Constants and Conversions

An energy splitting can be expressed in joules, electronvolts, hertz, inverse centimetres, or equivalent kelvin. The conversion follows

$$E=h\nu=\hbar\omega=hc\tilde\nu=k_B T_E.$$

Here $\nu$ counts cycles per second, $\omega=2\pi\nu$ is angular frequency, and $\tilde\nu$ is vacuum wavenumber. The quantity $T_E=E/k_B$ is an energy-equivalent temperature; it does not imply that the sample is in thermal equilibrium at that temperature.

Required background. Units and constants introduces dimensional analysis and uncertainty. For Hartree-unit multipliers, use Atomic units.

Measured constants below use the 2022 CODATA adjustment, checked on 14 October 2026. The SI Brochure defines $h,c,e,k_B$ exactly in SI; $m_e,\alpha$ and the magnetons retain measurement uncertainty. A truncated decimal for an exact quantity is not exact to all digits. Parentheses on a measured value give its standard uncertainty in the final quoted digits.

Energy, frequency, and wavelength

The same symbols are not used uniformly across every subfield. This page uses the following convention.

Quantity	Symbol	Definition	Preferred unit
energy	$E$	Hamiltonian eigenvalue difference or photon energy	$\mathrm J$ or $\mathrm{eV}$
cyclic frequency	$\nu$ or $f$	cycles per unit time	$\mathrm{Hz}$
angular frequency	$\omega$	$2\pi\nu$	$\mathrm{rad\,s^{-1}}$
vacuum wavelength	$\lambda_0$	$c/\nu$	$\mathrm m$ or $\mathrm{nm}$
vacuum wavenumber	$\tilde{\nu}$	$1/\lambda_0=\nu/c$	$\mathrm{m^{-1}}$ or $\mathrm{cm^{-1}}$
wavevector magnitude	$k$	$2\pi/\lambda_0$	$\mathrm{rad\,m^{-1}}$
energy-equivalent temperature	$T_E$	$E/k_{\mathrm B}$	$\mathrm K$

Two useful pairs are $\omega=2\pi\nu$ and $k=2\pi\tilde\nu$. Dimensional analysis alone cannot distinguish cycles from radians; the defining equation supplies the factor of $2\pi$.

Exact SI constants

The values in this table are exact in the present SI.

Constant	Symbol	SI value	AMO role
speed of light in vacuum	$c$	$299\,792\,458\ \mathrm{m\,s^{-1}}$	wavelength–frequency conversion
Planck constant	$h$	$6.626\,070\,15\times10^{-34}\ \mathrm{J\,s}$	energy per cycle frequency
reduced Planck constant	$\hbar$	$1.054\,571\,817\ldots\times10^{-34}\ \mathrm{J\,s}$	energy per angular frequency
elementary charge magnitude	$e$	$1.602\,176\,634\times10^{-19}\ \mathrm C$	electronvolt and electromagnetic coupling
Boltzmann constant	$k_{\mathrm B}$	$1.380\,649\times10^{-23}\ \mathrm{J\,K^{-1}}$	energy–temperature conversion

The symbol $e$ denotes a positive magnitude. The electron charge is $q_e=-e$. A Hamiltonian written with a particle charge $q$ must therefore use $q=-e$ for an electron; silently replacing $q$ by $e$ reverses electric and magnetic signs.

Measured constants

The following rounded values come from the 2022 CODATA adjustment. The parenthetical digits are one-standard-deviation uncertainties in the final quoted digits.

Constant	Symbol	2022 CODATA value
electron mass	$m_e$	$9.109\,383\,7139(28)\times10^{-31}\ \mathrm{kg}$
fine-structure constant	$\alpha$	$7.297\,352\,5643(11)\times10^{-3}$
inverse fine-structure constant	$\alpha^{-1}$	$137.035\,999\,177(21)$
Rydberg constant	$R_\infty$	$10\,973\,731.568\,157(12)\ \mathrm{m^{-1}}$
Bohr magneton	$\mu_{\mathrm B}$	$9.274\,010\,0657(29)\times10^{-24}\ \mathrm{J\,T^{-1}}$
nuclear magneton	$\mu_{\mathrm N}$	$5.050\,783\,7393(16)\times10^{-27}\ \mathrm{J\,T^{-1}}$

These constants are related, not independent entries in a bag of numbers. For example,

$$\alpha =\frac{e^2}{4\pi\epsilon_0\hbar c}, \qquad \mu_{\mathrm B} =\frac{e\hbar}{2m_e}, \qquad \mu_{\mathrm N} =\frac{e\hbar}{2m_p}.$$

After the 2019 SI revision, $\epsilon_0$ and $\mu_0$ are measured quantities because they inherit the uncertainty of $\alpha$. In particular, $\mu_0=4\pi\times10^{-7}\ \mathrm{N\,A^{-2}}$ is no longer exact.

