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Topic: Unit cell  (Read 2070 times)

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Offline nawinince

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Unit cell
« on: June 17, 2018, 04:05:28 AM »
1) Consider the unit cell of aluminum with an aluminum ion every corner and every face-centered site of the cube. Using the value of ionic radius of aluminum ion r = 0.143 nm, the length of each edge of the unit cell can be calculated as ... nm.
2) Using the atomic weight value of aluminum 27.0, the density of aluminum can be calculated as ... g/cm^3

***Calculate into 2 significant figure***

For 1), I draw a picture and give edge length = x, then I use equation x^2+x^2 = (4r)^2
and I got x =0.404404 but the answer is 0.41,  but if we consider into 2 sig figures it should be 0.40 right?

For 2), Valu = (4/3)*(22/7)*0.143*0.143*0.143
          mass = 27, then density = 27/Valu and I got 2700
what is my mistake?

Offline sjb

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Re: Unit cell
« Reply #1 on: June 17, 2018, 10:17:53 AM »
Both seem fine - for 2 perhaps consider units?

Offline Enthalpy

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Re: Unit cell
« Reply #2 on: June 18, 2018, 08:08:38 AM »
Not an answer to the original question, rather an observation:

Atomic and ionic radii have half a dozen different definitions with discrepancies like 30%. This results logically from atoms having a fuzzy limit, and this limit changing with the chemical bonds. For instance, LiH packs more H atoms per volume unit than pure solid H2.

So if you get a reasonable density from a computation based on atomic radii, it means that said radius was deduced from the density, or tweaked for the purpose of an assignment.

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