A material with a cubic structure has a density of 0.855 Mg/m³, an atomic mass of 39.09 g/mol, and a lattice parameter of 5.344 angstrom.  If one atom is located at each lattice point, determine the type of unit cell...use the formula thatbis derived from p=nA/Vc(NA)...it should be 2, a body centerd cubic

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.

Learn how to determine the type of unit cell for a material with a cubic structure using density and atomic mass calculations. This problem involves crystallography and density formulas, suitable for undergraduate level studies.

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