(a) Derive planar density expressions for BCC (100) and (110) planes in terms of the atomic radius \(R\). (b) Compute and compare planar density values for these same two planes for vanadium.

To find the planar density of the (100) plane, we need to determine how many atoms lie on this plane and the area of the plane. For the BCC structure, the (100) plane has only one effective atom contributing to the planar density. This is because the BCC structure has atoms at the corners and one at the center of the unit cell. In the (100) plane, the corner atoms only contribute 1/4 of their volume to the plane, and the center atom does not intersect the plane. The area of the (100) plane is a square with a side length equal to the edge length of the BCC unit cell. The BCC edge length can be found in terms of the atomic radius R by using the Pythagorean theorem for a right-angled triangle in which the legs are equal to the edge length and the hypotenuse is equal to 4R (the distance between the center atom and a corner atom). This gives us the following relationship: Edge Length\(= \sqrt{(4R)^2/2}\) Now we can find the area of the (100) plane: Area of (100) Plane \(= (\sqrt{(4R)^2/2})^2 = 8R^2\) The planar density of the BCC (100) plane can be calculated as follows: Planar Density of (100) Plane \(= \frac{Number\:of\:Atoms}{Area\:of\:Plane} = \frac{1}{8R^2} \)

For the BCC (110) plane, there are two effective atoms that contribute to the planar density, as there are two atoms on the (110) plane: one at the lower-left corner and half an atom at the top-right corner. The area of the (110) plane is a square with a side length equal to the diagonal of the unit cell. We can find the relation between the unit cell edge length (l) and the atomic radius (R) as follows: Diagonal Length \(= \sqrt{2}\cdot Edge\:Length = \sqrt{2}\cdot\sqrt{8R^2} = 4R\) Now we can find the area of the (110) plane: Area of (110) Plane \(= (4R)^2 = 16R^2\) The planar density of the BCC (110) plane can be calculated as follows: Planar Density of (110) Plane \(= \frac{Number\:of\:Atoms}{Area\:of\:Plane} = \frac{2}{16R^2} = \frac{1}{8R^2}\)

Now we can use the atomic radius of vanadium to compute the planar density of the (100) and (110) planes. The atomic radius of vanadium is 131 pm. Planar Density of (100) Plane for Vanadium \(= \frac{1}{8(131\cdot 10^{-12})^2}\) Planar Density of (110) Plane for Vanadium \(= \frac{1}{8(131\cdot 10^{-12})^2}\) Comparing both values, we can observe that both the (100) and (110) planes have the same planar density for vanadium in the BCC structure.