Gallium crystallizes in a simple cubic lattice. Sketch the unit cell. Determine the number of atoms present in the unit cell. The density of gallium is 5.904g/cm^3. Determine a value for the atomic (metallic) radius of gallium.
To sketch the unit cell of gallium, we need to understand the structure of a simple cubic lattice. In a simple cubic lattice, each lattice point is occupied by one atom. The atoms are positioned at the corners of the unit cell, with one atom per corner.
To determine the number of atoms present in the unit cell, we observe that there is only one atom at each corner of the unit cell. Therefore, we have 8 corner atoms, and each corner atom contributes 1/8th of its atom to the unit cell. So the total number of atoms in the unit cell is 8 × 1/8 = 1 atom.
Next, we can determine the atomic (metallic) radius of gallium. The density of gallium is given as 5.904 g/cm^3. The atomic radius can be calculated using the following formula:
Density = (Mass of unit cell) / (Volume of unit cell)
Since we have determined the number of atoms in the unit cell to be 1, the mass of the unit cell is equal to the atomic mass of gallium (which is approximately 69.72 g/mol). The volume of a simple cubic unit cell is given by:
Volume of unit cell = edge length^3
Rearranging the formula for density, we can solve for the edge length of the unit cell:
Edge length = (Mass of unit cell / (Density × Number of atoms in the unit cell))^(1/3)
Substituting the known values, we get:
Edge length = (69.72 g/mol / (5.904 g/cm^3 × 1))^(1/3)
Simplifying this expression, we find the edge length of the unit cell. Since each atom is positioned at the corner and the distance between opposite corners corresponds to two edge lengths, the atomic radius can be determined by dividing the edge length by 2.
Finally, let's calculate the value for the atomic (metallic) radius of gallium.
To sketch the unit cell of gallium, we need to consider that it crystallizes in a simple cubic lattice. In a simple cubic lattice, each lattice point represents one atom.
Since gallium crystallizes in a simple cubic lattice, the unit cell will consist of eight lattice points, one at each of the eight corners of the unit cell.
To determine the number of atoms present in the unit cell, we multiply the number of lattice points by the number of atoms per lattice point. In this case, there is one atom per lattice point, so the number of atoms in the unit cell is 8.
To determine the atomic (metallic) radius of gallium, we can use the information given about the density of gallium. The density of gallium is 5.904 g/cm^3. The density can be related to the atomic radius by the following equation:
Density = (molar mass * atomic mass constant) / (volume of the unit cell * Avogadro's number)
We can rearrange the equation to solve for the volume of the unit cell:
Volume of the unit cell = (molar mass * atomic mass constant) / (density * Avogadro's number)
The molar mass of gallium is 69.723 g/mol and the atomic mass constant is 1 g/mol. Avogadro's number is 6.0221 x 10^23 mol^-1.
Plugging these values into the equation, we get:
Volume of the unit cell = (69.723 g/mol * 1 g/mol) / (5.904 g/cm^3 * 6.0221 x 10^23 mol^-1)
Calculating the volume of the unit cell, we find:
Volume of the unit cell = 1.707 x 10^-23 cm^3
Since gallium crystallizes in a simple cubic lattice, the volume of the unit cell is equal to the edge length cubed:
Edge length = (Volume of the unit cell)^(1/3)
Plugging in the volume of the unit cell, we get:
Edge length = (1.707 x 10^-23 cm^3)^(1/3)
Calculating the edge length, we find:
Edge length ≈ 4.262 x 10^-8 cm
Finally, the atomic radius of gallium can be calculated by dividing the edge length by 2 (since it is a simple cubic lattice):
Atomic radius = (Edge length) / 2
Plugging in the edge length, we get:
Atomic radius ≈ 2.131 x 10^-8 cm
Therefore, the atomic (metallic) radius of gallium is approximately 2.131 x 10^-8 cm.