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?
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gukj
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.
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.
In every cell, there will be holes and in FCC, mostly will be tetrahedral and octahedral. Imagine, the hole is in between the center sphere and adjacent three spheres, it will be a tetrahedral hole and that's where we gonna put the "sphere" in, the size (radius) of the tetrahedral hole is between 0.225 R and 0.414 R, R represents the radius of spheres in FCC. Since we are asked for the maximum radius, 200 pm * 0.414 = 83pm.
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