In a body-centered cubic (BCC) unit cell of a | vedclass

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Question

(A) For a body-centered cubic $(BCC)$ unit cell,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $\sqrt{3} a = 4

Vedclass · Accepted Answer

$3.46 	imes 10^{-10} \ m$

Vedclass · Answer

(A) For a body-centered cubic $(BCC)$ unit cell,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $\sqrt{3} a = 4r$. The diameter of the atom $(d)$ is equal to $2r$. From the relation $\sqrt{3} a = 4r$,we can write $2r = \frac{\sqrt{3}}{2} a$. Given $a = 4 	imes 10^{-10} \ m$,the diameter $d = 2r = \frac{\sqrt{3}}{2} 	imes 4 	imes 10^{-10} \ m$. $d = 1.732 	imes 2 	imes 10^{-10} \ m = 3.464 	imes 10^{-10} \ m$.

In a body-centered cubic $(BCC)$ unit cell of a metal,the edge length is $4 \times 10^{-10} \ m$. The diameter of the atom will be:

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