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$\kappa$-(BEDT-TTF)$_2$Cu(NCS)$_2$ --- with chemical formula C$_{22}$H$_{16}$N$_{2}$S$_{18}$Cu$_{1}$ \cite{Urayama_1988} --- has a molar mass $m_{mole}$ of $832.98 \textrm{ g/mole}$.     The predominant stable isotope of hydrogen is ${}^{\ 1}_{\ 1}\textrm{H}$, which has a nuclear spin $I = 1/2$ (in units of $\hbar$) and a nuclear magnetic moment $\mu = 2.79278$ (in units of nuclear magnetons), resulting in a nuclear g-factor $g =\mu/I = 5.586$ \cite{Fuller_1976}. This can lead to a significant contribution to the magnetic field dependent nuclear specific heat at dilution refrigerator temperatures. \footnote{One way to correct for this term in H-rich organic conductors is to repeat the measurements using samples containing deuterated hydrogen ${}^{\ 2}_{\ 1}\textrm{H}$. This has a nuclear spin $I = 1$ and a nuclear magnetic moment $\mu = 0.85742$, resulting in a much smaller nuclear g-factor $g =I/\mu = 0.857$.}   ${}^{\ 63}_{\ 29}\textrm{Cu}$ and ${}^{\ 65}_{\ 29}\textrm{Cu}$ are the two most common stable Cu isotopes. ${}^{\ 63}_{\ 29}\textrm{Cu}$ has a nuclear spin $I = 3/2$ and a nuclear magnetic moment $\mu = 2.2228$ while ${}^{\ 65}_{\ 29}\textrm{Cu}$ has a nuclear spin $I = 3/2$ and a nuclear magnetic moment $\mu = 2.3812$ \cite{Fuller_1976}. Here, $\mu_N$ is the nuclear magneton, $g = \mu / I$ and the resulting energy splitting in applied field is $\Delta = g \mu_n H / k \textrm{ [K]}$.  Substituting $g_N = \mathrm{1.5}$ --- the weighted average nuclear g-factor for these two most common Cu isotopes \cite{Leyarovski_1988}--- into Eq.~\ref{eq:CurieConstant} gives a theoretical value $\lambda_{\mathrm{Cu}} = 3.93 \cdot 10^{-12} \textrm{ K}{\textrm{ m}}^3 {\textrm{ mol}}^{-1}$.Experimentally, \cite{Leyarovski_1988} find a value of $\lambda_{\mathrm{Cu}} = 4.03 \cdot 10^{-12} \textrm{ K}{\textrm{ m}}^3 {\textrm{ mol}}^{-1}$ upon fitting Eq.~\ref{eq:SchottkyTail} to their measurements of the specific heat of Cu taken between 0.3 K and 1 K in an applied field of 14 T. Similarly, the corresponding Curie Constant for H is $\lambda_{\mathrm{Cu}} = 10.91 \cdot 10^{-12} \textrm{ K}{\textrm{ m}}^3 {\textrm{ mol}}^{-1}$.   In our Jan 2015 heat capacity run, we used a 0.40 mg sample of $\kappa$-(BEDT-TTF)$_2$Cu(NCS)$_2$. In our July 2015 run, we used a 0.50 mg sample. The nuclear contribution of the Cu and H nuclei to the measured heat capacity of $\kappa$-(BEDT-TTF)$_2$Cu(NCS)$_2$ is predicted to be