Animated Solution for Chemistry - States of Matter: The radius of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell is
(Edge length is represented by 'a')
Select Answer:
Visualized Solution
BCCUnitCell
Body Centered Cubic (BCC) unit cell with edge length a.
RadiusofCornerAtom
For BCC structure, radius of corner atom:
R=43a
FittingtheSphere
Let the radius of the largest sphere fitting at the edge center be r.
EdgeGeometry
From the edge geometry:
a=R+2r+R
a=2(R+r)
Substitution
Substitute R=43a:
a=2(43a+r)
Solvingforr
2a=43a+r
r=2a−43a
Simplification
r=42a−3a
r=4a(2−3)
FinalCalculation
Using 3≈1.732:
r=4a(2−1.732)
r=40.268a
r=0.067a
TheWayForward
What is the radius of the largest sphere that can fit in the octahedral void of an FCC unit cell?
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The Sigma Insight: Solid State
Solution Diagram
The problem asks us to find the radius of the largest sphere that can fit perfectly at the center of an edge in a Body-Centered Cubic (BCC) unit cell. This is a beautiful exercise in spatial reasoning and basic geometry.
Visualizing the BCC Edge
Imagine you are looking at a BCC unit cell. In this structure, atoms are located at all eight corners of the cube, and one atom sits right in the center of the body.
The crucial constraint here is how these atoms touch. In a BCC crystal, the corner atoms touch the central body atom along the body diagonal. However, they do not touch each other along the edges of the cube. This leaves a small gap along every edge.
Let the edge length of the unit cell be a, and the radius of the corner atoms be R. From the geometry of the body diagonal, we know the relationship between R and a is:
R=43a
The Geometry of the Gap
Now, let's focus entirely on one edge of length a. At each end of this edge, there is a corner atom extending a distance R inwards.
We want to place a new, smaller sphere exactly at the center of this edge. Let's call its radius r. For this sphere to be the "largest" possible, it must perfectly touch the two corner atoms without pushing them apart.
If we look at the total length of the edge, it is composed of three segments:
1. The radius of the left corner atom (R)
2. The diameter of the small central sphere (2r)
3. The radius of the right corner atom (R)
Therefore, we can write the master equation for the edge length:
a=R+2r+R
a=2R+2r
a=2(R+r)
Final Calculation
Our goal is to find r in terms of a. Let's substitute our known value of R into the master equation:
a=2(43a+r)
Now, we carefully isolate r. First, divide both sides by 2:
2a=43a+r
Next, subtract the a term from the right side:
r=2a−43a
To combine these terms, find a common denominator of 4:
r=42a−3a
r=4a(2−3)
Finally, we substitute the approximate value of 3≈1.732:
r=4a(2−1.732)
r=40.268a
r=0.067a
The radius of the largest sphere that can fit at the edge center is 0.067a. This perfectly matches option (d).