The Rydberg constant describes the infinite-nuclear-mass Coulomb scale:

$$R_\infty =\frac{\alpha^2m_ec}{2h}, \qquad hcR_\infty =13.605\,693\,122\,990(15)\ \mathrm{eV}.$$

It should not be inserted as though it were the exact line wavenumber of a finite-mass atom. Reduced-mass, recoil, relativistic, radiative, nuclear-size, and hyperfine corrections separate an observed transition from the simple $R_\infty$ formula.

Energy conversion table

These compact entries are rounded for scale estimates. Their conversion factors follow from exact SI defining constants; the omitted digits do not represent experimental uncertainty. For a precision calculation, use the defining equations or the NIST energy equivalents.

Starting quantity	J	Hz	$\mathrm{cm^{-1}}$	K
$1\ \mathrm{eV}$	$1.6021766\times10^{-19}$	$2.4179892\times10^{14}$	$8065.5439$	$11604.518$
$1\ \mathrm{cm^{-1}}$	$1.9864459\times10^{-23}$	$2.9979246\times10^{10}$	$1$	$1.4387769$
$1\ \mathrm{Hz}$	$6.6260702\times10^{-34}$	$1$	$3.3356410\times10^{-11}$	$4.7992431\times10^{-11}$
$1\ \mathrm K$	$1.380649\times10^{-23}$	$2.0836619\times10^{10}$	$0.69503480$	$1$

For electronvolts and kelvin,

$$1\ \mathrm K =8.617\,333\,262\,145\times10^{-5}\ \mathrm{eV},$$

and therefore

$$1\ \mathrm{meV} \approx11.6045\ \mathrm K.$$

For wavenumbers,

$$1\ \mathrm{cm^{-1}} =29.979\,2458\ \mathrm{GHz}.$$

This exact frequency factor follows from $c$ and the exact relation $1\ \mathrm{cm^{-1}}=100\ \mathrm{m^{-1}}$.

The electronvolt is an energy unit: $1\ {\rm eV}=1.602176634\times10^{-19}\ {\rm J}$ exactly. To obtain a numerical energy in eV from joules, divide by that conversion factor. Do not divide an energy already in eV by the elementary charge again.

Wavenumbers

Spectroscopists often report an energy as a vacuum wavenumber:

$$\tilde{\nu} =\frac{1}{\lambda_0} =\frac{\nu}{c} =\frac{E}{hc}.$$

The common unit $\mathrm{cm^{-1}}$ is an inverse length, not “per centimetre of path traveled” and not an angular wavevector. Because $1\ \mathrm{cm^{-1}}=100\ \mathrm{m^{-1}}$,

$$E =hc\left(100\tilde{\nu}_{[\mathrm{cm^{-1}}]}\right)$$

when $\tilde{\nu}_{[\mathrm{cm^{-1}}]}$ is the numerical value quoted in $\mathrm{cm^{-1}}$.

For a vacuum wavelength expressed in nanometres,

$$\tilde{\nu}\,[\mathrm{cm^{-1}}] =\frac{10^7} {\lambda_0\,[\mathrm{nm}]}.$$

A reported wavelength must say whether it is a vacuum wavelength or an air wavelength. Frequency is continuous across a stationary interface, while the wavelength in a medium is

$$\lambda_{\mathrm{med}} =\frac{\lambda_0}{n(\nu)}.$$

At low precision, “589 nm” may be enough context. At precision-spectroscopy accuracy, the refractive-index model, pressure, temperature, humidity, and composition of the medium can matter.

Rates and spectral widths

For a stable ground state and an excited population decaying as $P_e(t)=P_e(0)e^{-\Gamma t}$, the weak-drive lifetime-limited Lorentzian has

$$\Delta\omega_{\rm FWHM}=\Gamma, \qquad \Delta\nu_{\rm FWHM}=\frac{\Gamma}{2\pi}.$$

The decay rate itself is in inverse seconds. Converting it from one time unit to another does not introduce $2\pi$; converting an angular linewidth to a cyclic linewidth does. Extra dephasing, driving, and motion can change the relation between lifetime and measured width. See Line-shape reference for those definitions.

