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(A) Sodium crystallizes in a body-centered cubic $(BCC)$ structure.In a $BCC$ unit cell,the body-centered atom lies at the center of the body diagonal

Vedclass · Accepted Answer

$3.66$

Vedclass · Answer

(A) Sodium crystallizes in a body-centered cubic $(BCC)$ structure.In a $BCC$ unit cell,the body-centered atom lies at the center of the body diagonal.The length of the body diagonal is $\sqrt{3}a$,where $a$ is the edge length of the cube.The distance between the body-centered atom and a corner atom is half of the body diagonal.Distance $= \frac{\sqrt{3}a}{2}$Given $a = 4.225 \, \mathring{A}$.Distance $= \frac{\sqrt{3} 	imes 4.225}{2} \approx \frac{1.732 	imes 4.225}{2} \approx \frac{7.3177}{2} \approx 3.6588 \, \mathring{A} \approx 3.66 \, \mathring{A}$.

The distance between the body-centered atom and a corner atom in sodium $(a = 4.225 \, \mathring{A})$ is ....... $\mathring{A}$ $[AIPMT \, 1995]$

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