Final answer:
The distance between the nearest parallel (100), (110), and (111) planes in a simple cubic lattice with a lattice constant of 5.63 Å are 5.63 Å, (5.63√2)/2 Å, and (5.63√3)/3 Å, respectively.
Step-by-step explanation:
To calculate the distance between the nearest parallel planes in a simple cubic lattice with a lattice constant of 5.63 Å, we use the lattice constant and the Miller indices that represent the orientation of the planes within the lattice.
- (100) planes: For the (100) planes, these are planes with normals along the x-axis. The spacing d between these planes is just the lattice constant, since each subsequent (100) plane is one lattice constant away. Thus, d100 = 5.63 Å.
- (110) planes: The (110) planes intersect the x and y axes at the lattice constant, creating a diagonal across the face of the unit cell. To find the distance between these planes, d110, we use the geometry of a square with side length 5.63 Å to find that the diagonal is √(5.63² + 5.63²), which is equal to 5.63√2 Å. Since the (110) planes are spaced at half the diagonal, we divide by 2, yielding d110 = (5.63√2)/2 Å.
- (111) planes: The (111) planes intersect all three axes at the lattice constant and form a plane that cuts through the body diagonal of the cube. The length of the body diagonal is √(5.63² + 5.63² + 5.63²) = 5.63√3 Å, and since there are three atomic layers within one body diagonal, we divide by 3 to get d111 = (5.63√3)/3 Å.
These calculations are fundamental to understanding the crystallographic structure of materials and are crucial for various applications in materials science and engineering.