Question

Metallic iron crystallizes in a body-centered cubic lattice, with one fe atom per lattice point. if the metallic radius of fe is 124 pm, what is the volume of the unit cell in pm3 and in cm3?

Answer 1

The volume of a unit cell is 4.3 × 10⁻¹⁰ pm³.

We need to start with the features of a face-centered cubic system in order to be able to determine the edge length of the unit cell.

As we know, a face-centered cubic system is characterized by a unit cell that has a total of 14 lattice points.

• one lattice point for every one of the eight corners of the unit cell.

• one lattice point for every one of the six faces of the unit cell.

Now, these lattice points will contain

• 1/8th of an atom in every corner lattice point.

• 1/2 of an atom in every face lattice point.

Assume that r is the edge length of the unit cell and that r is the radius of an iron atom. You'll see that we can represent x in terms of the cell's diagonal, d, using Pythagoras' Theorem.

        $d^2 = x^2+ x^2$

        $d^2=2x^2$

The cell's diagonal will be equal to one radius from the atom in the top corner, two from the atom in the face, and one from the atom in the lower corner.

$d= r+2r+r = 4r$

This means that we have,

$(4r)^2=2x^2$

$16.r^2 = 2x^2 \\x=\sqrt{8r^2}$

This is equivalent to

$x=2\sqrt{2} r = 2\sqrt{2} .124= 350.7 pm$

So, the edge length of the unit cell is equal to 350.7 pm.

The volume of a unit cell. As we know, the volume of a cube is given by the formula

V = $l$ × $l$ × $l$

$l$ = the edge length of the cube

Volume = (350.7)³ pm³

            = 4.3 × 10⁻¹⁰ pm³

            = 4.3e-30 cm³

Therefore, the volume of the unit cell is 4.3 × 10⁻¹⁰ pm³.

Learn more about unit cell here:

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