Pb crystal has a FCC lattice structure with an atomic radius of 1.8 ? and atomic

Question

Pb crystal has a FCC lattice structure with an atomic radius of 1.8 ? and atomic weight of 207.2. Assuming a closely packed lattice, (1) calculate the atomic packing fraction of Pb lattice; (2) calculate the number of atoms/cm3 ; (3) calculate the density in g/cm3 .

Pb crystal has a FCC lattice structure with an atomic radius of 1.8 A and atomic weight of 207.2. Assuming a closely packed lattice, (1) calculate the atomic packing fraction of Pb lattice; (2) calculate the number of atoms/cm3; (3) calculate the density in g/cm3 6.

Explanation / Answer

Radius = 1.8*10-8cm

Weight = 207.2g

1).   Packing fraction = (16*3.14*R3*20.5)/(3*32*R3) = .74

2).   No. of atoms/cm3 = no. of atoms/vol. of cube = 4/(32*R3/20.5) = 3.03*1022 atoms/cm3

3). Density = Atomic weight * No. of atoms/cm3 = 207.2*3.03*1022 g/cm3 = 627.816*1022g/cm3

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