Question

\(\alpha\) -form of iron exists in BCC form and \(\gamma\) -form of iron exists in FCC structure. Assuming that the distance between the nearest neighbours is the same in the two forms, the ratio of the density of \(\gamma\) form to that of \(\alpha\) -form is (a) \(4 \sqrt{2}: 3 \sqrt{3}\) (b) \(4 \sqrt{3}: 3 \sqrt{2}\) (c) \(\sqrt{3}: \sqrt{2}\) (d) \(2: 1\)

Short answer

\(4 \sqrt{2}: 3 \sqrt{3}\)

Step-by-step solution

Determine the relationship between the edge length and the nearest neighbor distance in BCC and FCC

For a BCC structure, the nearest neighbour distance (d_BCC) is related to the edge length of the cube (a_BCC) by the formula: \(d_{BCC} = \frac{\sqrt{3}}{2}a_{BCC}\).For an FCC structure, the nearest neighbour distance (d_FCC) is related to the edge length of the cube (a_FCC) by the formula: \(d_{FCC} = \frac{\sqrt{2}}{2}a_{FCC}\).

Express the edge lengths in terms of the uniform nearest neighbor distance

Since the nearest neighbour distance is the same for both BCC and FCC, we can equate the two expressions for nearest neighbour distances and express the edge lengths in terms of the nearest neighbour distance (d):\(d_{BCC} = d_{FCC} = d\)\(\frac{\sqrt{3}}{2}a_{BCC} = \frac{\sqrt{2}}{2}a_{FCC} = d\)Hence, we express the edge lengths as:\(a_{BCC} = \frac{2d}{\sqrt{3}}\)\(a_{FCC} = \frac{2d}{\sqrt{2}}\)

Determine the volume of the unit cells

The volume of a unit cell for both BCC and FCC is the cube of the edge length:For BCC:\(V_{BCC} = a_{BCC}^3 = \left(\frac{2d}{\sqrt{3}}\right)^3\)For FCC:\(V_{FCC} = a_{FCC}^3 = \left(\frac{2d}{\sqrt{2}}\right)^3\)

Calculate the number of atoms per unit cell

A BCC unit cell has 2 atoms per unit cell while FCC has 4 atoms per unit cell.

Relate the densities to the mass and volume of the unit cells

The density (\(\rho\)) is mass per unit volume. For a constant mass, the density ratio of FCC to BCC can be given by the inverse ratio of their volumes, as density is inversely proportional to volume for a given mass. Thus:\(\frac{\rho_{FCC}}{\rho_{BCC}} = \frac{V_{BCC}}{V_{FCC}}\cdot\frac{4}{2}\)where the factor 4/2 comes from the number of atoms per unit cell in FCC and BCC respectively.

Compute the density ratio

Substitute the volumes calculated in Step 3:\(\frac{\rho_{FCC}}{\rho_{BCC}} = \frac{\left(\frac{2d}{\sqrt{3}}\right)^3}{\left(\frac{2d}{\sqrt{2}}\right)^3}\cdot2\)Now compute the ratio:\(\frac{\rho_{FCC}}{\rho_{BCC}} = \frac{\sqrt{2}}{\sqrt{3}}\cdot2\)\(\frac{\rho_{FCC}}{\rho_{BCC}} = \frac{2\sqrt{2}}{\sqrt{3}}\)The simplest form of this ratio is:\(\frac{4\sqrt{2}}{3\sqrt{3}}\) when it's multiplied by \(\frac{2}{2}\) for simplification.

Key concepts

Crystal Structures

Crystal structures refer to the ordered arrangements of atoms, ions, or molecules in a crystalline material. The arrangement is so precise that it repeats itself in three-dimensional space and forms a structure known as a crystal lattice. There are various types of crystal lattices, but the most common ones in metals are the face-centered cubic (FCC) and body-centered cubic (BCC) structures.

In the FCC structure, atoms are located at each of the corners and the centers of all the cube faces. This arrangement is known for its high packing efficiency – meaning that it occupies a large percentage of the space within the lattice. FCC structures tend to have high ductility, which makes them useful in applications where materials need to be shaped without breaking.

The BCC structure, on the other hand, has a central atom and an atom at each corner of the cube, resulting in a less densely packed lattice compared to FCC. BCC metals generally have lower ductility but higher strength. Understanding these structures is crucial for applications in material science and engineering, especially when properties such as density and mechanical strength are important considerations.

Nearest Neighbour Distance

The nearest neighbour distance is the shortest distance between the centers of two adjacent atoms in a crystal lattice. This distance provides important information about the physical properties of the material. For instance, in the context of electrical conductivity, the shorter the distance, the easier it is for electrons to jump from one atom to another. Similarly, the nearest neighbour distance influences mechanical properties such as tensile strength and hardness.

In a BCC structure, the atoms at the corners of the cube connect with the atom at the body-center, forming a body diagonal across the cube. In an FCC structure, the closest neighbors are along the face diagonal. By understanding the geometric relations of these structures, as illustrated in the step-by-step solution, students can derive various relationships and calculate properties such as density, which can be counterintuitive due to the varying number of atoms contained within each unit cell of the different crystal structures.

Unit Cell Volume

The volume of a unit cell is fundamental in determining the physical properties of a crystal. The unit cell is the smallest repeating unit of the crystal lattice that exhibits the full symmetry of the crystal structure; it's like the building block of the crystal.

For cubic lattices, such as BCC and FCC, the volume is relatively simple to calculate because it is the cube of the edge length of the unit cell. However, it's essential to consider not just the volume but also the number of atoms present in each unit cell. BCC has 2 atoms per unit cell while FCC has 4, affecting density calculations significantly. The density of a material is not only a function of how much space the atoms take up (the volume), but also a function of how many atoms are there in that space. Hence, the density ratio can be surprising when comparing unit cells with different numbers of atoms within similar volumes, illustrating the importance of both volume and atom count in density calculations.