
The percentage of void space of a metallic element crystallising in a ABCABC type lattice pattern is:
Answer
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Hint:ABCABC type is a cubic close packing. It results in cubic close packing which resembles a face Centered cubic structure. It is an fcc lattice or cubic closest packed lattice.
Examples are, Silver, Copper Gold crystallizes in fcc lattices.
Complete step by step answer:
In fcc, the lattice points are at the eight corners of the unit cell plus the additional points at the centre of each face of the unit cell.
Number of atoms in the cubic unit cell will be,
$n = \theta \times \dfrac{1}{\theta } + 6 \times \dfrac{1}{2}\; = 1 = 3 = 4$
(Where, \[\theta \] is the number of corners and each atom at the corner. The number of faces of the cubic unit cell is 6 and each face share $\dfrac{1}{2}$ atoms in each unit cell)
So, \[n = 4\]
Also, atoms on face diagonals in FCC
\[4r = \surd 2a\] (where r is the radius of each atom)
$ \to r = \dfrac{a}{{2\sqrt 2 }}$
We know that the volume of cubic unit cell will be ${a^3}$ , as its side length is ‘a’
Total volume occupied $ = 4 \times \dfrac{4}{3}\pi {(r)^3} = 4 \times \dfrac{4}{3} \times \pi {a^3}16\sqrt 2 = \dfrac{{\pi {a^3}}}{{2\pi }}$
Volume occupied in percentage $ = (\dfrac{{\pi {a^3}}}{{3\sqrt 2 }}/{a^3}) \times 100 = 74\% $
Packaging efficiency \[ = 74\% \]
Percentage of void \[ = 26\% \]
Note: In cubic close-packed (ccp) or face-centred cubic (fcc) structure, the spheres of the third layer are not in alignment with either the first or the second layer. And when we name the first layer as A, second layer as B and the third layer is named as C, then this pattern of layer can be written as ABCABC…….
Examples are, Silver, Copper Gold crystallizes in fcc lattices.
Complete step by step answer:
In fcc, the lattice points are at the eight corners of the unit cell plus the additional points at the centre of each face of the unit cell.
Number of atoms in the cubic unit cell will be,
$n = \theta \times \dfrac{1}{\theta } + 6 \times \dfrac{1}{2}\; = 1 = 3 = 4$
(Where, \[\theta \] is the number of corners and each atom at the corner. The number of faces of the cubic unit cell is 6 and each face share $\dfrac{1}{2}$ atoms in each unit cell)
So, \[n = 4\]
Also, atoms on face diagonals in FCC
\[4r = \surd 2a\] (where r is the radius of each atom)
$ \to r = \dfrac{a}{{2\sqrt 2 }}$
We know that the volume of cubic unit cell will be ${a^3}$ , as its side length is ‘a’
Total volume occupied $ = 4 \times \dfrac{4}{3}\pi {(r)^3} = 4 \times \dfrac{4}{3} \times \pi {a^3}16\sqrt 2 = \dfrac{{\pi {a^3}}}{{2\pi }}$
Volume occupied in percentage $ = (\dfrac{{\pi {a^3}}}{{3\sqrt 2 }}/{a^3}) \times 100 = 74\% $
Packaging efficiency \[ = 74\% \]
Percentage of void \[ = 26\% \]
Note: In cubic close-packed (ccp) or face-centred cubic (fcc) structure, the spheres of the third layer are not in alignment with either the first or the second layer. And when we name the first layer as A, second layer as B and the third layer is named as C, then this pattern of layer can be written as ABCABC…….
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