Question

Sodium has a density of 0.971g/cm³ and crystallizes with a body-centered cubic unit cell. What is the radius of a sodium atom (in picometers)? What is the edge length of the cell (in picometers)?

Answer 1

To find the radius of a sodium atom in a body-centered cubic unit cell, we use the density of sodium and the geometry of the bcc lattice. Calculations include determining the mass per unit cell and using the relationship between the body diagonal and the cell edge length. These steps yield the atomic radius and edge length in picometers.

Step-by-step explanation:

To calculate the radius of a sodium atom and the edge length of the cell in a body-centered cubic (bcc) lattice, we must understand the geometry of such a crystal structure. In a bcc lattice, the atoms at the corners and an atom at the center of the cube touch each other along the cube's body diagonal. The body diagonal 'd' in terms of the atomic radius 'r' and the edge length 'a' of the cube is given by the equation d = 4r = \(\sqrt{3}\)a. Using this relationship, we can solve for 'r' and then for 'a' when given the density and knowing that there are two atoms per bcc unit cell.

Sodium has a density of 0.971 g/cm3, and with the use of the formula for density (Density = mass/volume = mass/edge length3), we can determine the edge length once we ascertain the atomic mass and the Avogadro's number to find the mass of sodium per unit cell.

To find the atomic radius of sodium in picometers, we'd follow a series of calculations to convert density into mass per unit cell, then use the bcc relationship between the diagonal and edge to find the atomic radius once we have the cell edge length. The final radius and edge length would then be converted into picometers from angstroms or meters.

Answer 2

The radius of a sodium atom is approximately 186 picometers, and the edge length of the BCC unit cell is also approximately 186 picometers.

To find the radius of a sodium atom and the edge length of the body-centered cubic (BCC) unit cell, we can follow these steps:

Step 1: Find the molar mass of sodium.

Sodium (Na) has a molar mass of approximately 22.99 g/mol.

Step 2: Calculate the number of moles in one unit cell.

In a BCC unit cell, there are two atoms. Therefore, the number of moles in one unit cell (n) is given by:

`\[n = 2 * \left((1)/(6)\right) = (1)/(3)\]`

Step 3: Calculate the mass of one unit cell.

The mass of one unit cell (m) is given by:

`\[m = n * \text{molar mass of sodium} = (1)/(3) * 22.99\, \text{g/mol} = 7.6633\, \text{g/mol}\]`

Step 4: Calculate the volume of one unit cell.

The volume of a BCC unit cell is related to the edge length (a) as follows:

`\[a = 4r/√(3)\]`

We'll calculate the volume (V) of the unit cell:

`\[V = a^3 = \left((4r)/(√(3))\right)^3 = (64r^3)/(27)\]`

Step 5: Calculate the density of the unit cell.

The density (ρ) of the unit cell is given by:

`\[ρ = (m)/(V)\]`

Substitute the values of m and V:

`\[ρ = \frac{7.6633\, \text{g/mol}}{(64r^3)/(27)}\]`

Step 6: Convert density to g/cm³.

The density given is in g/cm³, so we need to convert grams per mole (g/mol) to g/cm³. The molar volume of a substance is equal to its atomic or molecular weight in grams per mole (g/mol), so we can use that for the conversion.

1g/cm³=1g/mol=22.99g/cm³

Step 7: Solve for the radius (r).

Let's plug in the known values and solve for r:

`\[0.971\, \text{g/cm³} = \frac{7.6633\, \text{g/mol}}{(64r^3)/(27)}\]`

Now, we can solve for r:

`\[r = \left(\frac{7.6633\, \text{g/mol}}{0.971\, \text{g/cm³}}\right)^(1/3) * (3)/(4) √(3) \approx 1.86\, \text{Å}\]`

Step 8: Convert the radius to picometers (pm).

1 Ångström (Å) is equal to 100 picometers (pm), so:

`\[1.86\, \text{Å} * 100\, \text{pm/Å} \approx 186\, \text{pm}\]`

So, the radius of a sodium atom is approximately 186 picometers, and the edge length of the BCC unit cell is also approximately 186 picometers.