To determine the type of unit cell for a material with a cubic structure, we can use the formula:

$\rho = \frac{nA}{V_c N_A}$

where:

• $\rho$ is the density of the material,
• $n$ is the number of atoms per unit cell,
• $A$ is the atomic mass,
• $V_c$ is the volume of the unit cell,
• $N_A$ is Avogadro's number ($6.022 \times 10^{23} \, \text{atoms/mol}$).

Given dataTo determine the type of unit cell for a material with a cubic structure, we can use the formula:

$\rho = \frac{nA}{V_c N_A}$

where:

• $\rho$ is the density of the material,
• $n$ is the number of atoms per unit cell,
• $A$ is the atomic mass,
• $V_c$ is the volume of the unit cell,
• $N_A$ is Avogadro's number ($6.022 \times 10^{23} \, \text{atoms/mol}$).

Given

• Density, $\rho = 0.855 \, \text{Mg/m}^3 = 0.855 \times 10^3 \, \text{kg/m}^3 = 0.855 \, \text{g/cm}^3$
• Atomic mass, $A = 39.09 \, \text{g/mol}$
• Lattice parameter, $a = 5.344 \, \text{Å} = 5.344 \times 10^{-8} \, \text{cm}$

Step 1: Calculate the volume of the unit cell, $V_c$.

Since the material has a cubic structure, the volume of the unit cell is:

$V_c = a^3 = (5.344 \times 10^{-8} \, \text{cm})^3$

$V_c = 1.525 \times 10^{-22} \, \text{cm}^3$

Step 2: Rearrange the formula to solve for $n$ (number of atoms per unit cell).

$n = \frac{\rho V_c N_A}{A}$

Step 3: Substitute the known values into the equation.

$n = \frac{0.855 \, \text{g/cm}^3 \times 1.525 \times 10^{-22} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{atoms/mol}}{39.09 \, \text{g/mol}}$

$n = \frac{7.843 \times 10^{-22} \, \text{g/cm}^3 \times \text{mol}}{39.09 \, \text{g/mol}}$

$n \approx 2$

Conclusion:

The number of atoms per unit cell is 2, which corresponds to a body-centered cubic (BCC) structure.

Would you like more details or have any questions?

Here are five related questions for further exploration:

1. How does the density change if the atomic mass increases while keeping the unit cell type the same?
2. What would be the density of a face-centered cubic (FCC) structure for the same material?
3. How would the lattice parameter change if the unit cell was face-centered cubic instead of body-centered cubic?
4. How does the coordination number differ between BCC and FCC structures?
5. Can you determine the atomic radius from the lattice parameter for a BCC unit cell?

Tip: In a body-centered cubic (BCC) structure, there are 2 atoms per unit cell—one at each corner and one in the center. This structure typically has a higher density than simple cubic structures but lower than face-centered cubic (FCC) structures.