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

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SUMMARY

The volume of the hexagonal close-packed (HCP) unit cell for cobalt, with an atomic radius of 0.1253 nm and a c/a ratio of 1.623, was calculated to be approximately 0.0766 nm³. However, the correct volume is 0.0664 nm³, as the apothem was mistakenly treated as the hexagon radius. The discussion highlights the importance of accurately determining the apothem and understanding the geometric relationships in HCP structures, particularly how the c/a ratio influences nearest neighbor distances and volume calculations.

PREREQUISITES
  • Understanding of hexagonal close-packed (HCP) crystal structures
  • Familiarity with geometric calculations involving hexagons
  • Knowledge of solid state physics principles, particularly unit cell volume calculations
  • Proficiency in manipulating ratios, specifically c/a ratios in crystallography
NEXT STEPS
  • Study the geometric properties of hexagonal close-packed (HCP) structures
  • Learn about the calculation of unit cell volumes in different crystal systems
  • Explore the significance of the c/a ratio in determining atomic arrangements
  • Investigate the relationship between atomic radius and nearest neighbor distances in crystalline solids
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Students and professionals in materials science, solid state physics, and crystallography, particularly those focusing on crystal structure analysis and unit cell calculations.

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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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mistook the apothem for the hexagon radius.
 
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:

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