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Question
Chemistry
Posted 3 months ago
write down the formula for the relationship between the atom' radius and unit cell in 
(
a
)
a simple cubic
(
b
)
a body centered cubic 
(
c
)
face
-
centered cubic
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Answer from Sia
Posted 3 months ago
Solution
1
Simple Cubic (sc) Unit Cell: In a simple cubic lattice, the atoms are located at the corners of the cube. The edge length of the unit cell (aa) is equal to two times the radius of the atom (rr)
2
Body-Centered Cubic (bcc) Unit Cell: In a body-centered cubic lattice, there is an atom at each corner and one atom at the center of the cube. The diagonal of the cube is equal to four times the radius of the atom (4r4r), and the cube's edge length (aa) can be found using the Pythagorean theorem in three dimensions: a=3ra = \sqrt{3}r
3
Face-Centered Cubic (fcc) Unit Cell: In a face-centered cubic lattice, there is an atom at each corner and one atom at the center of each face. The face diagonal is equal to four times the radius of the atom (4r4r), and the edge length (aa) can be found using the Pythagorean theorem: a=22r2=2ra = \frac{2\sqrt{2}r}{\sqrt{2}} = 2r
1 Answer
a=2ra = 2r for simple cubic
2 Answer
a=4r3a = \frac{4r}{\sqrt{3}} for body-centered cubic
3 Answer
a=2ra = 2r for face-centered cubic
Key Concept
The relationship between the atom's radius and the unit cell edge length depends on the lattice type.
Explanation
For simple cubic, the edge length is twice the atomic radius. For body-centered cubic, the edge length is four times the atomic radius divided by the square root of three. For face-centered cubic, the edge length is twice the atomic radius.

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