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: Write the density formula for a cubic crystal.
Density is \(\rho = \dfrac{n M}{N_A a^3}\), where \(n\) is atoms per unit cell, \(M\) is atomic weight, \(N_A\) is Avogadro's number, and \(a\) is the lattice constant.

Step 2: Use n = 4 for FCC and solve for a.
\(a^3 = \dfrac{4 \times 26.98}{2.71 \times 6.023 \times 10^{23}} = \dfrac{107.92}{1.6322 \times 10^{24}} = 6.61 \times 10^{-23}\) cm\(^3\).
Taking the cube root gives \(a = 4.04 \times 10^{-8}\) cm \(= 0.404\) nm.

Step 3: Relate the atomic radius to a for FCC.
In FCC, atoms touch along the face diagonal, so \(r = \dfrac{a\sqrt{2}}{4}\).

Step 4: Compute the radius.
\(r = \dfrac{0.404 \times 1.4142}{4} = 0.143\) nm, which rounds to 0.14 nm.

Final Answer:
The atomic radius of the metal is close to 0.14 nm. \[ \boxed{r \approx 0.14 \text{ nm}} \]