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

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SUMMARY

The atomic radius of barium (Ba) in a body-centered cubic (BCC) lattice is calculated using the formula A = 4r/√3, where A is the unit cell edge length. Given a unit cell edge length of 502 pm, the correct atomic radius is determined to be 217 pm. The initial calculation using the Pythagorean theorem was incorrect, leading to an erroneous radius of 177 pm. The correct approach confirms that the atomic radius of Ba in this lattice structure is indeed 217 pm.

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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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