Calculate the packing efficiency in a simple cubic metal crystal.
(N/A) Simple cubic lattice:
In a simple cubic lattice,the particles are located only at the corners of the cube and touch each other along the edge.
Let the edge length of the cube be $a$ and the radius of each particle be $r$.
So,we can write: $a = 2r$.
Now,volume of the cubic unit cell $= a^3 = (2r)^3 = 8r^3$.
We know that the number of particles per unit cell is $1$.
Therefore,volume of the occupied unit cell $= \frac{4}{3} \pi r^3$.
Hence,packing efficiency $= \frac{\text{Volume of one particle}}{\text{Volume of cubic unit cell}} \times 100 \%$.
Packing efficiency $= \frac{\frac{4}{3} \pi r^3}{8r^3} \times 100 \% = \frac{\pi}{6} \times 100 \%$.
Using $\pi \approx 3.14159$,packing efficiency $\approx \frac{3.14159}{6} \times 100 \% \approx 52.36 \% \approx 52.4 \%$.
In a simple cubic lattice,the particles are located only at the corners of the cube and touch each other along the edge.
Let the edge length of the cube be $a$ and the radius of each particle be $r$.
So,we can write: $a = 2r$.
Now,volume of the cubic unit cell $= a^3 = (2r)^3 = 8r^3$.
We know that the number of particles per unit cell is $1$.
Therefore,volume of the occupied unit cell $= \frac{4}{3} \pi r^3$.
Hence,packing efficiency $= \frac{\text{Volume of one particle}}{\text{Volume of cubic unit cell}} \times 100 \%$.
Packing efficiency $= \frac{\frac{4}{3} \pi r^3}{8r^3} \times 100 \% = \frac{\pi}{6} \times 100 \%$.
Using $\pi \approx 3.14159$,packing efficiency $\approx \frac{3.14159}{6} \times 100 \% \approx 52.36 \% \approx 52.4 \%$.
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Similar Questions
In hexagonal systems of crystals,a frequently encountered arrangement of atoms is described as a hexagonal prism. Here,the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. $A$ space-filling model of this structure,called hexagonal close-packed $(HCP)$,is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally,the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be $r$.
$1.$ The number of atoms on this $HCP$ unit cell is
$(A)$ $4$ $(B)$ $6$ $(C)$ $12$ $(D)$ $17$
$2.$ The volume of this $HCP$ unit cell is
$(A)$ $24 \sqrt{2} r^3$ $(B)$ $16 \sqrt{2} r^3$
$(C)$ $12 \sqrt{2} r^3$ $(D)$ $\frac{64 r^3}{3 \sqrt{3}}$
$3.$ The empty space in this $HCP$ unit cell is
$(A)$ $74 \%$ $(B)$ $47.6 \%$ $(C)$ $32 \%$ $(D)$ $26 \%$
Give the answer for questions $1, 2$ and $3.$
$1.$ The number of atoms on this $HCP$ unit cell is
$(A)$ $4$ $(B)$ $6$ $(C)$ $12$ $(D)$ $17$
$2.$ The volume of this $HCP$ unit cell is
$(A)$ $24 \sqrt{2} r^3$ $(B)$ $16 \sqrt{2} r^3$
$(C)$ $12 \sqrt{2} r^3$ $(D)$ $\frac{64 r^3}{3 \sqrt{3}}$
$3.$ The empty space in this $HCP$ unit cell is
$(A)$ $74 \%$ $(B)$ $47.6 \%$ $(C)$ $32 \%$ $(D)$ $26 \%$
Give the answer for questions $1, 2$ and $3.$
What is the total number of tetrahedral voids in $0.6 \ mole$ of a compound that forms a $hcp$ structure?
DifficultMHT CET 2024
View SolutionWhat type of crystal structure from the following has $52.36 \%$ packing efficiency?
Calculate the void volume in an $fcc$ unit cell if the total volume of the unit cell is $6.4 \times 10^{-23} \text{ cm}^3$.
DifficultMHT CET 2026
View SolutionWhat is the void volume in the crystal lattice formed by a $BCC$ unit cell (in $\%$)?
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