Barium crystallizes in a body-centered cubic structure. (a) How many Ba atoms are contained in each unit cell? (b) How many nearest neighbors does each Ba atom possess? (c) Estimate the length of the unit cell edge, \(a\), from the atomic radius of barium \((0.215 \mathrm{nm}) .\) (d) Estimate the density of Ba metal at this temperature.

In a body-centered cubic unit cell, there is one atom at each corner of the cube and one atom at the center. There are 8 corners, each containing 1/8 of an atom since it is shared by 8 unit cells. So, the total number of barium atoms in each unit cell can be calculated as: Atoms in corners = 8 corners * 1/8 atom per corner = 1 atom Atoms in center = 1 atom Total barium atoms = 1 atom (corners) + 1 atom (center) = 2 atoms Therefore, there are 2 barium atoms per unit cell.

In a body-centered cubic structure, the atom at the center has 8 nearest neighbors at the corner of the cube, while an atom at the corner has 4 nearest neighbors along the edges connected to the corner and 1 center atom. So, each barium atom in the BCC structure has 8 nearest neighbors.

The unit cell edge length (a) can be estimated from the atomic radius (r) of barium. In a body-centered cubic structure, the length of the diagonal of the cube is equal to 4 times the atomic radius. We can use Pythagorean theorem to relate the length of the diagonal to the unit cell edge length: Diagonal = \( \sqrt{3a^2} \) Given that Diagonal = 4r, we can rewrite the equation as: \( 3a^2 = (4r)^2 \) Now, we can solve for the unit cell edge length 'a' using the given atomic radius of barium (0.215 nm): \( a = \sqrt{\frac{(4 \times 0.215 \mathrm{nm})^2}{3}} \) \( a \approx 0.502 \mathrm{nm} \) Therefore, the unit cell edge length (a) for barium is approximately 0.502 nm.

To estimate the density of barium, we can use the following formula: Density = \( \frac{Mass}{Volume} \) First, we need to find the mass of the unit cell. The mass of a single barium atom can be calculated using its molar mass (137.33 g/mol) and Avogadro's number (6.022 x 10^23 atoms/mol): Mass of one Ba atom = \( \frac{137.33 \mathrm{g/mol}}{6.022\times10^{23} \mathrm{atoms/mol}} \approx 2.28 \times 10^{-22} \mathrm{g} \) Since there are 2 barium atoms in each unit cell, the mass of the unit cell is twice the mass of a single barium atom: Mass of unit cell = 2 x Mass of one Ba atom = 2 x \( 2.28 \times 10^{-22} \mathrm{g} \approx 4.56 \times 10^{-22} \mathrm{g} \) Next, we need to find the volume of the unit cell. Since the unit cell is a cube with edge length 'a', the volume is: Volume of unit cell = \( a^3 \approx (0.502 \mathrm{nm})^3 \) Converting nm to cm to keep the volume consistent with the mass unit: Volume of unit cell = \( (0.502 \times 10^{-7} \mathrm{cm})^3 \approx 1.26 \times 10^{-21} \mathrm{cm}^3 \) Now we can find the density: Density = \( \frac{4.56 \times 10^{-22} \mathrm{g}}{1.26 \times 10^{-21} \mathrm{cm}^3} \approx 3.61 \mathrm{g/cm^3} \) Therefore, the estimated density of barium at this temperature is approximately 3.61 g/cm³.