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An element has a face-centred cubic (fcc) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is :
Option 1)
$\frac{a}{2}$
Option 2)
a
Option 3)
$\sqrt{2a}$
Option 4)
$\frac{3}{2}a$

Answers (1)

best_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 -

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

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

$b=4r.$

$a=2\sqrt{2}r.$

-

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 $\frac{a}{2}$

Option 1)
$\frac{a}{2}$
Option 2)

a

Option 3)

$\sqrt{2a}$

Option 4)

$\frac{3}{2}a$

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