}$$