What is the maximum number of hexagonal layers of atom in close packed planes that will lie within two imaginary parallel planes having a distance between them of 3 12 2 R in a copper crystal (fcc)? The atoms to be within the parallel considered planes if their center are on or with the two parallel planes.

The distance between two consecutive close-packed (hexagonal) layers in FCC is: d = 3 2 2 R Given the distance between the parallel planes is: D = 3 12 2 R = 6 d If the first and last layer centers lie on the two planes, then the number of layers is: n = 6 + 1 = 7

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ChemistrySolid StatePacking in lonic SolidsMedium2 minQB

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What is the maximum number of hexagonal layers of atom in close packed planes that will lie within two imaginary parallel planes having a distance between them of $\frac{12 2 R}{3}$ in a copper crystal (fcc)? The atoms to be within the parallel considered planes if their center are on or with the two parallel planes.

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Answer:

Solution:

The distance between two consecutive close-packed (hexagonal) layers in FCC is:

$d = \frac{2 2 R}{3}$

Given the distance between the parallel planes is:

$D = \frac{12 2 R}{3} = 6 d$

If the first and last layer centers lie on the two planes, then the number of layers is:

$n = 6 + 1 = 7$

Stream:JEESubject:ChemistryTopic:Solid StateSubtopic:Packing in lonic Solids

⏱ 2mℹ️ Source: QB

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