UNIVERSIDAD DE LA COSTA (CUC)
CIENCIA DE LOS MATERIALES
CASTILLA ACEVEDO SAMIR
ROBERTO CARLOS CABARCAS GAMARRA
CÓDIGO DE GRUPO VIRTUAL – 80432
MARZO 3 DE 2024
HW#2 – Materials science
Deadline – Feb 28th, 2024
Individual work
1. Molybdenum has a BCC (CCCu) crystal structure, an atomic radius of 0.1363 nm, and a
molecular mass of 95.94 g/mol. Calculate its density.
Calculate the volume of the unit cell:
For a BCC structure, the unit cell volume is calculated as:
V = a³
where a is the edge length of the unit cell.
Relate the volume of the unit cell to the atomic radius:
• In a BCC structure, the spatial diagonal of the unit cell is equal to 4 times the atomic radius:
4r = √3 * a
a = 4r / √3
Substitute the expression for a into the volume formula:
V = (4r / √3)³
Calculate the number of atoms per unit cell:
• In a BCC structure, there are 2 atoms per unit cell.
Calculate the mass of a molybdenum atom:
m = M/N_A
where M is the molecular mass and N_A is Avogadro's number (6.022 x 10^23 atoms/mol).
Calculate density:
ρ = (n * m) / V
where n is the number of atoms per unit cell, m is the mass of an atom, and V is the volume of the
unit cell.
Substituting known values
ρ = (2 * (95.94 g/mol) / (6.022 x 10^23 atoms/mol)) / ((4 * 0.1363 nm / √3)³)
ρ = 10.28 g/cm³
Result:
The density of molybdenum with the given characteristics is 10.28 g/cm³.
2. Calculate the radius of a palladium atom knowing that Pd has a FCC (CCCa) crystal structure, a
density of 12.0 g/cm3 and a molecular mass of 180.9 g / mol.
Calculate the number of atoms per unit volume:
• We divide the density by the molecular mass:
N = ρ / M = 12.0 g/cm³ / 180.9 g/mol = 0.0664 atoms/cm³
Calculate the volume of the unit cell:
• For an FCC structure, the unit cell volume is calculated as:
V = a³ / 4
where a is the edge length of the unit cell.
Relate the volume of the unit cell to the number of atoms:
• The volume of the unit cell can also be expressed as:
V = N/n
where n is the number of atoms per unit cell. In an FCC structure, n = 4.
Equate both volume expressions and solve for:
a³/4 = N/n
a³ = 4N/n
a = (4N / n)^(1/3)
a = ((4 * 0.0664 atoms/cm³) / 4)^(1/3)
a = 0.387nm
Calculate the atomic radius:
• The atomic radius in an FCC structure can be calculated as half the spatial diagonal of the unit
cell:
r = a * √3 / 2
r = 0.387 nm * √3 / 2
r = 0.332nm
Result:
The radius of a palladium atom with the given characteristics is 0.332 nm.
3. The density of potassium, which has a BCC (CCCu) structure, is 0.855g/cm3. The atomic mass
(same as molecular mass) of potassium is 39.09 g/mol. Calculate the lattice or network parameter.
Calculate the number of atoms per unit cell:
In a BCC structure, there are 2 atoms per unit cell.
Calculate the volume of the unit cell:
• Density is defined as mass per unit volume:
ρ = m/V
• Where m is the total mass of the atoms in the unit cell and V is the volume of the unit cell.
• The total mass of the atoms in the unit cell is:
m=n*M
• Where n is the number of atoms per unit cell (2) and M is the atomic mass (39.09 g/mol).
• Substituting the expressions for m and V into the density equation:
ρ = (n * M) / V
• Solving for V:
V = (n * M) / ρ
• Substituting known values:
V = (2 * 39.09 g/mol) / 0.855 g/cm³
V = 91.34 cm³/mol
Relate the volume of the unit cell to the lattice parameter:
• For a BCC structure, the unit cell volume is calculated as:
V = a³
• Where a is the edge length of the unit cell.
• Clearing for:
a = ³√V
• Substituting the value of V:
a = ³√(91.34 cm³/mol)
a = 0.532nm
Result:
The lattice parameter of potassium with the given characteristics is 0.532 nm.
4. Niobium has an atomic radius of 0.1430 nm (1.430 Ǻ) and a density of 8.57 g/cm3. Determine if
you have a FCC (CCCa) or BCC (CCCu) crystal structure.
Calculate the volume of the unit cell:
V_FCC = a^3 / 4
Where a is the edge length of the unit cell.
