Final answer:
The d spacing between the (100) planes of an FCC structure like aluminum is found using the atomic radius and the geometric relationships of an FCC lattice. Using the atomic radius of aluminum (0.143 nm), the edge length of the FCC unit cell is 0.404 nm, which is also the spacing between the (100) planes, d = 0.404 nm.
Step-by-step explanation:
The spacing (d) between the (100) planes in an FCC crystal structure like aluminum can be found using the atomic radius (r) and simple geometric relationships specific to the FCC lattice. In an FCC structure, the atoms are arranged at the corners and centers of each face of the cube.
For the (100) planes, which correspond to the faces of the cube, the distance between two adjacent planes is the same as the length of one edge of the cube. The atomic radius is half the distance between the nuclei of two adjacent atoms that are touching each other along the edge of the cube.
To calculate the edge length (a) of the FCC unit cell, we use the relationship that the diagonal of the face of the cube (which is √2 times longer than an edge) is equal to four times the atomic radius of the atom (d = 4r).
Therefore, a = 2r√2.
With an atomic radius for aluminum of 0.143 nm, the edge length of the unit cell is a = 2 * 0.143 nm * √2 = 0.404 nm.
Subsequently, the d spacing between the (100) planes is equal to this edge length, so d = a = 0.404 nm.