Solveeit Logo

Question

Question: Percentage of empty space in a simple cubic unit cell- A. 52% B. 42% C. 48% D. 68%...

Percentage of empty space in a simple cubic unit cell-
A. 52%
B. 42%
C. 48%
D. 68%

Explanation

Solution

A simple or primitive cubic unit cell is the simplest repeating unit in a simple unit structure, and the empty space is the area in a cubic unit cell where no atom is present. We should first find out the effective number of atoms present.

Complete Step by step answer: To determine the total amount of empty space present in the simple cubic cell, we will first find out the total amount of space occupied by atoms in simple cubic cells.
For us to determine the total space occupied by atoms, we will use the formula of packing efficiency,
\Rightarrow Packing efficiency (PE{{\text{P}}_{\text{E}}})  = z×volume of an atom Total volume of the cubic cell{\text{ = }}\dfrac{{{\text{z}} \times {\text{volume of an atom }}}}{{{\text{Total volume of the cubic cell}}}}
\Rightarrow $$$${{\text{P}}_{\text{E}}}= 1 ×43πr3a3 \times \dfrac{{\dfrac{4}{3}\pi {{\text{r}}^3}}}{{{{\text{a}}^3}}} [z = 1, as there are 8×188 \times \dfrac{1}{8} atoms (1 atom at each corner of the cube) present in per unit cell of the simple cubic cell]
\Rightarrow $$$${{\text{P}}_{\text{E}}}= 4πr33(2r)3\dfrac{{4\pi {{\text{r}}^3}}}{{3{{(2{\text{r)}}}^3}}} [the relation between edge length ‘a’ and ‘r’ (radius of an atom) in a simple cubic cell is a = 2r]
\Rightarrow $$$${{\text{P}}_{\text{E}}} = 4×3.143×8\dfrac{{4 \times 3.14}}{{3 \times 8}}
\Rightarrow $$$${{\text{P}}_{\text{E}}} = 12.5624\dfrac{{12.56}}{{24}}
\Rightarrow $$$${{\text{P}}_{\text{E}}} = 0.5233 ×\times100 (to convert the result into percentage)
\Rightarrow $$$${{\text{P}}_{\text{E}}}= 52 % (taking a round off figure to ease our solution)
Therefore, from the above result, we can conclude that the packing efficiency in a simple cubic cell unit is 52%. And to find out the empty space, we just need to subtract this result from total space i.e., 100
\Rightarrow100 – 52 = 48 %

Hence, our required answer is option (c). i.e., 48%

Note: We are writing the total number of atoms per unit cell, and not the total number of atoms in the unit cell. For Eg:- FCC has a total number of 14 atoms( 8 in each corner of the cube and 6 in the face of cubes) as a whole, but only 4 atoms per unit cell.