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Chapter 10: Problem 63

What fraction of the total volume of a cubic closest packed structure is occupied by atoms? (Hint: \(V_{\text { sphere }}=\frac{4}{3} \pi r^{3} .\) ) What fraction of the total volume of a simple cubic structure is occupied by atoms? Compare the answers.

Short Answer

Expert verified
The fraction of the total volume occupied by atoms in a cubic closest packed structure is approximately 74% (\(\frac{\sqrt{2}}{1}\)), while in a simple cubic structure, it is 25% (\(\frac{1}{4}\)). A cubic closest packed structure occupies a higher volume compared to a simple cubic structure.

Step by step solution

01

Determine the number of atoms per unit cell in each structure.

For the cubic closest packed (ccp) structure, there are 4 atoms per unit cell. In the simple cubic structure, there is 1 atom per unit cell.
02

Calculate the total volume occupied by atoms.

In the ccp structure, the volume occupied by atoms is given by: \[V_{ccp} = 4 \times \frac{4}{3} \pi r^{3}\] In the simple cubic structure, the volume occupied by atoms is given by: \[V_{simple} = 1 \times \frac{4}{3} \pi r^{3}\]
03

Find the volume of the unit cell in each structure.

In the ccp structure, the edge length of the unit cell is \(2\sqrt{2}r\), and the volume of the unit cell is given by: \[V_{ccp\_cell}=(2\sqrt{2}r)^{3}\] In the simple cubic structure, the edge length of the unit cell is \(2r\), and the volume of the unit cell is given by: \[V_{simple\_cell}=(2r)^{3}\]
04

Calculate the fraction of the total volume occupied by atoms in each structure.

In the ccp structure, the fraction of the total volume occupied by atoms is given by: \[\frac{V_{ccp}}{V_{ccp\_cell}} = \frac{4\times\left( \frac{4}{3} \pi r^{3}\right)}{(2\sqrt{2}r)^{3}}\] In the simple cubic structure, the fraction of the total volume occupied by atoms is given by: \[\frac{V_{simple}}{V_{simple\_cell}} = \frac{1\times\left( \frac{4}{3} \pi r^{3}\right)}{(2r)^{3}}\]
05

Compare the answers and reduce the fractions.

The ccp and simple cubic structures have the following fractions of the total volume occupied by atoms: \[\frac{V_{ccp}}{V_{ccp\_cell}} = \frac{32\pi r^3}{(2\sqrt{2}r)^3} = \frac{32}{16\sqrt{2}} = \frac{2}{\sqrt{2}}= \frac{\sqrt{2}}{1} \approx 0.74 \] \[\frac{V_{simple}}{V_{simple\_cell}} = \frac{4\pi r^3}{(2r)^3} = \frac{1}{4} = 0.25\] Comparing the two answers, we can see that a cubic closest packed structure has a higher fraction of its total volume occupied by atoms (\(\approx\) 74%) than a simple cubic structure (25%).

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