Temperature as an Energy Unit

The mapping

$$T_E=\frac{E}{k_{\mathrm B}}$$

answers: “What temperature has thermal energy $k_{\mathrm B}T$ equal to this energy scale?” It does not assert that a single photon, level splitting, trap depth, or recoil energy is itself a thermodynamic temperature.

Examples include:

• a trap depth $U_0/k_{\mathrm B}=1\ \mathrm{mK}$;
• a recoil scale $E_r/k_{\mathrm B}=200\ \mathrm{nK}$;
• a hyperfine splitting expressed as an equivalent kelvin; and
• a vibrational quantum compared with $k_{\mathrm B}T$ to estimate thermal occupation.

For two nondegenerate states separated by $\Delta E>0$ in thermal equilibrium,

$$\frac{p_e}{p_g} =e^{-\Delta E/(k_{\mathrm B}T)}.$$

For manifolds of degeneracies $g_e,g_g$, their total population ratio has an additional factor $g_e/g_g$. Here the ratio depends on both the gap-equivalent temperature $T_E=\Delta E/k_{\mathrm B}$ and the actual ensemble temperature $T$. Conflating them erases the statistical-mechanical content.

Negative binding energies require care. Converting $E=-13.6\ \mathrm{eV}$ to a negative kelvin is algebraically possible, but a thermal comparison usually uses the positive ionization energy $|E|/k_{\mathrm B}$.

Magnetic Energy Scales

The Bohr and nuclear magnetons convert magnetic fields into electron-scale and nuclear-scale energies:

$$\mu_{\mathrm B} =\frac{e\hbar}{2m_e}, \qquad \mu_{\mathrm N} =\frac{e\hbar}{2m_p}.$$

Useful frequency forms are

$$\begin{aligned} \frac{\mu_{\mathrm B}}{h} &=13.996\,244\,9171(44)\ \mathrm{GHz\,T^{-1}},\\ \frac{\mu_{\mathrm N}}{h} &=7.622\,593\,2188(24)\ \mathrm{MHz\,T^{-1}}. \end{aligned}$$

Since $1\ \mathrm G=10^{-4}\ \mathrm T$,

$$\frac{\mu_{\mathrm B}}{h} =1.399\,624\,49171\ \mathrm{MHz\,G^{-1}}.$$

For a weak-field hyperfine Zeeman shift,

$$\Delta E =g_F\mu_{\mathrm B}B\,\Delta m_F,$$

so the shift in hertz is

$$\Delta\nu =g_F\frac{\mu_{\mathrm B}}{h}B\,\Delta m_F.$$

The sign depends on $g_F$, the magnetic-field direction, and the definition of the magnetic quantum number. The magneton supplies a scale, not the full magnetic moment. Nuclear moments also include species-dependent nuclear $g$ factors and may be quoted with signs.

Dipole Moments and Electric Fields

Molecular electric dipoles are commonly reported in debye. IUPAC defines the debye as a non-SI unit with

$$1\ \mathrm D \approx3.335\,64\times10^{-30}\ \mathrm{C\,m}.$$

The 2022 CODATA atomic unit of electric dipole moment is

$$ea_0 =8.478\,353\,6198(13)\times10^{-30}\ \mathrm{C\,m} \approx2.54175\ \mathrm D.$$

For a matrix element $d$ coupled to an electric field amplitude $\mathcal E$,

$$E_{\mathrm{int}}\sim d\mathcal E.$$

A useful laboratory scale is

$$\frac{(1\ \mathrm D)(1\ \mathrm{kV\,cm^{-1}})}{h} \approx503.412\ \mathrm{MHz}.$$

This is a scale estimate, not a promise of a linear Stark shift. A parity eigenstate has no first-order diagonal electric dipole unless degeneracy, state mixing, orientation, or another symmetry-breaking mechanism permits it. For an optical transition, the relevant $d$ is a transition matrix element and field conventions may distinguish peak, root-mean-square, and complex amplitudes.

