Height of a Hexagonal Closing Packing Unit Cell
$begingroup$
According to my book, the dimensions of a HCP unit cell is $2r$,$2r$, $2.83r$.
How in the world is the height $2.83r$?
The length and width are obviously $2r$ because there the base is a rhombus and the atoms at each corner of the rhombus are touching.
However, the height has to to account for the extra atom in between the top and bottom layers of the rhombus. Check diagram:
I was able to calculate (what I thought to be the answer) of the height of this unit by finding the height of a tetrahedron of length $2r$: $2frac{sqrt{6}}{3}$. In other words, the height from the base of the cell (the plane that cuts the $4$ atoms of the lower layer in half) to the plane through the midpoint of the middle atom is approximately $1.6r$. Therefore, from the middle atom to the top plane (cuts $4$ atoms of top layer in half), it should be be double the original calculation: approximately $3.2r$.
But my book says height is $2.83$...
Wikipedia seems to corroborate my answer for the height (z-axis).
What did I do wrong/conceptualize incorrectly?
P.S. I think Wikipedia is wrong for the $y$ component of the first row of the second layer. It should be $(r, r+ rfrac{sqrt{3}}{2}, r + 2frac{sqrt{6}}{3})$ because it is in between row 1 and 2 of layer 1, which have a $y$ difference of $rsqrt{3}$, so between them would be $rfrac{sqrt{3}}{2}$. Could someone confirm my suspicions?
packing-problem
edited Sep 26 '17 at 21:35
achille hui
96k5132258
asked Apr 2 '16 at 1:26
NovaNova
145112
$endgroup$
add a comment |
$begingroup$
According to my book, the dimensions of a HCP unit cell is $2r$,$2r$, $2.83r$.
How in the world is the height $2.83r$?
The length and width are obviously $2r$ because there the base is a rhombus and the atoms at each corner of the rhombus are touching.
However, the height has to to account for the extra atom in between the top and bottom layers of the rhombus. Check diagram:
I was able to calculate (what I thought to be the answer) of the height of this unit by finding the height of a tetrahedron of length $2r$: $2frac{sqrt{6}}{3}$. In other words, the height from the base of the cell (the plane that cuts the $4$ atoms of the lower layer in half) to the plane through the midpoint of the middle atom is approximately $1.6r$. Therefore, from the middle atom to the top plane (cuts $4$ atoms of top layer in half), it should be be double the original calculation: approximately $3.2r$.
But my book says height is $2.83$...
Wikipedia seems to corroborate my answer for the height (z-axis).
What did I do wrong/conceptualize incorrectly?
P.S. I think Wikipedia is wrong for the $y$ component of the first row of the second layer. It should be $(r, r+ rfrac{sqrt{3}}{2}, r + 2frac{sqrt{6}}{3})$ because it is in between row 1 and 2 of layer 1, which have a $y$ difference of $rsqrt{3}$, so between them would be $rfrac{sqrt{3}}{2}$. Could someone confirm my suspicions?
packing-problem
edited Sep 26 '17 at 21:35
achille hui
96k5132258
asked Apr 2 '16 at 1:26
NovaNova
145112
$endgroup$
$begingroup$
Here is the wikipedia link because I can't have more than two links in my post for some reason: ttps://en.wikipedia.org/wiki/Close-packing_of_equal_spheres#Simple_hcp_lattice
$endgroup$
– Nova
Apr 2 '16 at 2:44
add a comment |
$begingroup$
According to my book, the dimensions of a HCP unit cell is $2r$,$2r$, $2.83r$.
How in the world is the height $2.83r$?
The length and width are obviously $2r$ because there the base is a rhombus and the atoms at each corner of the rhombus are touching.
