A metal crystalizes in a BCC lattice with edge of unit cell 10 3 A ˚ . Then distance in A ˚ between two nearest neighbours is?

The distance between two nearest neighbors in a BCC lattice is given by 2 3 a , where a is the edge length of the unit cell. Given a = 10 3 Å. Distance = 2 3 × 10 3 Å Distance = 2 3 × 10 Å Distance = 2 30 Å Distance = 15 Å.

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ChemistrySolid StateCubic CrystalsMedium2 minQB

ChemistryMediuminteger

A metal crystalizes in a BCC lattice with edge of unit cell $103 \overset{˚}{A}$ . Then distance in $\overset{˚}{A}$ between two nearest neighbours is?

Answer:

Solution:

The distance between two nearest neighbors in a BCC lattice is given by $\frac{3 a}{2}$ , where $a$ is the edge length of the unit cell.

Given $a = 103$ Å.

Distance = $\frac{3 \times 10 3}{2}$ Å

Distance = $\frac{3 \times 10}{2}$ Å

Distance = $\frac{30}{2}$ Å

Distance = $15$ Å.

Stream:JEESubject:ChemistryTopic:Solid StateSubtopic:Cubic Crystals

⏱ 2mℹ️ Source: QB

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