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(D) In a face-centred cubic $(FCC)$ unit cell, the atoms touch along the face diagonal.The length of the face diagonal is $\sqrt{2}a$, where $a$ is th

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$609.6$

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(D) In a face-centred cubic $(FCC)$ unit cell, the atoms touch along the face diagonal.The length of the face diagonal is $\sqrt{2}a$, where $a$ is the edge length.The distance between the centers of two nearest neighbor atoms (shortest separation of nuclei) is equal to half of the face diagonal, which is $\frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.Given $a = 862 \, pm$.Shortest separation $= \frac{862}{1.414} \approx 609.6 \, pm$.

$A$ metal $X$ crystallises in a face-centred cubic arrangement with the edge length $862 \, pm$. What is the shortest separation of any two nuclei of the atom? ........... $\, pm$

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