Question

Titanium metal has a body-centered cubic unit cell. The density of titanium is \(4.50 \mathrm{g} / \mathrm{cm}^{3} .\) Calculate the edge length of the unit cell and a value for the atomic radius of titanium. (Hint: In a body-centered arrangement of spheres, the spheres touch across the body diagonal.)

Short answer

Using the given density of titanium, we first find the mass of one titanium atom, the volume of the unit cell, and then the edge length of the unit cell by calculating the cube root of the volume. With the edge length, we calculate the body diagonal using the Pythagorean theorem and finally, we determine the atomic radius of titanium as one-fourth of the body diagonal.

Step-by-step solution

Calculate the volume of the unit cell

To start, we will first need to get the total volume of the unit cell using the formula: \(Density (\rho) = \frac{Mass}{Volume}\) \(Volume = \frac{Mass}{\rho}\) In a body-centered cubic unit cell, there are 2 atoms per unit cell. The molar mass of titanium is 47.87 g/mol and we know that the density is given as 4.50 g/cm³.

Determine the mass of one titanium atom

To get the mass of 1 titanium atom present in the unit cell, we'll use Avogadro's number (6.022 x 10²³ particles/mol): \(Mass_{Ti-atom} = \frac{Molar\,Mass_{Ti}}{Avogadro's\,number}\) \(Mass_{Ti-atom} = \frac{47.87\,g/mol}{6.022 \times 10^{23}\,particles/mol}\)

Find the volume of the unit cell

Now, we can find the volume of the unit cell by plugging the mass of 1 titanium atom into the density formula: \(Volume_{Unit\,Cell} = \frac{2 \times Mass_{Ti-atom}}{Density}\)

Calculate the edge length of the unit cell

We know the volume of a cube is equal to \(edge^3\). Therefore, by solving for the edge length of the unit cell we get: \(Edge^3 = Volume_{Unit\,Cell}\) \(Edge = \sqrt[3]{Volume_{Unit\,Cell}}\)

Calculate the body diagonal of the unit cell

Let d be the body diagonal of the unit cell and a be the edge length. Using the Pythagorean theorem, we have: \(d^2 = a^2 + a^2 + a^2\) \(d^2 = 3a^2\)

Calculate the atomic radius of titanium

In a body-centered cubic unit cell, the atoms touch across the body diagonal. Therefore, the body diagonal is equal to the sum of four atomic radii (2 on each end and 2 in the center): \(d = 4 \times atomic\,radius\) Now, we can calculate the atomic radius: \(atomic\,radius = \frac{d}{4}\) By following these step-by-step calculations, we will find the edge length of the unit cell and the atomic radius of titanium.

Key concepts

Density

In the study of crystallography, understanding the density of a material is crucial. Density is determined by how much mass is contained in a given volume. For a body-centered cubic (BCC) unit cell like titanium, knowing the density allows us to find other properties, such as the unit cell's edge length.

The formula used is \[\text{Density} (\rho) = \frac{\text{Mass}}{\text{Volume}}\] Given the density of titanium as 4.50 g/cm³, we can rearrange this formula to find the volume if we know the mass.

Finding the mass involves more than simply weighing a single atom, since it's impractical at such small scales. Instead, we use the molar mass of titanium (47.87 g/mol) and Avogadro's number (6.022 x 10²³ particles/mol) to determine the mass of one titanium atom. This information helps us to calculate the mass of two atoms per unit cell, as a BCC contains two such atoms.

• Use the molar mass to find individual atom mass: \[\text{Mass of one Ti atom} = \frac{47.87 \text{ g/mol}}{6.022 \times 10^{23} \text{ particles/mol}}\]
• Calculate the volume of the unit cell with known mass and density. \[\text{Volume of Unit Cell} = \frac{2 \times \text{Mass of one Ti atom}}{\text{Density}}\]

Atomic Radius

The atomic radius is a fundamental property that dictates the size of an atom. In a BCC unit cell, the atoms are arranged such that they touch across the body diagonal. This specific arrangement helps us deduce the atomic radius when the unit cell's dimensions are known.

For titanium in a BCC formation, the body diagonal of the cube includes contributions from four atomic radii. Therefore, it's important to first calculate the diagonal using the edge length previously determined:

• Express the body diagonal in terms of the edge length: \[d = \sqrt{3} \times a\]
• Knowing that the body diagonal equals four radii (as atoms touch across the diagonal), we find: \[\text{Atomic Radius} = \frac{d}{4}\]

This calculation is key as it translates the macroscopic measurements into atomic-scale dimensions, revealing just how compactly those titanium atoms are nestled within their crystal lattice.

Pythagorean Theorem

The Pythagorean Theorem, a pivotal concept in mathematics, is not just for solving triangles but also serves an essential role in determining lengths within geometric structures like a cube. In this exercise, this theorem helps to find the diagonal of a body-centered cubic cell, linking it to the atomic radius.

We start by realizing that each cube's diagonal can be described in terms of its edges. For any cube, the relationship is represented as:

• The traditional right triangle formulation: \[c^2 = a^2 + a^2 + a^2\]
• This simplifies to the body diagonal formula: \[c^2 = 3a^2\]
• Thus, the diagonal (c) is: \[c = a \sqrt{3}\]

The utility of the Pythagorean Theorem here is in its ability to bridge known edge lengths with unknown diagonal lengths, which are necessary to compute the atomic radius. Understanding this theorem's application in varying contexts is vital for solving many real-world problems in physics and chemistry.