The unit cell of a crystal can be described in term of its edge length, l, and t

Question

The unit cell of a crystal can be described in term of its edge length, l, and the number of atoms it contains. The edge length defines the volume of a cubic unit cell, V, as V=l3 The number of atoms per unit cell depends on the type of unit cell. For cubic unit cells, the position (body, face, or corner) of the atoms determines the fraction of the atoms that are completely contained within the unit cell.

Based on the number of atoms per unit cell and the mass of the atom, the mass m of the unit cell can be calculated. The density of the unit cell and the material as a whole can be determined from the mass mand the volume V of the unit cell as.

density=mV

The usual units of density are grams per cubic centimeter or g/cm3.

Nickel, Ni, has a face-centered cubic structure(Figure 1) with an edge length of 352 pm. What is the density of this metal? Use the periodic table as needed.

Express your answer with the appropriate units.

Explanation / Answer

Step 1 : To calculate volume of the unit cell

Edge length, a = 352pm = 352X10-12 m Volume, V = a3 = (352X10-12 m)3   V =  = 4.36X10-29 m3

Step 2: To calculate mass of an atom of Nickel

Molar mass of Nickel = 58.69 g/mol 1 mole = 6.023X1023 atoms So, mass of 6.023X1023 atoms of Nickel = 58.69g mass of 1 atom of Nickel = 58.69g / (6.023X1023) = 9.744X10-23 g

Step 3 : To calculate density of the unit cell / Nickel metal

No. of atoms per fcc unit cell , n = 4 Mass of 4 atoms of Nickel,m = mass of 1 atom of Ni X 4 = (9.744X10-23 g) X 4 m = 38.977 X 10-23 g

Density = m/V = (38.977 X 10-23 g ) / ( 4.36X10-29 m3) = 8.94 X 106 g/m3Density = = 8.94 g/cm3    ( 1m3 = 106 cm3)

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