Question

The length of the unit cell edge of a bcc lattice metal is 352 pm. Calculate radius of the atom of the metal.

Answer 1

Hint There is a formula to calculate the length of the unit cell in bcc lattice and it is as follows.
\[r=\dfrac{\sqrt{3}}{4}a\]
Here r = radius of the atom
a = length of the unit cell edge

Complete step by step answer:
- In the question it is asked to calculate the radius of the atom of metal bcc lattice metal.
- In the question it is given that the length of the unit cell edge of a bcc lattice metal is 352 pm.
- We know that in a bcc unit cell the metal atoms touch each other through the body diagonal.
- Therefore, if the radius of the metal atom is ‘r’ then body diagonal = r + 2r + r = 4r.
- The body diagonal for cube = \[\sqrt{\text{3}}\text{ }\!\!\times\!\!\text{ edge length = }\sqrt{3}a\] .
- Then $4r=\sqrt{3}a$ or
Radius of the metal atom in bcc = $\dfrac{\sqrt{3}}{4}a$
- Here we know that a = 352 pm
- Substitute the ‘a’ value in the above formula to get the radius of the metal atom.
- Then radius of the metal atom is
\[\dfrac{\sqrt{3}}{4}a=\dfrac{\sqrt{3}}{4}\times 352=152.4pm\] .

- Therefore the radius of the atom of the metal is 152.4 pm.

Note: bcc lattice means Body centered cubic structure. In the bcc lattice one point is in the center of the unit cell other than the eight corner points. Sodium, potassium and chromium atoms form bcc lattice crystal structure.