Question

A metal has an FCC crystal structure with a density of \(2.71\) g/cm\(^3\) and atomic weight of \(26.98\) g/mol. Avogadro's number is \(6.023 \times 10^{23}\). The atomic radius of the metal is ________ nm (rounded off to 2 decimal places).

Hint:

Find the lattice constant from the density formula first, then use the FCC face diagonal rule to get the radius.

Answer 1

Correct Answer: 0.14

Step 1: Start from the mass packed into one unit cell.
An FCC cell holds 4 atoms, so its mass is $4M/N_A$ grams.

Step 2: Get the cell edge from density.
Volume of the cell is mass over density, $a^3 = \dfrac{4(26.98)}{(2.71)(6.023 \times 10^{23})} = 6.61 \times 10^{-23}$ cm$^3$, so $a = 4.04 \times 10^{-8}$ cm.

Step 3: Convert the edge to nanometers.
$a = 0.404$ nm.

Step 4: Use the FCC geometry rule that atoms touch along the face diagonal, $4r = a\sqrt{2}$.
$r = a\sqrt{2}/4 = (0.404)(1.4142)/4 = 0.143$ nm.

Final Answer:
The atomic radius comes out to about 0.14 nm. \[ \boxed{r \approx 0.14 \text{ nm}} \]