Calorie, or more specifically, thermochemical calorie, is defined as an exact number of $4.184\ \texttt{joule}$, i.e. $$1\ \texttt{cal} \equiv 4.184\ \texttt{J}$$ The conversion between mass $m$ and energy $E$, is
$$ E = mc^2 $$ where $m$ is the mass, and light speed in vacuum $c$ is $299\ 792\ 458\ $ meter per second ($\texttt{m}\cdot \texttt{s}^{\texttt{-1}}$). In addition,
$$ E = h\nu = h\frac{c}{\lambda} $$
where $\nu$ is frequency (in the unit of hertz, $\texttt{Hz}$ or $\texttt{s}^{\texttt{-1}}$), $\lambda$ is wavelength, and $h$ is Planck constant, $6.626\ 070\ 15 \times 10^{-34}\ \texttt{J}\cdot \texttt{s}$.

The energy of $1\ \texttt{Hz}$ (or $1\ \texttt{s}^\texttt{-1}$), is then equivalent to $$ E = h\nu = ( 6.62607015 \times 10^{-34}\ \texttt{J}\cdot \texttt{s} ) \times ( 1\ \texttt{s}^\texttt{-1} ) = 6.62607015 \times 10^{-34}\ \texttt{J} = 6.62607015 \times 10^{-31}\ \texttt{kJ} $$ Note that the unit $\texttt{J}$ here, is the energy in the unit of $\texttt{joule per particle}$ (or $\texttt{ per atom}$, or $\texttt{ per molecule}$, depending on the system being described). Multiplied by the Avogadro's constant $N_A,\ 6.022\ 140\ 76 \times 10^{23}\ \texttt{mol}^\texttt{-1}$, we have the energy conversion of $1 \ \texttt{Hz} $ to $3.990\ 312\ 712 \times 10^{-10} \ \texttt{J}\cdot \texttt{mol}^{\texttt{-1}} $ ($\texttt{joule per mole}$) or $3.990\ 312\ 712 \times 10^{-13} \ \texttt{kJ}\cdot \texttt{mol}^{\texttt{-1}} $ ($\texttt{kilojoule per mole}$) as: $$ \begin{align} E & = (6.62607015 \times 10^{-34}\ \texttt{J}) \times (6.02214076 \times 10^{23}\ \texttt{mol}^\texttt{-1}) \\ & = 3.990312712 \times 10^{-10} \ \texttt{ J}\cdot \texttt{mol}^{\texttt{-1}} \\ & = 3.990312712 \times 10^{-13} \ \texttt{kJ}\cdot \texttt{mol}^{\texttt{-1}} \end{align} $$ Further divided by $4.184$, it is $\ 9.537\ 076\ 272 \times 10^{-14} \ \texttt{kcal}\cdot \texttt{mol}^{\texttt{-1}} $ ($\texttt{kilocalorie per mole}$).

The multiplicative inverse (reciprocal) of wavelength, $\lambda^{-1}$, is another frequently used equivalence to energy in atomic and molecular spectroscopy. The reciprocal of centimeter, $\texttt{cm}^\texttt{-1}$, is defined as $\texttt{wavenumber}$ and intepreted as number of waves per centimeter. In that sense, energy with $1\ \texttt{Hz}$ of frequency is equivalent to $3.335\ 640\ 951 \times 10^{-11}\ \texttt{cm}^{\texttt{-1}}$, by $$ E = h\nu = h\frac{c}{\lambda} $$ $$ \frac{1}{\lambda} = \frac{\nu}{c} = \frac{ 1\ \texttt{s}^{\texttt{-1}} }{ 299792458 \times 10^2\ \texttt{cm}\cdot \texttt{s}^{\texttt{-1}} } = 3.335640951 \times 10^{-11}\ \texttt{cm}^{\texttt{-1}}$$

The gas constant $R$, is $8.314\ 462\ 618\ \texttt{J}\cdot \texttt{mol}^{\texttt{-1}}\cdot \texttt{K}^{\texttt{-1}} $ (or $0.082\ 057\ 338\ \texttt{L}\cdot \texttt{atm}\cdot\texttt{mol}^{\texttt{-1}}\cdot \texttt{K}^{\texttt{-1}} $). Divided by the Avogadro's constant $N_A$, we have: $$ \frac{R}{N_A} = \frac {8.314462618\ \texttt{J}\cdot \texttt{mol}^{\texttt{-1}}\cdot \texttt{K}^{\texttt{-1}}}{6.02214076 \times 10^{23}\ \texttt{mol}^\texttt{-1}} = 1.380649\times 10^{-23}\ \texttt{J}\cdot \texttt{K}^{\texttt{-1}} = k_B $$ $k_B$ is the Boltzmann constant. We can say that $R$ is used to describe the properties of 1 $\texttt{mol}$ of gas, while $k_B$ is used to describe the properties of one single gas molecule. By the relation of $$ E = k_B T $$ where $T$ is temperature in the unit of Kelvin $(\texttt{K})$, we can express the energy equivalence in the unit of absolute temperature. Electromagnetic wave with $1\ \texttt{Hz}$ of frequency is then converted to $4.799\ 243\ 073 \times 10^{-11}\ \texttt{K} $ via: $$ T = \frac{E}{k_B} = \frac {6.62607015 \times 10^{-34}\ \texttt{J}}{1.380649\times 10^{-23}\ \texttt{J}\cdot \texttt{K}^{\texttt{-1}}} = 4.799243073 \times 10^{-11}\ \texttt{K} $$

The electron volt ($\texttt{eV}$) is one of the units of energy defined by the amount of energy gain or lost by the charge of one single electron moved across an electric potential difference of 1 $\texttt{volt}$. Since that 1 $\texttt{volt}$ = 1 $\texttt{J}\cdot \texttt{C}^{\texttt{-1}}$ ($\texttt{joule per coulomb}$) and the charge of the electron is $-1.602\ 176 \ 634 \times 10^{-19}\ \texttt{C}$, we have $$ 1\ \texttt{eV} = (1.602176634 \times 10^{-19}\ \texttt{C}) \times (1\ \texttt{J}\cdot \texttt{C}^{\texttt{-1}}) = 1.602176634 \times 10^{-19} \texttt{J}\ \ (\texttt{joule per electron}) $$ Multiplied by the Avogadro's constant $N_A$, $$ 1\ \texttt{eV} = \frac{(1.602176634 \times 10^{-19}\ \texttt{J}) \times (6.02214076 \times 10^{23}\ \texttt{mol}^\texttt{-1})}{10^3} = 96.485332\ \texttt{kJ} \cdot \texttt{mol}^\texttt{-1} $$

The $\texttt{hartree}$ energy (named after Douglas Rayner Hartree), also denoted as $\texttt{Eh}$ or atomic unit in energy ($\texttt{a.u.}$), is a large unit in energy frequently used in computational chemistry. It is suggested by NIST CODATA (2018) with a conversion factor of:
$$ 1\ \texttt{hartree} = 27.211\ 386\ 245\ 988\ (53)\ \texttt{eV} $$