Chemistry - Solid State Question with Solution | TestHub

ChemistrySolid StateCubic CrystalsMedium2 minQB

ChemistryMediuminteger

An element A (molecular weight $80 gm / mole$ ) crystallize in BCC and FCC lattice. If density of FCC lattice is 4 $gm / cm^{3}$ and distance between next nearest neighbour in BCC is d than what is $32 \times 1 0^{7} d$ (in cm).

Answer:

Solution:

$FCC 4 r = a_{1} 2$

$a_{1} = \frac{4 r}{2}$

Stream:JEESubject:ChemistryTopic:Solid StateSubtopic:Cubic Crystals

⏱ 2mℹ️ Source: QB

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