Lithium forms BCC structure having an edge le | vedclass

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Question

(B) For a $BCC$ (Body-Centered Cubic) unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $

Vedclass · Accepted Answer

$152$

Vedclass · Answer

(B) For a $BCC$ (Body-Centered Cubic) unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $\sqrt{3} a = 4 r$. Given that $a = 351 \ pm$, we can calculate $r$ as follows: $r = \frac{\sqrt{3} 	imes 351}{4} \approx \frac{1.732 	imes 351}{4} \approx 151.98 \ pm$. Rounding this value gives $r \approx 152 \ pm$.

Lithium forms $BCC$ structure having an edge length of unit cell $351 \ pm$. What is the atomic radius of lithium (in $pm$)?

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