Wavelength Shortcuts

For vacuum wavelength in nanometres,

$$\begin{aligned} \nu\,[\mathrm{THz}] &=\frac{299\,792.458} {\lambda_0\,[\mathrm{nm}]},\\ E\,[\mathrm{eV}] &=\frac{1\,239.841\,984\,332\ldots} {\lambda_0\,[\mathrm{nm}]},\\ \tilde{\nu}\,[\mathrm{cm^{-1}}] &=\frac{10^7} {\lambda_0\,[\mathrm{nm}]}. \end{aligned}$$

These are compact forms of $E=hc/\lambda_0$, not independent empirical rules. Keeping the wavelength unit in the denominator is safer than memorizing a bare value such as “1240.”

For spectroscopy, frequency is usually the cleaner invariant quantity. Wavelength becomes nonlinear under uncertainty propagation:

$$\lambda_0=\frac{c}{\nu}, \qquad \frac{\delta\lambda_0}{\lambda_0} \approx-\frac{\delta\nu}{\nu}$$

for small changes. The minus sign says that a higher frequency has a shorter wavelength.

Worked examples

A 780 nm photon

Take a vacuum wavelength $\lambda_0=780\ \mathrm{nm}$. The cyclic frequency is

$$\nu =\frac{c}{\lambda_0} =3.84349\times10^{14}\ \mathrm{Hz} =384.349\ \mathrm{THz}.$$

The photon energy is

$$E =h\nu =2.54672\times10^{-19}\ \mathrm J =1.58954\ \mathrm{eV}.$$

The vacuum wavenumber and energy-equivalent temperature are

$$\tilde{\nu} =12\,820.5\ \mathrm{cm^{-1}}, \qquad \frac{E}{k_{\mathrm B}} =18\,445.9\ \mathrm K.$$

The last number does not mean a 780 nm laser beam has a temperature of $18\,446\ \mathrm K$. It states the single-photon energy in kelvin units.

A Rabi-frequency convention

Suppose a resonant drive is reported as

$$\frac{\Omega}{2\pi}=10\ \mathrm{MHz}.$$

Then

$$\Omega =2\pi\times10^7\ \mathrm{rad\,s^{-1}},$$

and the resonant $\pi$-pulse duration is

$$t_\pi =\frac{\pi}{\Omega} =50\ \mathrm{ns}.$$

Reading “$10\ \mathrm{MHz}$” as $\Omega=10^7\ \mathrm{rad\,s^{-1}}$ would instead give $t_\pi\approx314\ \mathrm{ns}$. The physics did not change; the unstated convention did.

Precision and uncertainty

Conversion does not create information. If a measured line is $384.230(5)\ \mathrm{THz}$, reporting a wavelength with fifteen digits is misleading even though a calculator can produce them.

For $y=y(x)$ and a small standard uncertainty $u(x)$,

$$u(y) \approx \left|\frac{dy}{dx}\right|u(x).$$

Thus for $\lambda=c/\nu$,

$$u(\lambda) \approx \frac{c}{\nu^2}u(\nu).$$

If several measured constants enter, use their covariance matrix:

$$u^2(y) = \sum_{i,j} \frac{\partial y}{\partial x_i} \frac{\partial y}{\partial x_j} \operatorname{cov}(x_i,x_j).$$

CODATA values are correlated because they come from a joint least-squares adjustment. Treating them as independent can overstate or understate a precision uncertainty. For routine AMO scale estimates, rounded values are appropriate. For precision metrology, retrieve the current recommended values, uncertainties, and correlation coefficients from the official database.

For a precision comparison, keep the constant release, frequency convention, wavelength medium, and uncertainty attached to the number. The worked conversion can have many digits even when the physical input does not.

Common pitfalls

Inverse centimetres need a factor of 100. Before using SI $h,c$ in $E=hc\tilde\nu$, write $1\ {\rm cm^{-1}}=100\ {\rm m^{-1}}$.

The Rydberg constant is not an energy. $R_\infty$ is an inverse length; the infinite-mass Rydberg energy is $hcR_\infty=E_h/2$. Finite-mass atomic lines also require reduced-mass and other physical corrections.

A magneton gives a scale. The moment also contains the particle or state’s $g$ factor and sign. The Bohr and nuclear magnetons are different units, using $m_e$ and $m_p$ respectively.

Exercises

A visible photon

Treat $589.0\ \mathrm{nm}$ as a vacuum wavelength. Convert it to terahertz, electronvolts, inverse centimetres, and energy-equivalent kelvin. Keep four significant figures.

