Answer
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Hint:Packing efficiency is the percentage of total space covered by the particles, it explains how strong solid packing is present in the particles. In simple cubic and body centred cube the fraction of voids are higher due to less efficient packing.
Complete step by step answer:
Solids consist of particles which are arranged in some units, such that repeating arrangement of units will lead to solid formation. In all these structures, some voids are created as all the particles are of similar volumes and sizes.
Thus, the percentage of total space covered by the particles is defined as packing efficiency or fraction (PE $\% $). It is this number which determines how close or effective the solid is packed. It is given by the formula as follows:
Packing Efficiency $ = \dfrac{{Volume\;occupied\;by\;the\;atom\operatorname{s} \;in\;a\;unit\;cell}}{{Total\;volume\;of\;unit\;cell}} \times 100$
In this equation, space filled by the particles may be written as the product of the number of particles per unit cell (N) and volume of each particle. Also, space present in the unit cell of side ‘a’ then volume of the solid cube
Volume (V) $ = $ ${{{a}}^{{3}}}$
If particles are assumed to be hard spheres with radius r then volume of particle will be
Volume (V) $ = $ $\dfrac{{{4}}}{{{3}}}{{\pi }}{{{r}}^{{3}}}$
Now, we can write the packing efficiency as follows:
Packing efficiency $ = $ $\dfrac{{{{N \times }}\dfrac{{{4}}}{{{3}}}{{\pi }}{{{r}}^{{3}}}}}{{{{{a}}^{{3}}}}}{{ \times 100}}$
Now, for different types of arrangement in cubic system, number of particles present in a unit cell and volume of the unit cell are as follows:
For simple cube, as shown in figure, N $ = $ 2 and a $ = $ 2r
For bcc, N $ = $ 2, and $\sqrt {{3}} {{a = 4r}}$
For fcc, N $ = $ 4 and $\sqrt {{2}} {{a = 4r}}$
Thus, substituting the above values in equation of PE $\% $, we obtain as follows:
Simple cube, PE $\% $=54 $\% $
Body centred cube, PE $\% $= 68 $\% $
Face centred cube, PE $\% $ =74 $\% $
Hexagonal close packing, PE $\% $=74 $\% $
Thus, the correct answer is (B) hcp and ccp structures have the efficient close packing.
Note:
Determine the number of particles in the unit cell with the help of contributions of each particle in the cube.
Always remember to substitute the value of the side of the unit cell ‘a’ in terms of radius of sphere ‘r’.
The voids are classified into two types namely octahedral and tetrahedral voids.
Complete step by step answer:
Solids consist of particles which are arranged in some units, such that repeating arrangement of units will lead to solid formation. In all these structures, some voids are created as all the particles are of similar volumes and sizes.
Thus, the percentage of total space covered by the particles is defined as packing efficiency or fraction (PE $\% $). It is this number which determines how close or effective the solid is packed. It is given by the formula as follows:
Packing Efficiency $ = \dfrac{{Volume\;occupied\;by\;the\;atom\operatorname{s} \;in\;a\;unit\;cell}}{{Total\;volume\;of\;unit\;cell}} \times 100$
In this equation, space filled by the particles may be written as the product of the number of particles per unit cell (N) and volume of each particle. Also, space present in the unit cell of side ‘a’ then volume of the solid cube
Volume (V) $ = $ ${{{a}}^{{3}}}$
If particles are assumed to be hard spheres with radius r then volume of particle will be
Volume (V) $ = $ $\dfrac{{{4}}}{{{3}}}{{\pi }}{{{r}}^{{3}}}$
Now, we can write the packing efficiency as follows:
Packing efficiency $ = $ $\dfrac{{{{N \times }}\dfrac{{{4}}}{{{3}}}{{\pi }}{{{r}}^{{3}}}}}{{{{{a}}^{{3}}}}}{{ \times 100}}$
Now, for different types of arrangement in cubic system, number of particles present in a unit cell and volume of the unit cell are as follows:
For simple cube, as shown in figure, N $ = $ 2 and a $ = $ 2r
For bcc, N $ = $ 2, and $\sqrt {{3}} {{a = 4r}}$
For fcc, N $ = $ 4 and $\sqrt {{2}} {{a = 4r}}$
![seo images](https://www.vedantu.com/question-sets/7160c134-0587-4c20-a886-9b8206161f70560892420654255461.png)
![seo images](https://www.vedantu.com/question-sets/788abe5a-3afe-41fb-8d48-a696a24196612140892038607066096.png)
![seo images](https://www.vedantu.com/question-sets/e1594f97-183c-41e1-9768-5594d09b14674853202862480653864.png)
Thus, substituting the above values in equation of PE $\% $, we obtain as follows:
Simple cube, PE $\% $=54 $\% $
Body centred cube, PE $\% $= 68 $\% $
Face centred cube, PE $\% $ =74 $\% $
Hexagonal close packing, PE $\% $=74 $\% $
Thus, the correct answer is (B) hcp and ccp structures have the efficient close packing.
Note:
Determine the number of particles in the unit cell with the help of contributions of each particle in the cube.
Always remember to substitute the value of the side of the unit cell ‘a’ in terms of radius of sphere ‘r’.
The voids are classified into two types namely octahedral and tetrahedral voids.
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