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Rydberg constant

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The Rydberg constant, named after physicist Johannes Rydberg, is a physical constant that appears in the Rydberg formula. It was discovered when measuring the spectrum of hydrogen, and builds upon results from Anders Jonas Ångström and Johann Balmer.

The "infinite" Rydberg constant is often simply called the "Rydberg constant" and is essentially the (cyclical) wavenumber of the photon emitted when a Hydrogen atom decays from n = infinity (unbound electron and proton) directly into the ground state, n = 1. Thus it also represents the minimum wavenumber a single photon must have in order to completely free the electron of a hydrogen atom in the ground state.

The Rydberg constant is one of the most well-determined physical constants, with a relative experimental uncertainty of less than 7 parts per trillion. The ability to measure it directly to such a high precision confirms the proportions of the values of the other physical constants that define it, and can thus be used to stringently test physical theories such as quantum electrodynamics.

Each chemical element has its own Rydberg constant. For all Hydrogen-like atoms (atoms with a single electron in their outermost orbit) the Rydberg constant $R_M \$ can be derived from the "infinity" Rydberg constant, as follows:

$R_M = \frac{R_\infty}{1+m_e/M} \$

where,

$R_M \$ is the Rydberg constant for a certain atom with one electron with the rest mass $m_e \$

$M \$ is the mass of its atomic nucleus.

The "infinity" Rydberg constant is (according to 2002 CODATA results):

$R_\infty = \frac{m_e e^4}{(4 \pi \epsilon_0)^2 \hbar^3 4 \pi c} = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c} = 1.0973731568525(73) \cdot 10^7 \,\mathrm{m}^{-1}$

where,

$\hbar \$ is the reduced Planck's constant,

$m_e \$ is the rest mass of the electron,

$e \$ is the elementary charge,

$c \$ is the speed of light in vacuum, and

$\epsilon_0 \$ is the permittivity of free space.

This constant is often used in atomic physics in the form of an energy:

$h c R_\infty = 13.6056923(12) \,\mathrm{eV} \equiv 1 \,\mathrm{Ry} \$

Additional recommended knowledge

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1 Alternate expressions
2 Rydberg constant for hydrogen
3 Derivation of Rydberg constant
4 See also
5 References

Alternate expressions

The Rydberg constant can also be expressed as the following equations.

$R_\infty = \frac{\alpha^2 m_e c}{4 \pi \hbar} = \frac{\alpha^2}{2 \lambda_e} \$

and

$h c R_\infty = \frac{h c \alpha^2}{2 \lambda_e} = \frac{h f_C \alpha^2}{2} = \frac{\hbar \omega_C}{2} \alpha^2 \$

where

$h \$ is Planck's constant,

$c \$ is the speed of light in a vacuum,

$\alpha \$ is the fine-structure constant,

$\lambda_e \$ is the Compton wavelength of the electron,

$f_C \$ is the Compton frequency of the electron,

$\hbar \$ is the reduced Planck's constant, and

$\omega_C \$ is the Compton angular frequency of the electron.

Rydberg constant for hydrogen

Substituting the 2002 CODATA value for the electron-proton mass ratio, $m_e / m_p = 5.446 170 2173(25) \cdot 10^{-4} \$ , into the general formula for the Rydberg constant for any Hydrogen-like element $R_M \$ , we find the Rydberg constant for hydrogen, $R_H \$ .

$R_H = 10 967 758.341 \pm 0.001\,\mathrm{m}^{-1} \$

Substituting $R = R_H \$ into the Rydberg formula for the Hydrogen-like atoms, we can obtain the emission spectrum of hydrogen,

$\frac{1}{\lambda_{\mathrm{vac}}} = R_{\mathrm{H}} Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Where

λ vac

is the wavelength of the light emitted in vacuum,

R H

is the Rydberg constant for hydrogen,

n 1

and

n 2

are integers such that

n 1 < n 2

Z is the atomic number, which is 1 for hydrogen.

Derivation of Rydberg constant

The Rydberg constant for hydrogen can be derived using Bohr's condition, centripetal force, electric force, and electric potential energy of an electron in orbit around a proton (corresponding to the case for the hydrogen atom).

Bohr's condition,
The angular momentum of the electron can only have certain discrete values:
$L = m_e v r = n \frac{h}{2 \pi} = n \hbar$

where n = 1,2,3,… (some integer) and is called the principal quantum number, h is Planck's constant, and $\hbar=h/(2\pi)$ .
$r \$ is the radius of the electron's orbit
Force necessary to maintain circular motion (a.k.a. centripetal force),
$F_\mathrm{centripetal}= \frac{m_ev^2}{r} \$
where
$m_e \$ is the rest mass of the electron, and $v \$ is the electron's velocity
Electric Force of Attraction between an electron and a proton
$F_\mathrm{electric}= \frac{e^2}{4 \pi \epsilon_0 r^2 } \$
where
$e \$ is the elementary charge,
$\epsilon_0 \$ is the permittivity of free space.
The expression for the total electric potential energy of an electron some distance $r$ from a proton is $E_\mathrm{total} = - \frac {e^2}{ 8 \pi \epsilon_0 r} \$

To begin, we take Bohr's primary condition and solve it in terms of the electron's permitted orbital velocity $v$ :

$v = \frac {n h}{2 \pi r m_e} \$

Since the electric force attracting the electron to the nucleus is the (centripetal) force driving the electron into a circular orbit around the proton, we can set $F centripetal = F electric$ to obtain

$\frac{m_e v^2}{r} = \frac{e^2}{4 \pi \epsilon_0 r^2 } \$

Substitute our previous expression for the electron orbital velocity $v \$ in and solve for $r \$ to obtain

$r = \frac{n^2 h^2 \epsilon_0 }{ \pi m_e e^2} \$

This value of $r$ supposedly represents the only allowed values for the orbital radius of an electron in orbit around a proton assuming the Bohr condition holds for the wave nature of the electron. If we now substitute $r$ into the expression for the electric potential energy of an electron some distance from a proton and we get

$E_\mathrm{total} = \frac{- m_e e^4}{8 \epsilon_0^2 h^2}. \frac{1}{n^2} \$

Therefore a change in energy in an electron changing from one value of $n$ to another is

$\Delta E = \frac{ m_e e^4}{8 \epsilon_0^2 h^2} \left( \frac{1}{n_\mathrm{initial}^2} - \frac{1}{n_\mathrm{final}^2} \right) \$

We simply change the units to wavelength $\left( \frac{1}{ \lambda} = \frac {E}{hc} \rightarrow \Delta{E} = hc \Delta \left( \frac{1}{\lambda}\right)\right) \$ and we get

$\Delta \left( \frac{1}{ \lambda}\right) = \frac{ m_e e^4}{8 \epsilon_0^2 h^3 c} \left( \frac{1}{n_\mathrm{initial}^2} - \frac{1}{n_\mathrm{final}^2} \right) \$