Answer:
124.45 pm
Explanation:
Given that:
Metallic nickel crystallizes in a face-centered cubic lattice;
The diagram below shows an illustration of that;
and the edge length of the unit cell is found to be 352 pm;
we are tasked to find the metallic radius of Ni in pm?
In order to do that; from the diagram; we can use the Pythagoras theorem to determine the value for the radius.
So; we have:
(4r)² = d² + d²
16r² = 2d²
r² = [tex]\frac{2d^2}{16}[/tex]
[tex]r^2= \frac{d^2}{8}[/tex]
[tex]r= \sqrt{\frac{d^2}{8} }[/tex]
[tex]r=\frac{\sqrt{d^2} }{\sqrt{8} }[/tex]
[tex]r =\frac{d}{\sqrt{2*4} }[/tex]
[tex]r= \frac{d}{2\sqrt{2} }[/tex]
So since d (the edge length of the unit cell) = 352 pm
then metallic radius of Ni in pm will be:
[tex]r= \frac{d}{2\sqrt{2} }[/tex]
[tex]r= \frac{352}{2\sqrt{2} }[/tex]
r = 124.4507935 pm
r ≅ 124.45 pm
∴ the metallic radius of Ni in pm = 124.45 pm
