The correct answer is (1) \(2\sqrt{2r}, \frac{4r}{\sqrt{3}}, 2r\)
Let's go into more detail about the edge lengths of the unit cells for FCC (face-centered cubic), BCC (body-centered cubic), and simple cubic structures in terms of the radius (\(r\)) of the spheres constituting these unit cells.
1. FCC (Face-Centered Cubic):
- In an FCC structure, each corner of the cube has a sphere, and there is an additional sphere at the center of each face.
- The edge length (\(a_{\text{FCC}}\)) can be determined by considering the diagonal of the face of the cube. The diagonal is equal to twice the radius of the sphere (\(2r\)).
- The body diagonal is the hypotenuse of a right triangle formed by two radii. By applying the Pythagorean theorem, the body diagonal is given by \(2\sqrt{2}r\).
- Therefore, \(a_{\text{FCC}} = \frac{2\sqrt{2}r}{\sqrt{2}} = 2\sqrt{2}r\).
2. BCC (Body-Centered Cubic):
- In a BCC structure, each corner of the cube has a sphere, and there is an additional sphere at the center of the cube.
- The edge length (\(a_{\text{BCC}}\)) is related to the body diagonal of the cube. The body diagonal is equal to the space diagonal of a cube with side length \(2r\).
- The space diagonal (\(d_{\text{BCC}}\)) is given by \(d_{\text{BCC}} = \sqrt{3} \cdot a_{\text{BCC}}\), where \(a_{\text{BCC}}\) is the edge length of the BCC unit cell.
- For BCC, \(d_{\text{BCC}} = 4r\) (twice the radius of the sphere).
- Solving for \(a_{\text{BCC}}\), we get \(a_{\text{BCC}} = \frac{4r}{\sqrt{3}}\).
3. Simple Cubic:
- In a simple cubic structure, each corner of the cube has a sphere.
- The edge length (\(a_{\text{SC}}\)) is simply twice the radius of the sphere (\(2r\)), as it is the distance between two spheres along one edge of the cube.
In summary, the correct order of edge lengths for FCC, BCC, and simple cubic structures in terms of the radius \(r\) is:
\[2\sqrt{2}r, \frac{4r}{\sqrt{3}}, 2r\]
So, the correct answer is: 1. \(2\sqrt{2}r, \frac{4r}{\sqrt{3}}, 2r\) |
