Chapter 11, Problem 11.88QE

(a)

Interpretation Introduction

Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

Answer to Problem 11.88QE

The spacing between the layers in crystal of $\text{KI}$ is $3.526\times {10}^{-10}\text{m}$.

Explanation of Solution

Wavelength of X-ray is given as $1.790\times {10}^{-10}\text{m}$.  The angle of diffraction is given as $14.7$ degrees.

Bragg equation can be used to determine the spacing between the layers.  Bragg equation is given as shown below;

    $d=\frac{n\lambda }{2\sin \theta }$        (1)

Where,

    $n$ is given as 1.

    $d$ is the distance between layers.

    $\lambda$ is the wavelength of X-ray.

    $\theta$ is the angle of diffraction.

Substituting the values in equation (1), the spacing between the layers can be calculated as shown below;

    $\begin{array}{c}d=\frac{1\times 1.790\times {10}^{-10}\text{m}}{2(\sin 14.7°)}\\ =\frac{1.790\times {10}^{-10}\text{m}}{2(0.2538)}\\ =3.526\times {10}^{-10}\text{m}\end{array}$

Therefore, the spacing between layers is $3.526\times {10}^{-10}\text{m}$.

(b)

Interpretation Introduction

Interpretation:

Density of $\text{KI}$ has to be calculated.

Answer to Problem 11.88QE

Density of potassium iodide is $3.144{\text{g/cm}}^{\text{3}}$.

Explanation of Solution

The crystal is of $\text{KI}$.  The spacing between the layers is $3.526\times {10}^{-10}\text{m}$.  The edge length of the cube is given as twice of the length of spacing between the layers.  Therefore, the edge length is calculated as follows;

    $\begin{array}{c}\text{Edge}\text{length}=2\times 3.526\times {10}^{-10}\text{m}\\ =7.052\times {10}^{-10}\text{m}\end{array}$

The unit cell of potassium iodide is face centered cubic cell.  Therefore, there are eight corners and thus one atom is shared by eight other unit cells that are present.  One atom is present at the center of each face of the cell.  Therefore, the total number of atoms present is calculated as shown below;

    $6\text{faces}\times \frac{\text{1}\text{atom}}{\text{2}\text{faces}}+8\text{corners}\times \frac{\text{1}\text{atom}}{\text{8}\text{corners}}\text{ = 4}\text{atoms}$

The volume of the cubic cell can be calculated using the edge length.  This can be done as shown below;

    $\begin{array}{c}\text{Volume}={(\text{edge}\text{length})}^3\\ ={(7.052\times {10}^{-10}\text{m})}^3\\ =350.70\times {10}^{-30}{\text{m}}^3\end{array}$

Volume of cubic cell can be converted into centimeters as shown below using conversion factors;

    $\begin{array}{c}350.70\times {10}^{-30}{\text{m}}^3=350.70\times {10}^{-30}{\text{m}}^3\times \frac{100{\text{cm}}^{\text{3}}}{1{\text{m}}^{\text{3}}}\\ =3.507\times {10}^{-22}{\text{cm}}^{\text{3}}\end{array}$

Molar mass of $\text{KI}$ is $\text{166}\text{.00}\text{g/mol}$.  Avogadro’s number is $6.022\times {10}^{23}\text{atoms}$.  Therefore, the mass of $\text{KI}$ can be calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{of}\text{crystal}=\text{number}\text{of}\text{atoms×}\frac{\text{1}}{\text{Avogadro's}\text{number}}\text{×}\text{molar}\text{mass}\\ \text{= 4}\times \frac{1}{6.022\times {10}^{23}\text{atoms}}\times 166.00\text{g/mol}\\ \text{= 110}\text{.26}\times {\text{10}}^{\text{-23}}\text{g}\end{array}$

Therefore, mass of $\text{KI}$ crystal is $\text{110}\text{.26}\times {\text{10}}^{\text{-23}}\text{g}$.

Density of the potassium iodide crystal can be calculated using the density formula as follows;

    $\begin{array}{c}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ =\frac{\text{110}\text{.26}\times {\text{10}}^{\text{-23}}\text{g}}{3.507\times {10}^{-22}{\text{cm}}^{\text{3}}}\\ =31.44\times {10}^{-1}{\text{g/cm}}^3\\ =3.144{\text{g/cm}}^{\text{3}}\end{array}$

Thus, density of potassium iodide is calculated as $3.144{\text{g/cm}}^{\text{3}}$.