Question

The density of a FCC unit cell is 6.5 g/cm³. If the mass of a single atom is 60 g/mol, the diagonal length of the face {100} is ________Å. (Round off to two decimal places) (Use NA = 6.022 × 10²³)

Answer 1

Correct Answer: 5.55

The face-centered cubic (FCC) unit cell has 4 atoms per unit cell. The density formula for an FCC unit cell is: d = (n m) / (a³ NA), where n is the number of atoms per unit cell (4 for FCC), m is the molar mass, a is the edge length, and NA is Avogadro's number. Given that d = 6.5 g/cm³ and m = 60 g/mol, we need to calculate a.
Rearranging the density formula gives: a³ = (n m) / (d NA). Substituting the known values:
a³ = (4 × 60) / (6.5 × 6.022 × 10²³)
Solve for a:
a³ = (240) / (39.143 × 10²³) 
a³ = 6.13 × 10^-23 cm³
a = (6.13 × 10^-23)^1/3 cm
Convert a to Angstroms (1 cm = 10⁸ Å):
a = (3.92 × 10^-8) cm × 10⁸ Å/cm = 3.92 Å
For the diagonal length of the face {100}, use diagonal length = a√2:
diagonal length = 3.92 × 1.414 Å = 5.54 Å
Checking if it falls within the expected range (5.55,5.55):
The calculated diagonal length of 5.54 Å is within the expected rounding to two decimal places.