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

0.228 cm is the smallest distance between two copper atoms

Step-by-step solution

Description of cubes  

The atomic mass number of copper is A = 64

Consider, that the bigger cube consists of x cubes.

the distance of side of the cubes is$\alpha$.

For instance, total smaller cubes is $x^3$and its volume should be

$V={\left(xa\right)}^3=x^3{a}^3,$

Result of finding the smallest distance between two copper atoms 

The mass of N atoms or n moles of copper will be

$m=n{M}_{cu}=\frac{N}{N_A}M{C}_u\Rightarrow m=\frac{x^3}{N_A}{M}_{Cu}$

Mass of copper atoms will be

$p=\frac{m}{v}\Rightarrow m=pV\Rightarrow m=p{x}^3{a}^3$

The expression for the distance $\alpha$is:

$\frac{x^3}{N_A}{M}_{Cu}=m=p{x}^3{a}^3\Rightarrow \frac{M_{Cu}}{N_A}=p{a}^3\Rightarrow a=\sqrt[3]{\frac{M_{Cu}}{N_{Ap}}}$

For all elements, the molar mass M in grams is effectively equal to atomic mass number A.

After substituting $M_{Cu}=64g/mol=0.06kg/mol,N_A=6.02.{10}^{23},andp=8960kg/m^3,$

The result is

$a=\sqrt[3]{\frac{0.064}{6.02.{10}^{23}.8960}}=2.28.{10}^{-10}m$

Please note, for the range of lengths, the units of A, pm, nm are effectively common. Therefore, for these units, the length will be

$a=0.228nm=228pm=2.28A$