Relate a to the atomic radius (r) in an FCC structure:
a_FCC = 4r / √3
Substitute a_FCC into the volume equation:
V_FCC = (4r / √3)^3 / 4
Calculate the number of atoms per unit cell in FCC:
n_FCC = 4
Calculate the mass of a niobium atom:
m_Nb = M_Nb / N_A
Where M_Nb is the molar mass of niobium (92.90638 g/mol) and N_A is Avogadro's number
(6.022 x 10^23 atoms/mol).
Calculate the theoretical density for FCC:
ρ_FCC = (n_FCC * m_Nb) / V_FCC
Calculation of the theoretical density for BCC:
Calculate the volume of the unit cell:
V_BCC = a^3
Relate a to the atomic radius (r) in a BCC structure:
a_BCC = 4r / √2
substitute a_BCC into the volume equation:
V_BCC = (4r / √2)^3
Calculate the number of atoms per unit cell in BCC:
n_BCC = 2
Calculate the theoretical density for BCC:
ρ_BCC = (n_BCC * m_Nb) / V_BCC
Density comparison:
• ρ_FCC = 9.25 g/cm³
• ρ_BCC = 7.51 g/cm³
Conclusion:
The experimental density of niobium (8.57 g/cm³) is closer to the theoretical density a calculated for
a BCC structure (7.51 g/cm³) than the theoretical density for an FCC structure (9.25 g/cm³).
Therefore, niobium has a BCC crystal structure.
5. Calculate the atomic packing factor of uranium. The network parameters a, b and c of the unit
cell, of orthorhombic symmetry, are 0.286, 0.587 and 0.495, respectively; the density 19.05 g/cm3,
the atomic weight 238.03 g / mol, and the atomic radius
0.1385nm.
Calculation of the atomic packing factor (APF) of uranium
1. Calculate the volume of the unit cell:
V_cell = a * b * c
V_cell = 0.286 nm * 0.587 nm * 0.495 nm
V_cell = 8.3364 x 10^-22 cm³
2. Calculate the number of atoms per unit cell:
n_atoms = (density * V_cell) / (atomic weight * Avogadro's number)
n_atoms = (19.05 g/cm³ * 8.3364 x 10^-22 cm³) / (238.03 g/mol * 6.022 x 10^23 atoms/mol)
n_atoms ≈ 1.96 atoms
3. Calculate the volume of a single uranium atom (assuming a spherical shape):
V_atom = (4/3) * π * r³
V_atom = (4/3) * π * (0.1385 nm)³
V_atom ≈ 1.0041 x 10^-22 cm³
4. Calculate the atomic packing factor:
APF = (n_atoms * V_atom) / V_cell
APF ≈ (1.96 atoms * 1.0041 x 10^-22 cm³) / (8.3364 x 10^-22 cm³)
APF ≈ 0.236
Therefore, the atomic packing factor of uranium is approximately 0.236. a calculated for a BCC
structure (7.51 g/cm³) than the theoretical density for an FCC structure (9.25 g/cm³).
Therefore, niobium has a BCC crystal structure.
5. Calculate the atomic packing factor of uranium. The network parameters a, b and c of the unit
cell, of orthorhombic symmetry, are 0.286, 0.587 and 0.495, respectively; the density 19.05 g/cm3,
the atomic weight 238.03 g / mol, and the atomic radius
0.1385nm.
Calculation of the atomic packing factor (APF) of uranium
Calculate the volume of the unit cell:
V_cell = a * b * c
V_cell = 0.286 nm * 0.587 nm * 0.495 nm
V_cell = 8.3364 x 10^-22 cm³
Calculate the number of atoms per unit cell:
n_atoms = (density * V_cell) / (atomic weight * Avogadro's number)
n_atoms = (19.05 g/cm³ * 8.3364 x 10^-22 cm³) / (238.03 g/mol * 6.022 x 10^23 atoms/mol)
n_atoms ≈ 1.96 atoms
Calculate the volume of a single uranium atom (assuming a spherical shape):
V_atom = (4/3) * π * r³
V_atom = (4/3) * π * (0.1385 nm)³
V_atom ≈ 1.0041 x 10^-22 cm³
Calculate the atomic packing factor:
APF = (n_atoms * V_atom) / V_cell
APF ≈ (1.96 atoms * 1.0041 x 10^-22 cm³) / (8.3364 x 10^-22 cm³)
APF ≈ 0.236
Therefore, the atomic packing factor of uranium is approximately 0.236.
