Question

Sodium metal crystallizes in body-centred cubic lattice with cell edge \(=4.29 \AA\) What is the radius of sodium atom?

Short answer

The radius of a sodium atom is 1.86 Å.

Step-by-step solution

Understand the structure

Sodium metal crystallizes in a body-centered cubic (BCC) lattice. In a BCC structure, the unit cell has an atom at each corner and one atom in the center of the cube.

Cell edge length

Given that the cell edge length (also called lattice parameter) is 4.29 Å.

Relationship between cell edge and radius in BCC

In a BCC lattice, the body-diagonal of the cube is equal to four times the atomic radius. The body-diagonal can be calculated using the Pythagorean theorem: \(d = \sqrt{3} \times \text{cell edge}\).

Calculate the body-diagonal

Using the given cell edge length, calculate the body-diagonal length: \[d = \sqrt{3} \times 4.29 \text{Å} = 7.4255 \text{Å} \].

Solve for the atomic radius

Since the body-diagonal is equal to four times the atomic radius, the formula is: \[4r = 7.4255 \text{Å}\]. Solve for \(r\): \[r = \frac{7.4255 \text{Å}}{4} = 1.856375 \text{Å}\].

Key concepts

body-centered cubic structure

In a body-centered cubic (BCC) structure, the atoms are arranged in a unique way. This type of lattice is one of the most common crystal structures for metals. Here's what you need to know:

Each unit cell in a BCC structure has atoms located at each corner of a cube and one atom right in the center of the cube.
The corner atoms are shared with adjacent unit cells, meaning each corner atom is part of eight different cubes.
The atom in the center is completely within its unit cell.

This configuration results in a more compact structure, promoting strength and stability in materials like sodium metal.

atomic radius calculation

Calculating the atomic radius in a BCC lattice might seem tricky at first. But with the right steps, it's quite simple. Let's break it down:

The radius of an atom in a BCC structure is determined by the relationship between the cell edge (lattice parameter) and the body diagonal of the cube.
Using the Pythagorean theorem, we find that the body diagonal is \(\root 3} \times \text{cell edge}\).

For sodium, given the cell edge of 4.29 Å, the body diagonal is calculated as \(\root 3} \times 4.29 \text{Å} = 7.4255 \text{Å}\).

Knowing that the body diagonal spans the distance of four atomic radii (i.e., \(4r\), we then solve for the atomic radius: \(4r = 7.4255 \text{Å}\) which gives \(r = \frac{7.4255 \text{Å}}{4} = 1.856375 \text{Å}\).

lattice parameter relationship

Understanding the relationship between the lattice parameter and atomic radius in a BCC lattice is crucial. Here's how they connect:

The lattice parameter, or cell edge length, provides a direct way to find distances between atoms within the crystal structure.
In a BCC lattice, the body diagonal is defined by the equation \(d = \root 3} \times \text{cell edge}\), where 'd' is the body diagonal.
The radius of an atom in this structure is then derived by knowing that this body diagonal equals four times the atomic radius (i.e., \(\root 3} \times \text{cell edge} = 4r\).

This relationship allows us to accurately determine the size of atoms packed in a solid, providing insights into material properties and behavior.