Final answer:
The atomic radius of calcium in a face-centered cubic (FCC) structure can be calculated using the given edge length and the diagonal relationship in FCC lattices. The density confirms the mass distribution in the unit cell considering the number of atoms per cell and the cell's volume.
Step-by-step explanation:
The student has asked for the atomic radius of calcium in a face-centered cubic (FCC) structure, given its density is 1.54 g/ml. We start by noting that in an FCC lattice, each calcium atom touches its neighbors across the face diagonal, hence the diagonal's length (d) is four times the atomic radius (4r). Now, using the Pythagorean theorem where the face diagonal forms the hypotenuse of a right triangle with the cell edge length, we use the given edge length (a) of 558.8 pm to calculate r.
a = edge length = 558.8 pm
d = face diagonal = 4r
Using the Pythagorean theorem for the face diagonal gives us:
(√2) × a = d
Substituting d for 4r, we have:
(√2) × 558.8 pm = 4r
Therefore,
r = √2 × 558.8 pm / 4
r = (√2 / 4) × 558.8 pm
Rearranging and solving this gives us the value of r.
The density of calcium can be confirmed by taking the mass of calcium atoms in a unit cell divided by the volume of the cell. Since an FCC cell has 1 atom from the corners and 3 from the faces, contributing to a total of 4 atoms per unit cell, we can use the atomic mass of calcium and Avogadro's number to find the mass of these atoms. The density formula then can be expressed as: Density = (Mass of atoms per unit cell)/(Volume of the unit cell).