Question

What is the Relation Between Edge Length 'a' and Radius of Atom 'R' for BCC Lattice?

Answer 1

Body-Centered Cubic Crystal Structure:

• A body-centered cubic unit cell structure is made up of atoms arranged in a cube with one atom in the center and one atom shared by each of the cube's four corners.
• Eight more unit cells share the atom in the corners of the cube.
• As a result, every corner atom stands in for one-eighth of an atom.

Step 1: Defining the quantities

• Typically, $\mathrm{a}$ is used to represent the length of a cell edge.
• Body diagonal refers to the path a cube takes from one corner to the opposite corner (say$\mathrm{bd}$).
• Face diagonal is equal to $\mathrm{fd}$

Step 2: Using Pythogoras' theorem

• ${\mathrm{bd}}^2={\mathrm{fd}}^2+{\mathrm{a}}^2$
• ${\mathrm{bd}}^2=2{\mathrm{a}}^2+{\mathrm{a}}^2$
• ${\mathrm{bd}}^2=3{\mathrm{a}}^2$

Step 3: Relation Between Edge Length '$\mathrm{a}$‘ and Radius of Atom ’$\mathrm{R}$'

• Atoms touch each other along the body diagonal, let's say$\mathrm{bd}$.
• As a result, the body diagonal has a length equal to four times the atom's radius, $\mathrm{R}$, or $\mathrm{bd}=4\mathrm{R}$.
• The Pythagorean theorem can be used to determine the relationship between $\mathrm{a}$ and $\mathrm{R}$:
  ${\left(4\mathrm{R}\right)}^2=3\mathrm{a}$
• Therefore, $4\mathrm{R}=\sqrt{3\mathrm{a}}$

Hence the relationship between Edge Length 'A' and Radius of Atom 'R' for BCC Lattice Is $4\mathrm{R}=\sqrt{3\mathrm{a}}$