Question 1.16: THE COPPER (FCC) CRYSTAL Consider the FCC unit cell of the c...
THE COPPER (FCC) CRYSTAL Consider the FCC unit cell of the copper crystal shown in Figure 1.40.
a. How many atoms are there per unit cell?
b. If R is the radius of the Cu atom, show that the lattice parameter a is given by a=R 2 \sqrt{2}.
c. Calculate the atomic packing factor (APF) defined by
APF =\frac{\text { Volume of atoms in unit cell }}{\text { Volume of unit cell }}
d. Calculate the atomic concentration (number of atoms per unit volume) in Cu and the density of the crystal given that the atomic mass of Cu is 63.55 g mol^{−1} and the radius of the Cu atom is 0.128 nm.
a. There are four atoms per unit cell. The Cu atom at each corner is shared with eight other adjoining unit cells. Each Cu atom at the face center is shared with the neighboring unit cell. Thus, the number of atoms in the unit cell =8 \text { corners }\left(\frac{1}{8} \text { atom }\right)+6 \text { faces }\left(\frac{1}{2} \text { atom }\right)= 4 \text {atoms}
b. Consider the unit cell shown in Figure 1.40 and one of the cubic faces. The face is a square of side a and the diagonal is \sqrt{a^{2}+a^{2}} \text { or } a \sqrt{2} . The diagonal has one atom at the center of diameter 2R, which touches two atoms centered at the corners. The diagonal, going from corner to corner, is therefore R + 2R + R. Thus, 4 R=a \sqrt{2} and a=4 R / \sqrt{2}=R 2 \sqrt{2} . Therefore, a = 0.3620 nm.
c. APF =\frac{(\text { Number of atoms in unit cell }) \times(\text { Volume of atom })}{\text { Volume of unit cell }}
=\frac{4 \times \frac{4}{3} \pi R^{3}}{a^{3}}=\frac{\frac{4^{2}}{3} \pi R^{3}}{(R 2 \sqrt{2})^{3}}=\frac{4^{2} \pi}{3(2 \sqrt{2})^{3}}=0.74
d. In general, if there are x atoms in the unit cell, the atomic concentration is
n_{ at }=\frac{\text { Number of atoms in unit cell }}{\text { Volume of unit cell }}=\frac{x}{a^{3}}
Thus, for Cu
n_{ at }=\frac{4}{\left(0.3620 \times 10^{-7} cm \right)^{3}}=8.43 \times 10^{22} cm ^{-3}
There are x atoms in the unit cell, and each atom has a mass of M_{at} ∕ N_{A} grams. The density \rho is
\rho=\frac{\text { Mass of all atoms in unit cell }}{\text { Volume of unit cell }}=\frac{x\left(\frac{M_{ at }}{N_{A}}\right)}{a^{3}}
that is,
\rho=\frac{n_{ at } M_{ at }}{N_{A}}=\frac{\left(8.43 \times 10^{22} cm ^{-3}\right)\left(63.55 g mol ^{-1}\right)}{6.022 \times 10^{23} mol ^{-1}}=8.9 g cm ^{-3}
Note that the expression \rho=\left(n_{ at } M_{ at }\right) / N_{A} is particularly useful in finding the atomic concentration n_{at} from the density since the latter can be easily measured or available in various data resources.