Volume of HCP unit cell from radius + c/a ratio

In summary, the homework statement states that cobalt has an HCP crystal structure, an atomic radius of 0.1253 nm, and a c/a ratio of 1.623. The Attempt at a Solution states that the hexagon area is 1.88401e-1 and the volume is 7.66272e-2. My answer is 6.64e-2 nm3.
  • #1
physicsnnewbie
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Homework Statement


Cobalt has an HCP (hexagonal close packed) crystal structure, an atomic radius of 0.1253 nm and a c/a ratio of 1.623. Compute the volume of the unit cell for cobalt.


Homework Equations


area of hexagon = apothem*perimeter/2


The Attempt at a Solution


calculate 1 of the 6 sides (2 atomic radii per side):
2*.1253 = 2.506e-1

calculate perimeter:
2.506e-1*6 = 1.5036e+0

calculate height:
2.506e-1*1.623 = 4.06724e-1

calculate hexagon area (apothem = 2 radii):
1.5036e+0*2.506e-1 / 2 = 1.88401e-1

calculate volume:
1.88401e-1*4.06724e-1 = 7.66272e-2

my answer:
7.66272e-2

correct answer:
6.64e-2 nm3
 
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  • #2
mistook the apothem for the hexagon radius.
 
  • #3
Hi... I'm currently studying for my first test on Solid State Physics and looking for some info I found your question... I agree that considering the right expression for the apothem one gets to the answer proposed... Nevertheless, I think there is a consideration to make in the problem...

I think they are asking for six times the volume of the parallelipiped formed by the vectors

a1 = (a,0,0)
a2 = (a/2,a*sqr(3)/2,0)
b3 = a1/3 + a2/3 + (0,0,c/2)

. In other words, just the volume proposed by the previous answers, 3*sqr(3)*a*a*c/2.

But, due to the geometry of the hcp the distance to the nearest neighbors depends on c/a (It is not always a). One finds that this distance is the minimum between a and sqr(a*a/3 +c*c/4) (This last value is just the norm of the vector b3 I used before). Then if c/a > sqr(8/3) the closest neighbors are just the 6 ones in the xy planes (the hexagon around the point). But if c/a < sqr(8/3) (around 1.63299) then the closest neighbors are 6 different points (3 in the upper plane and 3 in the lower plane). So, for this case the radius given for the "sphere-atom" wouldn't be a/2 but 0.5*sqr(a*a/3 +c*c/4) (in order to avoid overlapping). Using this fact and the value of c/a you find a and then replacing in the volume you'll get a slightly different answer (6.7179e-2 nm^3). Am I making a mistake? If I am, I apologize for any slip either in the solution or in my English... it's my first post. Good luck.
 
Last edited:

What is the formula for calculating the volume of a HCP unit cell?

The formula for calculating the volume of a HCP unit cell is V = (4/3)πr^3 + (c/a)^2 * √3/2 * a^3, where r is the radius of the atom and c/a is the ratio of the height to the base of the unit cell.

How do you determine the radius of an atom in a HCP unit cell?

The radius of an atom in a HCP unit cell can be determined using X-ray crystallography or other methods such as electron diffraction. It can also be estimated using the empirical radius ratio rules for ionic compounds.

What is the significance of the c/a ratio in the calculation of HCP unit cell volume?

The c/a ratio is significant because it represents the relative lengths of the two lattice vectors in a HCP unit cell. This ratio affects the overall shape and volume of the unit cell, and plays a role in determining the density and packing efficiency of the crystal structure.

Can the volume of a HCP unit cell be calculated without knowing the c/a ratio?

No, the volume of a HCP unit cell cannot be accurately calculated without knowing the c/a ratio. This is because the c/a ratio affects the overall shape and volume of the unit cell, and is an essential component of the formula for calculating the volume.

How does the volume of a HCP unit cell compare to other crystal structures?

The volume of a HCP unit cell is typically smaller than other crystal structures, such as FCC or BCC. This is due to the close packing of atoms in the HCP structure, which results in a higher packing efficiency and a smaller unit cell volume compared to other structures.

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