What Went Wrong with My Calculation for Iridium Atomic Radii?

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SUMMARY

The discussion centers on the calculation of the atomic radii of iridium, specifically within a face-centered cubic (FCC) unit cell with an edge length of 3.833 Å. The initial calculation used the Pythagorean theorem incorrectly, leading to an incorrect radius of 2.710 Å instead of the correct value of 1.355 Å. The error was identified as a misunderstanding of how many atomic radii fit into the face diagonal, which requires dividing by 2 due to the presence of two atoms along that diagonal. The correct approach confirms that the atomic radius of iridium is indeed 1.355 Å.

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Taryn
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hey just wondering where I went wrong

Iridium Crystallizes in a face centred cubic unit call that has an edge length of 3.833A
The atom in the cantre of the face is in contact with the corner atoms

calculate the atomic radii of the irridium atom
I did this
using pythag, (3.833^2)+(3.833^2)=c^2
b^2= 29.384
b=5.42
Then I divided it by 2 as it was the radius and got 2.710
But I am assuming that this is wrong due to the qu asking for the atomic radii. So I divide it by 192.272 and I still get the wrong answer.
The answer is meant to be 1.355Angstrom

So if anyone could tell me where I went wrong I would be much appreciative
Thanks
 
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Hello Taryn,

try to revisualise one face. How many atom-radii will fit into the length of the face-diagonal? You are close to the correct answer. :smile:

Regards,

nazzard
 
Last edited:
figured it out, u divide by 2 as the atomic radii is being asked for and there are two atoms. So simple!
 

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