Palladium is a relatively rare catalytic metal that crystallizes in an FCC latti
ID: 886041 • Letter: P
Question
Palladium is a relatively rare catalytic metal that crystallizes in an FCC lattice. The atomic mass of Pd is 106.42. Calculate the mass of a unit cell.
_______ au
Given the density (12.02 g/cm3), calculate the size of the unit cell and the radius of a Pd atom.
size _______________ cm^3 / unit cell
Radius _______________ pm
The surface is the catalytically active portion of a palladium particle. Calculate the number of palladium atoms in 1 cm2 of surface assuming that the surface consists of a simple truncation of the bulk crystal structure.
_________ atoms/cm^2
Explanation / Answer
FCC unit cell has 4 atoms per cell
Atomic mass = (Cell mass / 4) * Avogadro's number
106.42 * 4 / (6.022 x 10^23) = Cell mass
Mass of unit cell = 7.07 x 10^-22 au
Volume / unit cell = Mass per unit cell / Density = 7.07 x 10^-22 / 12.02 = 5.88 x 10^-23 cm^3
edge-length(a) = cuberoot(V) = cuberoot( 5.88 x 10^-23) = 3.89 x 10^-8 cm
r = a / (2 * squareroot(2))
r = 3.89 x 10^-8 / ( 2 * 0.301) = 1.37 x 10^-8 cm = 13748 pm
if the radius = 1.37 x 10^-8 cm then the diameter is 2.74 x 10^-8 cm.
area of one side of the unit cell = Diameter^2 = 7.51 x 10^-16 cm2. In FCC there are 1 atom per side i.e 1 atom / Area of one side
1 cm2 * (1 atoms / 7.51 x 10^-16 cm2) = 1.33 x 10^15 atoms / cm2
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