However, the height has to to account for the extra atom in between the top and bottom layers of the rhombus. Check diagram:
I was able to calculate (what I thought to be the answer) of the height of this unit by finding the height of a tetrahedron of length $2r$: $2frac{sqrt{6}}{3}$. In other words, the height from the base of the cell (the plane that cuts the $4$ atoms of the lower layer in half) to the plane through the midpoint of the middle atom is approximately $1.6r$. Therefore, from the middle atom to the top plane (cuts $4$ atoms of top layer in half), it should be be double the original calculation: approximately $3.2r$.
But my book says height is $2.83$...
Wikipedia seems to corroborate my answer for the height (z-axis).
What did I do wrong/conceptualize incorrectly?
P.S. I think Wikipedia is wrong for the $y$ component of the first row of the second layer. It should be $(r, r+ rfrac{sqrt{3}}{2}, r + 2frac{sqrt{6}}{3})$ because it is in between row 1 and 2 of layer 1, which have a $y$ difference of $rsqrt{3}$, so between them would be $rfrac{sqrt{3}}{2}$. Could someone confirm my suspicions?
packing-problem
edited Sep 26 '17 at 21:35
achille hui
96k5132258
asked Apr 2 '16 at 1:26
NovaNova
145112
$endgroup$
According to my book, the dimensions of a HCP unit cell is $2r$,$2r$, $2.83r$.
How in the world is the height $2.83r$?
The length and width are obviously $2r$ because there the base is a rhombus and the atoms at each corner of the rhombus are touching.
However, the height has to to account for the extra atom in between the top and bottom layers of the rhombus. Check diagram:
I was able to calculate (what I thought to be the answer) of the height of this unit by finding the height of a tetrahedron of length $2r$: $2frac{sqrt{6}}{3}$. In other words, the height from the base of the cell (the plane that cuts the $4$ atoms of the lower layer in half) to the plane through the midpoint of the middle atom is approximately $1.6r$. Therefore, from the middle atom to the top plane (cuts $4$ atoms of top layer in half), it should be be double the original calculation: approximately $3.2r$.
But my book says height is $2.83$...
Wikipedia seems to corroborate my answer for the height (z-axis).
What did I do wrong/conceptualize incorrectly?
P.S. I think Wikipedia is wrong for the $y$ component of the first row of the second layer. It should be $(r, r+ rfrac{sqrt{3}}{2}, r + 2frac{sqrt{6}}{3})$ because it is in between row 1 and 2 of layer 1, which have a $y$ difference of $rsqrt{3}$, so between them would be $rfrac{sqrt{3}}{2}$. Could someone confirm my suspicions?
packing-problem
packing-problem
edited Sep 26 '17 at 21:35
achille hui
96k5132258
asked Apr 2 '16 at 1:26
NovaNova
145112
edited Sep 26 '17 at 21:35
achille hui
96k5132258
asked Apr 2 '16 at 1:26
NovaNova
145112
edited Sep 26 '17 at 21:35
achille hui
96k5132258
edited Sep 26 '17 at 21:35
achille hui
96k5132258
edited Sep 26 '17 at 21:35
achille hui
96k5132258
96k5132258
asked Apr 2 '16 at 1:26
NovaNova
145112
asked Apr 2 '16 at 1:26
NovaNova
145112
asked Apr 2 '16 at 1:26
NovaNova
145112
145112
$begingroup$
Here is the wikipedia link because I can't have more than two links in my post for some reason: ttps://en.wikipedia.org/wiki/Close-packing_of_equal_spheres#Simple_hcp_lattice
$endgroup$
– Nova
Apr 2 '16 at 2:44
add a comment |
$begingroup$
Here is the wikipedia link because I can't have more than two links in my post for some reason: ttps://en.wikipedia.org/wiki/Close-packing_of_equal_spheres#Simple_hcp_lattice
$endgroup$
– Nova
Apr 2 '16 at 2:44
$begingroup$
Here is the wikipedia link because I can't have more than two links in my post for some reason: ttps://en.wikipedia.org/wiki/Close-packing_of_equal_spheres#Simple_hcp_lattice
$endgroup$
– Nova
Apr 2 '16 at 2:44
$begingroup$
Here is the wikipedia link because I can't have more than two links in my post for some reason: ttps://en.wikipedia.org/wiki/Close-packing_of_equal_spheres#Simple_hcp_lattice
$endgroup$
– Nova
Apr 2 '16 at 2:44
add a comment |
1 Answer
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With spheres the height is indeed $4sqrt{6}/3$ (actually closer to 3.3 than to 3.2) times the radius of each sphere as you figured. Does the book explain how it got its ratio?
