The atomic radius the density and the atomic weight for thre

The atomic radius, the density, and the atomic weight for three hypothetical metals are listed below. Determine their crystal structure (simple cubic metal, BCC metal or FCC metal).

Solution

Answer Material A :-

Test for simple Cubic

Simply consider 1 cm^3 material. its mass is then 3.2g,

and it contains: (3.2/21.55)mol = 0.1484 mol.

Since one mole atom is 6.022e23 atoms, this 1 cm^3 material contains 8.936e22 atoms.
If the material is in simple cubic structure, since a simple cubic unit cell holds one atom, this 1 cm^3 material must contain 8.936e22 unit cells, and thus the side length of the unit cell must be (1/8.936e22)^(1/3) = 2.23675773e-8 = 2.236. Since a unit cell of the side length of 2.236 can not contain an atom with diameter 2.44. The material can not be in simple cubic.

Therefore material A is not Simple cubic.

Test for BCC

If the material is in body-centered cubic structure, since a body-centered cubic unit cell holds two atoms, this 1 cm^3 material must contain 8.936e22/2 = 4.468e22 unit cells, and thus the side length of the unit cell must be (1/4.468e22)^(1/3) = 2.81813815e-8 cm = 2.818.

Since the longest diagonal of this unit cell is 2.818*3 = 4.88 is exactly twice of the atomic diameter, the material CAN be in body-centered cubic structure.

Test for FCC

If the material is in face-centered cubic structure, since a face-centered cubic unit cell holds four atoms, this 1 cm^3 material must contain 8.936e22/4 = 2.234e22 unit cells, and thus the side length of the unit cell must be (1/2.234e22)^(1/3) = 3.55063157e-8.Since the face diagonal of this unit cell is 3.550*2 = 5.020 is much big than twice of the atomic diameter,Therefore the material is not FCC.Hence, this material is in body-centered cubic structure.

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Answer Material B

Test for simple Cubic

Assuming 1 Cm^3 material, Mass will comes to be 4.8 g.

It contains no. Of mol =4.8/45.8=0.1048 mol.

Since one mole atom is 6.022e23 atoms, this 1 cm^3 material contains 6.311e22 atoms.

Therefore side length of the unit cell can be calculated as (1/6.311e22)^(1/3) = 2.51169715e-8 = 2.511

Since a unit cell of the side length of 2.511 can not contain an atom with diameter 2.74. The material can not be in simple cubic.

Test for BCC

If the material is in body-centered cubic structure, since a body-centered cubic unit cell holds two atoms, this 1 cm^3 material must contain 6.022e23/2 = 3.011e22 unit cells, and thus the side length of the unit cell must be (1/3.011e22)^(1/3) = 3.21437406e-8= 3.214 .Since the longest diagonal of this unit cell is 3.2143 = 5.5668 is exactly twice of the atomic diameter.

Hence the material is in BCC structure.

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Answer Material B

Test for simple Cubic

Assuming 1 Cm^3 material, Mass will comes to be 4.1 g.

It contains no. Of mol =4.1/61.47=0.0666 mol.

Since one mole atom is 6.022e23 atoms, this 1 cm^3 material contains 4.010e22 atoms.

Therefore side length of the unit cell can be calculated as (1/4.010e22)^(1/3) = 2.92158511e-8 = 2.921

Since a unit cell of the side length of 2.921 containing an atom with diameter of 0.292 approx.. So it could be Simple cubic.