Q. 7.58 - An Introduction to Thermal Physics

Question

The speed of sound in copper is $3560 m/s$. Use this value to calculate its theoretical Debye temperature. Then determine the experimental Debye temperature from Figure 7.28, and compare.

Step-by-Step Solution

Answer

Hence, the  Debye Temperature from experimental Data (graph) $T_D=359.9176 \mathrm{K}$.

Step 1. Given information

We have,$\text{ the speed of sound in copper is }3560 \mathrm{m}/\mathrm{s}$

$\text{ The Debye temperature is }{T}_D=\frac{h{C}_s}{2k_B}·{\left(\frac{6N}{\pi V}\right)}^{\frac{1}{3}}$.

Step 2. Putting the value of h   ,   V   ,   N   , C S   , k B .

$V=7.11{ \mathrm{cm}}^3/\mathrm{mol}$

$C_s=3560 \mathrm{m}/\mathrm{s}=3560\times 100 \mathrm{cm}/\mathrm{s}$

$N=6.022\times {10}^{23}\text{ atoms }/\mathrm{mol}$

$k_B=1.381\times {10}^{-23} \mathrm{J}/\mathrm{K}$

$h=6.626\times {10}^{-34} \mathrm{J}·\mathrm{s}$

substituting all the above value in Debye Temperature we get

$T_D=\frac{6.626\times {10}^{-34} \mathrm{J}·\mathrm{s}\times 3560\times 100 \mathrm{cm}/\mathrm{s}}{\left(2\times 1.381\times {10}^{-23} \mathrm{J}/\mathrm{K}\right)}{\left[\frac{6\times 6.022\times {10}^{23}/\mathrm{mol}}{\pi \times 7.11{ \mathrm{cm}}^3/\mathrm{mol}}\right]}^{\frac{1}{3}}$

$\therefore T_D=465.3381 \mathrm{K}$

Step 3. Calculating the experimental Debye Temperature.

Take any two points on experimental data calculate the slope [Approximately] Approximately take two points

Approximately take two points 

$A(0,0.75) B(18,1.5)$

$\frac{\Delta \left(\frac{C_V}{T}\right)}{\Delta \left(T^2\right)}=\frac{1.5-0.75}{18-0}$

$=0.0417$

Step 4. Finding the slope of the graph.

$\text{ Slope: }\frac{\Delta \left(\frac{C_V}{T}\right)}{\Delta \left(T^2\right)}=\frac{12{\pi }^{4}N{k}_B}{5T_D^3}$

$\frac{12{\pi }^{4}N{k}_B}{5T_D^3}=0.0417 \mathrm{mJ}/{\mathrm{K}}^4$

then , the Debye Temperature

$T_D={\left[\frac{12{\pi }^{4}N{k}_B}{5\times 0.0417\times {10}^{-3} \mathrm{J}/{\mathrm{K}}^4}\right]}^{\frac{1}{3}}$

     $={\left(\frac{12{\pi }^4\times 6.022\times {10}^{23}\times 1.381\times {10}^{-23}}{5\times 0.0417\times {10}^{-3}}\right)}^{\frac{1}{3}}$

     $=359.9176 \mathrm{K}$