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Question

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

Vedclass · Accepted Answer

$7.951 	imes 10^{-23} \ cm^3$

Vedclass · Answer

(C) For a $bcc$ (body-centered cubic) unit cell,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3}a$. Therefore,$a = \frac{4r}{\sqrt{3}}$. Given $r = 1.86 	imes 10^{-8} \ cm$. $a = \frac{4 	imes 1.86 	imes 10^{-8}}{\sqrt{3}} \approx \frac{7.44 	imes 10^{-8}}{1.732} \approx 4.2956 	imes 10^{-8} \ cm$. The volume of the unit cell $(V)$ is $a^3$. $V = (4.2956 	imes 10^{-8} \ cm)^3 \approx 79.28 	imes 10^{-24} \ cm^3 = 7.928 	imes 10^{-23} \ cm^3$. Rounding to the nearest option,the correct answer is $7.951 	imes 10^{-23} \ cm^3$.

Calculate the volume of a $bcc$ unit cell if the radius of an atom present in it is $1.86 \times 10^{-8} \ cm$.

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