Question

An element has fcc structure with cell edge a. The distance between the centres of two nearest tetrahedral voids

Answer 1

Answer:

No. of atoms(z) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- wherein

Relation between radius of constituent particle, r and edge length, a for face centered cubic unit cell -

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AB^{2}=b^{2}=a^{2}+a^{2}.

b=4r.

a=2\sqrt{2}r.

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We know that, in fcc cell unit, the tetrahedral voids are located at body diagonal at a distance \frac{\sqrt{3}}{4}a from the corners.

Now, (Tv= tetrahedral voids)

Now, \triangle ACE & \triangle BCD are similar triangle,

\therefore \frac{AE}{AC}=\frac{BD}{BC}

\Rightarrow \frac{a\times 2}{\sqrt{3}a}=\frac{BD\times 4}{\sqrt{3}a}

\therefore BD=\frac{a}{2}

\therefore distance between two nearest tetrahedral voids is \frac{a}{2}

Explanation:

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