Question

At 1115 degrees Celcius, where iron is still a solid (melting point 1538 degrees Celcius), the unit cell for the most stable crystal lattice of the metal is face- centered cubic (fcc) with an edge length of 362 pm. What is the atomic radius of iron at this temperature?

Answer 1

Answer: The atomic radius of iron is 128 pm.

Explanation:

To calculate the radius of the metal having FCC crystal lattice, the relationship between edge length and radius follows:

$4r=\sqrt{2}a$

Where,

a = edge length = 362 pm

r = atomic radius of iron = ?

Plugging values in above equation, we get:

$r=\frac{\sqrt{2}\times 362}{4}\\\\a=128pm$

Hence, the atomic radius of iron is 128 pm.