How Is the Atomic Radius of Barium Calculated in a BCC Lattice?

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The discussion focuses on calculating the atomic radius of barium in a body-centered cubic (BCC) lattice with a unit cell edge length of 502 pm. Initially, the user incorrectly applied the Pythagorean theorem, leading to an atomic radius calculation of 177 pm. Upon realizing the mistake, they corrected their approach using the formula A = 4r/√3, which accurately yielded the atomic radius of 217 pm. The correct answer aligns with the answer key, confirming that the atomic radius of barium in this lattice structure is indeed 217 pm. The importance of using the appropriate formula for BCC lattices is emphasized.
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Homework Statement


Barium metal crystallizes in a body centered cubic lattice. The unit cell edge length is 502pm. Calculate the atomic radius of Ba.

a) 251 pm
b) 217 pm
c) 177 pm
d) 125 pm
e) 63 pm


Homework Equations


Pythagorean Theorem


The Attempt at a Solution


A^2 + B^2 = C^2
C^2 = 2A^2
C^2 = 2(502 pm)^2
C^2 = 504008 pm
C = 709.94

C = 4r
r = C/4
r = 709.94 / 4
r = 177 pm, choice C



But my answer key says the correct answer is B, 217pm. What am I doing wrong?
 
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Ah... never mind. I was using the wrong equation. :)

A = 4r/√3
4r = (502 pm)(√3)
4r = 869.5 pm
r = 217 pmThanks anyway!
 
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