The edge length of a diomond unit cell is ( 2 3 = 1.225 )

Chemistry - Solid State Question with Solution | TestHub

ChemistrySolid StateCubic CrystalsMedium2 minQB

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Passage / Comprehension

In a diamond structure, carbon atoms form an FCC lattice and also occupy 50% of the tetrahedral voids.

The carbon-carbon bond length is 173.2 nm. (Use $N_{A} = 6 \times 1 0^{23}$ , $3 = 1.732$ , $2 = 1.414$ )

The edge length of a diomond unit cell is $(\frac{3}{2} = 1.225)$

Answer:

Solution:

In a diamond structure, the bond length is given by $d = \frac{3}{4} a$ .

Given the bond length $d = 173.2 p m$ .

Therefore, we can calculate $a$ as:

$a = \frac{4 \times 173.2}{3}$

$a = \frac{4 \times 173.2}{1.732}$

Solving this gives $a \approx 400 p m$ .

Stream:JEESubject:ChemistryTopic:Solid StateSubtopic:Cubic Crystals

⏱ 2mℹ️ Source: QB

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