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
95.6k5130257
asked Apr 2 '16 at 1:26
NovaNova
145111
$endgroup$
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
95.6k5130257
asked Apr 2 '16 at 1:26
NovaNova
145111
$endgroup$
$begingroup$
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
95.6k5130257
asked Apr 2 '16 at 1:26
NovaNova
145111
$endgroup$
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
95.6k5130257
asked Apr 2 '16 at 1:26
NovaNova
145111
edited Sep 26 '17 at 21:35
achille hui
95.6k5130257
asked Apr 2 '16 at 1:26
NovaNova
145111
edited Sep 26 '17 at 21:35
achille hui
95.6k5130257
edited Sep 26 '17 at 21:35
achille hui
95.6k5130257
edited Sep 26 '17 at 21:35
achille hui
95.6k5130257
95.6k5130257
asked Apr 2 '16 at 1:26
NovaNova
145111
asked Apr 2 '16 at 1:26
NovaNova
145111
asked Apr 2 '16 at 1:26
NovaNova
145111
145111
$begingroup$
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
1
active
oldest
votes
$begingroup$
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
12.2k12036
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1724056%2fheight-of-a-hexagonal-closing-packing-unit-cell%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
12.2k12036
$endgroup$
$begingroup$
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
12.2k12036
$endgroup$
$begingroup$
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
12.2k12036
$endgroup$
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
12.2k12036
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.2k12036
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.2k12036
answered Apr 2 '16 at 2:09
Oscar LanziOscar Lanzi
12.2k12036
12.2k12036
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1724056%2fheight-of-a-hexagonal-closing-packing-unit-cell%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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