Copper crystallises in a face-centred cubic l | vedclass

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(D) In a face-centred cubic $(fcc)$ lattice, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $r = \frac{\sqrt{2

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(D) In a face-centred cubic $(fcc)$ lattice, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $r = \frac{\sqrt{2}a}{4}$.Given, $a = 361 	ext{ pm}$.Substituting the value of $a$ in the formula:$r = \frac{\sqrt{2} 	imes 361}{4} \approx \frac{1.414 	imes 361}{4} \approx 127.6 	ext{ pm}$.Rounding to the nearest integer, the radius is $128 	ext{ pm}$.

Copper crystallises in a face-centred cubic lattice with a unit cell length of $361 \text{ pm}$. What is the radius of copper atom in $\text{pm}$?

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