74.4k views
1 vote
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?

User Spinlock
by
4.1k points

1 Answer

6 votes

Answer: The atomic radius of iron is 128 pm.

Step-by-step explanation:

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


4r=√(2)a

Where,

a = edge length = 362 pm

r = atomic radius of iron = ?

Plugging values in above equation, we get:


r=(√(2)* 362)/(4)\\\\a=128pm

Hence, the atomic radius of iron is 128 pm.

User Lusitanian
by
4.0k points