Solution

The frequency is

$$\nu =\frac{299\,792.458} {589}\ \mathrm{THz} =508.9855\ \mathrm{THz}.$$

The photon energy is

$$E =\frac{1\,239.841984}{589}\ \mathrm{eV} =2.104995\ \mathrm{eV}.$$

The wavenumber is

$$\tilde{\nu} =\frac{10^7}{589}\ \mathrm{cm^{-1}} =16\,977.93\ \mathrm{cm^{-1}},$$

and

$$T_E =\frac{E}{k_{\mathrm B}} =24\,427.45\ \mathrm K.$$

To four significant figures:

$$509.0\ \mathrm{THz}, \quad 2.105\ \mathrm{eV}, \quad 1.698\times10^4\ \mathrm{cm^{-1}}, \quad 2.443\times10^4\ \mathrm K.$$

A π pulse

A laboratory note states “Rabi frequency $5\ \mathrm{MHz}$.” Compute the $\pi$-pulse time under each of these interpretations:

1. $\Omega/(2\pi)=5\ \mathrm{MHz}$;
2. $\Omega=5\times10^6\ \mathrm{rad\,s^{-1}}$.

What should the note have written?

Solution

For a resonant two-level system,

$$t_\pi=\frac{\pi}{\Omega}.$$

Under the first interpretation,

$$\Omega =2\pi(5\times10^6)\ \mathrm{rad\,s^{-1}},$$

so

$$t_\pi =\frac{1}{2(5\times10^6)} =100\ \mathrm{ns}.$$

Under the second interpretation,

$$t_\pi =\frac{\pi}{5\times10^6} =0.6283\ \mathrm{\mu s}.$$

The answers differ by $2\pi$. The note should have written either $\Omega/(2\pi)=5\ \mathrm{MHz}$ or $\Omega=5\ \mathrm{Mrad\,s^{-1}}$.

Dipole-field coupling

Find the frequency scale $d\mathcal E/h$ for $d=2.0\ \mathrm D$ and $\mathcal E=5.0\ \mathrm{kV\,cm^{-1}}$. State one reason this need not equal an observed first-order Stark shift.

Solution

Using

$$\frac{(1\ \mathrm D)(1\ \mathrm{kV\,cm^{-1}})}{h} \approx503.412\ \mathrm{MHz},$$

gives

$$\frac{d\mathcal E}{h} \approx (2.0)(5.0)(503.412\ \mathrm{MHz}) =5.034\ \mathrm{GHz}.$$

This is the bare interaction scale. A parity eigenstate can have zero first-order diagonal dipole moment; rotational averaging, matrix-element geometry, degeneracy, state mixing, and field polarization can all modify the observed shift.

Exact constants and rounded decimals

Classify each statement as correct or incorrect.

1. $1\ \mathrm{eV}$ in joules has experimental uncertainty.
2. The displayed decimal for $\hbar$ may be truncated even though $\hbar$ is derived exactly from $h$.
3. $\mu_0=4\pi\times10^{-7}\ \mathrm{N\,A^{-2}}$ is exact in the current SI.
4. A CODATA parenthesis such as $(11)$ marks uncertainty in the final digits.

Solution

1. Incorrect. The elementary charge is fixed exactly, so $1\ \mathrm{eV}=1.602\,176\,634\times10^{-19}\ \mathrm J$ is exact.
2. Correct. $\hbar=h/(2\pi)$ is derived exact, but its decimal expansion is nonterminating.
3. Incorrect. Since the 2019 SI revision, $\mu_0$ inherits uncertainty through the measured fine-structure constant.
4. Correct. For a measured CODATA value, parenthetical digits give the one-standard-deviation uncertainty in the corresponding final digits.

References

• Bureau International des Poids et Mesures. The International System of Units (SI), 9th edition, updated 2026. doi:10.59161/AUEZ1291.
• International Union of Pure and Applied Chemistry. “Debye.” Compendium of Chemical Terminology, 5th edition (2025). doi:10.1351/goldbook.D01533.
• Mohr, P. J., D. B. Newell, B. N. Taylor, and E. Tiesinga. “CODATA recommended values of the fundamental physical constants: 2022.” Reviews of Modern Physics 97, 025002 (2025). doi:10.1103/RevModPhys.97.025002.
• National Institute of Standards and Technology. The 2022 CODATA Recommended Values of the Fundamental Physical Constants, web version 9.0 (2024). Database and energy equivalents, accessed 14 October 2026.