What is the radius of the largest sphere that can be placed at the center of a face-centered cubic unit cell of a cubic closest-packed array of spheres if the spheres have radius of 2.0 x 10^2 pm?
In a face-centred cubic unit cell, the atoms in the cell take up 74% of its total volume. The rest is just empty space.
In this cell, each eight corners contains one-eighth of an atom. Each six faces contain one-half of an atom. The unit cell has an equivalent of four atoms.
The atoms-- which are referred to as spheres in the question, have a volume of 4/3*pi*r^3. Substituting 2.0*10^2 pm for r and then multiplying by 4 since there are 4 spheres, gives a number that is roughly 134 041 286.6 pm^3. The atoms in the face-centred unit cell take up that much space.
The calculated volume is only 74% of the total volume of the cell. Dividing 134 041 286.6 pm^3 by 0.74 gives a number that is roughly 181 136 873.7 pm^3. That's the total volume of the unit cell.
The cubed root of volume of the unit cell will give us the measure of one side length. Doing this calculation yields something like 565.8 pm.
The maximum diameter of a sphere that can be placed in such a cell is one side length, minus the radii of the two half-spheres at each end. The maximum diameter is 565.8-200-200 = 165.8pm.
To get the radius of this sphere, just divide the diameter by two. Rounding for significant digits, we get 83pm.
Thus, the maximum radius of such a sphere is 83pm.
gukj
To find the radius of the largest sphere that can be placed at the center of a face-centered cubic (FCC) unit cell, we need to consider the arrangement and size of the spheres in the unit cell.
In an FCC arrangement, each sphere in the middle layer is surrounded by six spheres in the same layer, three spheres above it, and three spheres below it in adjacent layers. This arrangement is also called a cubic closest-packed (CCP) array.
Given that the spheres in the CCP array have a radius of 2.0 x 10^2 pm, we can calculate the radius of the largest sphere that can fit in the center of the unit cell by considering the distances between the spheres.
Let's calculate the distance between the center of a sphere in the middle layer and one of its neighboring spheres in the same layer:
The distance between the centers of two adjacent spheres in the same layer can be determined using the relationship:
d = 2r
where d is the distance between the centers and r is the radius of each sphere.
In this case, d = 2(2.0 x 10^2 pm) = 4.0 x 10^2 pm.
Now, let's consider the distance between the center of a sphere in the middle layer and a neighboring sphere in an adjacent layer. To calculate this distance, we can use the relationship:
d' = √(3d^2)
where d' is the distance between the centers and d is the distance between the centers of two adjacent spheres in the same layer.
Substituting the known values, we have:
d' = √(3(4.0 x 10^2 pm)^2)
= √(3(16.0 x 10^4 pm^2))
= √(48.0 x 10^4 pm^2)
= 6.928 x 10^2 pm
Since the largest sphere will be tangent to three adjacent spheres in the same layer and one sphere in each of the three adjacent layers, the distance between the center of the sphere in the middle layer and any neighboring sphere in an adjacent layer is equal to 6.928 x 10^2 pm.
Finally, the radius of the largest sphere that can be placed at the center of the FCC unit cell is half of this distance:
Radius = 6.928 x 10^2 pm / 2
= 3.464 x 10^2 pm
Therefore, the radius of the largest sphere is 3.464 x 10^2 pm.