Chapter 15, Problem 15.1IA

Interpretation Introduction

Interpretation:

Thermal expansion coefficient of diamond on heating from $100\text{}Kto300K$ has to be calculated.

Concept Introduction:

Thermal expansion is given by the following formula.

  $\alpha =\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_P$

Where, $\alpha$ is the expansion coefficient, $V$ is the initial volume, $\partial V$ is the change in volume and $\partial T$ is the change in temperature.  Also, pressure must be constant during expansion.

Bragg’s equation is given by the following formula.

  $n\lambda =2d\sin \theta$

Where,

    $n=$ order of reflection

    $\lambda =$ wavelength of the light

    $d=$ inter planar distance

    $\theta =$ angle of reflection

Inter planar distance can be calculated using the following formula.

  $d=\frac{1}{\sqrt{{\left(\frac{h}{a}\right)}^2+{\left(\frac{k}{b}\right)}^2+{\left(\frac{l}{c}\right)}^2}}$

Where, $a,bandc$ are edge lengths of the unit cell and $h,kandl$ are the miller indices of the planes.

Answer to Problem 15.1IA

The thermal expansion coefficient of diamond is calculated as $5.012\times 10^{-5}{K}^{-1}.$

Explanation of Solution

Inter planar distances corresponds to the shift in reflections can be calculated using Bragg’s equation.

$d_1$ corresponds to ${\theta }_1$ can be calculated as follows,

Given, ${\theta }_{1}is{22.0404}^o$ and $\lambda is154.0562pm.$

  $\begin{array}{c}{d}_1=\frac{n\lambda }{2\sin {\theta }_1}\\ \\ =\frac{1\times 154.0562pm}{2\times \sin \left({22.0403}^o\right)}\\ \\ =205.2pm\end{array}$

$d_2$ corresponds to ${\theta }_2$ can be calculated as follows,

Given, ${\theta }_{2}is{21.9664}^o$ and $\lambda is154.0562pm.$

  $\begin{array}{c}{d}_2=\frac{n\lambda }{2\sin {\theta }_2}\\ \\ =\frac{1\times 154.0562pm}{2\times \sin \left({21.9664}^o\right)}\\ \\ =205.9pm\end{array}$

Diamond has a cubic unit cell so edge lengths will be same.

  $a=b=c$

Therefore the edge length $a_1$ corresponds to $d_1$ can be calculated as follows,

Given that $h=k=l=1.$

  $\begin{array}{c}{d}_1=\frac{1}{\sqrt{{\left(\frac{h}{a_1}\right)}^2+{\left(\frac{k}{a_1}\right)}^2+{\left(\frac{l}{a_1}\right)}^2}}\\ \\ =\frac{a_1}{\sqrt{1+1+1}}\\ \\ =\frac{a_1}{\sqrt{3}}\end{array}$

That is, $d_1=\frac{a_1}{\sqrt{3}}=205.2pm.$

Therefore, $a_1=205.2pm\times \sqrt{3}=355.4pm.$

Therefore the edge length $a_2$ corresponds to $d_2$ can be calculated as follows,

Given that $h=k=l=1.$

  $\begin{array}{c}{d}_2=\frac{1}{\sqrt{{\left(\frac{h}{a_2}\right)}^2+{\left(\frac{k}{a_2}\right)}^2+{\left(\frac{l}{a_2}\right)}^2}}\\ \\ =\frac{a_2}{\sqrt{1+1+1}}\\ \\ =\frac{a_2}{\sqrt{3}}\end{array}$

That is, $d_2=\frac{a_2}{\sqrt{3}}=205.9pm.$

Therefore, $a_2=205.9pm\times \sqrt{3}=356.6pm.$

So the initial volume $V_1$ can be calculated as follows,

  $\begin{array}{c}{V}_1=a_1^{}\times a_1^{}\times a_1^{}\\ \\ =355.4pm\times 355.4pm\times 355.4pm\\ \\ =4.489\times 10^{7}p{m}^3\end{array}$

And the final volume $V_2$ can be calculated as follows,

  $\begin{array}{c}{V}_2=a_2^{}\times a_2^{}\times a_2^{}\\ \\ =356.6pm\times 356.6pm\times 356.6pm\\ \\ =4.534\times 10^{7}p{m}^3\end{array}$

So the change in volume $\partial V$ can be calculated as below,

  $\begin{array}{c}\partial V=V_2-V_1\\ \\ =\text{4}{\text{.534×10}}^{\text{7}}{\text{pm}}^{\text{3}}-\text{ 4}{\text{.489×10}}^{\text{7}}{\text{pm}}^{\text{3}}\\ \\ =4.5\times 10^{5}p{m}^3\end{array}$

The change in temperature $\partial T$ can be calculated as below,

  $\begin{array}{c}\partial T=T_2-T_1\\ \\ =300K-100K\\ \\ =200K\end{array}$

Therefore the thermal expansion coefficient can be calculated by substituting these values in the following equation.

  $\begin{array}{c}\alpha =\frac{1}{V_1}{\left(\frac{\partial V}{\partial T}\right)}_P\\ \\ =\frac{1}{\text{4}{\text{.489×10}}^{\text{7}}\cancel{{\text{pm}}^{\text{3}}}}\left(\frac{\text{4}{\text{.5×10}}^{\text{5}}\cancel{{\text{pm}}^{\text{3}}}}{\text{200}\text{K}}\right)\\ \\ =5.012\times 10^{-5}{K}^{-1}.\end{array}$

The thermal expansion coefficient of diamond is calculated as $5.012\times 10^{-5}{K}^{-1}.$