(B) For an $fcc$ lattice, the atoms touch along the face diagonal.The relation between the edge length $(a)$ and the radius $(r)$ is given by $4r = a

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(B) For an $fcc$ lattice, the atoms touch along the face diagonal.The relation between the edge length $(a)$ and the radius $(r)$ is given by $4r = a \sqrt{2}$.The diameter $(d)$ of the atom is $2r$.Therefore, $d = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.Given $a = 407 \ pm$, we have $d = \frac{407}{1.414} \approx 287.8 \ pm$.

Class 12 Chemistry Solid State Crystal structure and Coordination number

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Gold has a face-centred cubic $(fcc)$ lattice with an edge length of the unit cube of $407 \ pm$. Assuming the closest packing, the diameter of the gold atom is .............. $pm$.

AIIMS 2011 All AIIMS

A
$576.6$
B
$287.8$
C
$352.5$
D
$704.9$

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Solid State

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Crystal structure and Coordination number

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AIIMS 2011 Paper All AIIMS years →

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Gold has a face-centred cubic $(fcc)$ lattice with an edge length of the unit cube of $407 \ pm$. Assuming the closest packing, the diameter of the gold atom is .............. $pm$.

Similar Questions

In zinc blende structure,the coordination number of $Zn^{2+}$ ion is

In an end-centered unit cell,the body diagonal contains two additional atoms. What is the total number of atoms present in the unit cell?

In a body-centered cubic $(BCC)$ unit cell,the atom at the body center is shared by how many unit cells?

$A$ compound is formed by elements $A$ and $B$. Atoms of $A$ are present at the corners of the cube,and atoms of $B$ are present at the body center. What is the formula of the compound?

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