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But what are the ramifications of this century-long proof? Does it have the same universe-unlocking abilities that Harriet hoped for? Not really, but the sphere packing problem does certainly have some real-world applications. Consider the element silver, which has a covalent radius of 145 picometers and molar mass of 107.87 grams / mol (National Center for Biotechnology Information). If one was in possession of a silver bar, what would the density be? First, the universe is always minimizing potential energy, and at an atomic level this often means interacting atoms get as close as electromagnetically possible— they seek the maximum packing density. Because atoms can be considered spheres, this means that one could find the density of a silver atom using its mass and geometry, and then simply multiply by Kepler’s packing coefficient. As an equation, using conversion factors convert moles to atoms with Avagadro’s number and picometers to centimeters, this is: | But what are the ramifications of this century-long proof? Does it have the same universe-unlocking abilities that Harriet hoped for? Not really, but the sphere packing problem does certainly have some real-world applications. Consider the element silver, which has a covalent radius of 145 picometers and molar mass of 107.87 grams / mol (National Center for Biotechnology Information). If one was in possession of a silver bar, what would the density be? First, the universe is always minimizing potential energy, and at an atomic level this often means interacting atoms get as close as electromagnetically possible— they seek the maximum packing density. Because atoms can be considered spheres, this means that one could find the density of a silver atom using its mass and geometry, and then simply multiply by Kepler’s packing coefficient. As an equation, using conversion factors convert moles to atoms with Avagadro’s number and picometers to centimeters, this is: | ||
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<math>\eta_{silver}\ =\ \frac{m_{atom}}{V_{atom}}\cdot\eta_{spheres}=\frac{m_{atom}}{\frac{4}{3}\pi r^{3}}\cdot\eta_{spheres}</math> | <math>\eta_{silver}\ =\ \frac{m_{atom}}{V_{atom}}\cdot\eta_{spheres}=\frac{m_{atom}}{\frac{4}{3}\pi r^{3}}\cdot\eta_{spheres}</math> | ||
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This value could be checked with the real density, which is accepted as 10.501 grams per cubic centimeter. Despite all of the complexity of molecular interactions, this simple constant is able to give an incredibly close value for the densities of kinds of elemental solids. | This value could be checked with the real density, which is accepted as 10.501 grams per cubic centimeter. Despite all of the complexity of molecular interactions, this simple constant is able to give an incredibly close value for the densities of kinds of elemental solids. | ||
Revision as of 02:49, 6 December 2020
Density of Silver
Author: Eli LeChien
An Application in Chemistry
But what are the ramifications of this century-long proof? Does it have the same universe-unlocking abilities that Harriet hoped for? Not really, but the sphere packing problem does certainly have some real-world applications. Consider the element silver, which has a covalent radius of 145 picometers and molar mass of 107.87 grams / mol (National Center for Biotechnology Information). If one was in possession of a silver bar, what would the density be? First, the universe is always minimizing potential energy, and at an atomic level this often means interacting atoms get as close as electromagnetically possible— they seek the maximum packing density. Because atoms can be considered spheres, this means that one could find the density of a silver atom using its mass and geometry, and then simply multiply by Kepler’s packing coefficient. As an equation, using conversion factors convert moles to atoms with Avagadro’s number and picometers to centimeters, this is:
$ \eta_{silver}\ =\ \frac{m_{atom}}{V_{atom}}\cdot\eta_{spheres}=\frac{m_{atom}}{\frac{4}{3}\pi r^{3}}\cdot\eta_{spheres} $
$ \eta_{silver}=\frac{\ \frac{107.87\ \frac{g}{mol}}{\left(6.022\ \cdot\ 10^{23}\ \frac{atoms}{mol}\right)}}{\frac{4}{3}\pi\cdot\left(145\ pm\ \cdot\ 10^{-10}\ \frac{cm}{pm}\right)^{3}}\cdot\frac{\pi}{3\sqrt{2}} $
$ \eta_{silver}\approx10.39\ \frac{g}{cm^{3}} $
This value could be checked with the real density, which is accepted as 10.501 grams per cubic centimeter. Despite all of the complexity of molecular interactions, this simple constant is able to give an incredibly close value for the densities of kinds of elemental solids.
