91Ó°ÊÓ

The total heat capacity at low temperature is equal to the sum of the electronic heat capacity lattice vibrational heat capacity.

$\text{Here,}\mathrm{γ}=\frac{{\mathrm{Ï€}}^{2}N{k}_B^2}{2{\mathrm{Î\mu }}_F},\mathrm{Î\pm }=\frac{12N{\mathrm{Ï€}}^4{k}_B}{5T_D^3}\text{and}T\text{is the temperature.}$

At low temperature, the electronic contribution to the heat capacity is directly proportional to the temperature.

The contribution of lattice vibrations to the heat capacity has a cubic dependence on temperature at the lower temperature.

$\text{Firstly,rearranging the equation}C=\mathrm{γ}T+\mathrm{Î\pm }{T}^3\text{for}\frac{C}{T}\text{.}$

$\frac{C}{T}=\mathrm{γ}+\mathrm{Î\pm }{T}^2$

$\text{Here,}\mathrm{Î\pm }\text{is the slope on}\frac{C}{T}\text{versus}{T}^2\text{plot and}\mathrm{γ}\text{is the intercept.}$

role="math" localid="1647619231371" $\text{The slope of the graph between}\frac{C}{T}\text{versus}{T}^2\text{of the Copper is as follows:}$

$\mathrm{Î\pm }=\frac{0.9\mathrm{mJ}/{\mathrm{K}}^2}{18{\mathrm{K}}^2}$

$=5×{10}^{-5}\mathrm{J}/{\mathrm{K}}^4$

$\text{The intercept}\mathrm{γ}\text{for the graph between}\frac{C}{T}\text{versus}{T}^2\text{of the Copper is as follows:}$

$\mathrm{γ}=0.7\mathrm{mJ}/{\mathrm{K}}^2$

The temperature at which the electronic and the lattice vibration contributions of the heat capacities can be calculated by equating the electronic contribution of the heat capacity to the lattice vibration heat capacity.

$C_{\text{electronic}}=C_{\text{vibration}}$

$\mathrm{γ}T=\mathrm{Î\pm }{T}^3$

$T^2=\frac{\mathrm{γ}}{\mathrm{Î\pm }}$

$T=\sqrt{\frac{\mathrm{γ}}{\mathrm{Î\pm }}}$

$\text{Substitute}0.7\mathrm{mJ}/{\mathrm{K}}^2\text{for}\mathrm{γ}\text{and}5×{10}^{-5}\mathrm{J}/{\mathrm{K}}^4\text{for}\mathrm{Î\pm }\text{.}$

$T=\sqrt{\frac{0.7\mathrm{mJ}/{\mathrm{K}}^2\left(\frac{{10}^{-3}\mathrm{J}}{1\mathrm{mJ}}\right)}{5×{10}^{-5}\mathrm{J}/{\mathrm{K}}^4}}$

$=3.7\mathrm{K}$

$\text{At this temperature, both heat capacities are as follows:}$

$C_{\text{electronic}}=\mathrm{γ}T$

$\text{Substitute}0.7\mathrm{mJ}/{\mathrm{K}}^2\text{for}\mathrm{γ}\text{and}3.7\mathrm{K}\text{for}T$

$C_{\text{electronic}}=\left(\left(0.7\mathrm{mJ}/{\mathrm{K}}^2\right)\left(\frac{{10}^{-3}\mathrm{J}}{1\mathrm{mJ}}\right)\right)(3.7\mathrm{K})$

$=0.0026\mathrm{J}/\mathrm{K}$.

The table we have,

$\begin{array}{|lll|}\hline T(\text{in K)}& C_{\text{clectronic}}=\mathrm{γ}T& C_{\text{vibation}}=\mathrm{Î\pm }{T}^3\\ 0& 0& 0\\ 1& 0.0007& 0.00005\\ 2& 0.0014& 0.0004\\ 3& 0.0021& 0.00135\\ 4& 0.0028& 0.0032\\ 5& 0.0035& 0.00625\\ \hline\end{array}$

So, the required plot we have is shown below