deletions | additions       diff --git a/section_Application_to_Specific_Heat__.tex b/section_Application_to_Specific_Heat__.tex  index 87733f9..a1ae145 100644  --- a/section_Application_to_Specific_Heat__.tex  +++ b/section_Application_to_Specific_Heat__.tex 
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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} --- contains 16 H atoms and 1 Cu atom per mole.     \subsection{H nuclear magnetic moment}  The predominant stable isotope of hydrogen is ${}^{\ 1}_{\ 1}\textrm{H}$, which has a nuclear spin $I = 1/2$ and a nuclear magnetic moment $\mu = 2.79278 \mu_N$, 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 $\mu = 0.85742 \mu_N$, resulting in a much smaller nuclear g-factor $g =I/\mu = 0.857$.}  \subsection{Cu contribution} nuclear magnetic moment}  ${}^{\ 63}_{\ 29}\textrm{Cu}$ 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 $\mu = 2.2228 \mu_N$ while ${}^{\ 65}_{\ 29}\textrm{Cu}$ has a nuclear spin $I = 3/2$ and a nuclear magnetic moment $\mu = 2.3812 \mu_N$ \cite{Fuller_1976}. Here, $\mu_N$ is the nuclear magneton, $g = \mu / I$ and the resulting energy splitting in applied field is $\Delta = g \mu_n H / k$ (expressed in temperature units of Kelvin). Numerically, the nuclear magneton $\mu_N = \frac{h e}{4 \Pi M_p c} = 5.051 10^{-27} \textrm{ J/T}$ is a factor of 1836.1 smaller than the Bohr magneton $\mu_B$.   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.   \subsection{H contribution}  The predominant stable isotope of hydrogen is ${}^{\ 1}_{\ 1}\textrm{H}$, which has a nuclear spin $I = 1/2$ and a nuclear magnetic moment $\mu = 2.79278 \mu_N$, resulting in a nuclear g-factor $g =\mu/I = 5.586$ \cite{Fuller_1976}. This can lead \subsection{Contributions  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 $\mu = 0.85742 \mu_N$, resulting in a much smaller nuclear g-factor $g =I/\mu = 0.857$.}   \subsection{nuclear Schottky correction due to H and Cu} Heat Capacity}  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. Assuming a molar mass of $832.98 \mathrm{g/mole}$ and 1 Cu atom per mole, the nuclear contribution to the measured heat capacity of $\kappa$-(BEDT-TTF)$_2$Cu(NCS)$_2$ is predicted to be  \begin{equation}  \label{eq:NuclearCuContribution}