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)

 

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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