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.7k12136
$endgroup$
$begingroup$
I updated the question with the highlighted excerpt from my book.
$endgroup$
– Nova
Apr 2 '16 at 2:42
$begingroup$
So the book explains nothing. Looks like someone there was drinking and deriving.
$endgroup$
– Oscar Lanzi
Apr 2 '16 at 3:29
add a comment |
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1 Answer
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1 Answer
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$begingroup$
With spheres the height is indeed $4sqrt{6}/3$ (actually closer to 3.3 than to 3.2) times the radius of each sphere as you figured. Does the book explain how it got its ratio?
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.7k12136
$endgroup$
$begingroup$
I updated the question with the highlighted excerpt from my book.
$endgroup$
– Nova
Apr 2 '16 at 2:42
$begingroup$
So the book explains nothing. Looks like someone there was drinking and deriving.
$endgroup$
– Oscar Lanzi
Apr 2 '16 at 3:29
add a comment |
$begingroup$
With spheres the height is indeed $4sqrt{6}/3$ (actually closer to 3.3 than to 3.2) times the radius of each sphere as you figured. Does the book explain how it got its ratio?
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.7k12136
$endgroup$
$begingroup$
I updated the question with the highlighted excerpt from my book.
$endgroup$
– Nova
Apr 2 '16 at 2:42
$begingroup$
So the book explains nothing. Looks like someone there was drinking and deriving.
$endgroup$
– Oscar Lanzi
Apr 2 '16 at 3:29
add a comment |
$begingroup$
With spheres the height is indeed $4sqrt{6}/3$ (actually closer to 3.3 than to 3.2) times the radius of each sphere as you figured. Does the book explain how it got its ratio?
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.7k12136
$endgroup$
With spheres the height is indeed $4sqrt{6}/3$ (actually closer to 3.3 than to 3.2) times the radius of each sphere as you figured. Does the book explain how it got its ratio?
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.7k12136
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.7k12136
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.7k12136
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.7k12136
12.7k12136
$begingroup$
I updated the question with the highlighted excerpt from my book.
$endgroup$
– Nova
Apr 2 '16 at 2:42
$begingroup$
So the book explains nothing. Looks like someone there was drinking and deriving.
$endgroup$
– Oscar Lanzi
Apr 2 '16 at 3:29
add a comment |
$begingroup$
I updated the question with the highlighted excerpt from my book.
$endgroup$
– Nova
Apr 2 '16 at 2:42
$begingroup$
So the book explains nothing. Looks like someone there was drinking and deriving.
$endgroup$
– Oscar Lanzi
Apr 2 '16 at 3:29
$begingroup$
I updated the question with the highlighted excerpt from my book.
$endgroup$
– Nova
Apr 2 '16 at 2:42
$begingroup$
I updated the question with the highlighted excerpt from my book.
$endgroup$
– Nova
Apr 2 '16 at 2:42
$begingroup$
So the book explains nothing. Looks like someone there was drinking and deriving.
$endgroup$
– Oscar Lanzi
Apr 2 '16 at 3:29
$begingroup$
So the book explains nothing. Looks like someone there was drinking and deriving.
$endgroup$
– Oscar Lanzi
Apr 2 '16 at 3:29
add a comment |
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$begingroup$
Here is the wikipedia link because I can't have more than two links in my post for some reason: ttps://en.wikipedia.org/wiki/Close-packing_of_equal_spheres#Simple_hcp_lattice
$endgroup$
– Nova
Apr 2 '16 at 2:44