Chemistry - Solid State Question with Solution | TestHub

ChemistrySolid StateCubic CrystalsMedium2 minQB

ChemistryMediummultiple choice

The HCP and CCP structure of a given element. (Given radius of element is same in both structures)

Answer:

A, B, C, D

Solution:

$d_{FCC} = \frac{4 M}{N _{A} a ^{3}} = \frac{4 M}{N _{A} ( \frac{4 R}{2} ) ^{3}}$

$d_{HCP} = \frac{6 M}{N _{A} 24 2 r ^{3}}$

Both are same.

Stream:JEESubject:ChemistryTopic:Solid StateSubtopic:Cubic Crystals

⏱ 2mℹ️ Source: QB

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