Chemical Sciences: A Manual for CSIR-UGC National Eligibility Test for Lectureship and JRF/Rydberg constant

The Rydberg constant, named after physicist Johannes Rydberg, is a physical constant that apperas in the Rydberg formula. It was discovered when measuring the spectrum of hydrogen, and building upon results from Anders Jonas Ångström and Johann Balmer. Each chemical element has its own Rydberg constant, which can be derived from the "infinity" Rydberg constant.

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.

For a series of discrete spectral lines emitted by atmoic hydrogen,

${\displaystyle {\frac {1}{\lambda }}=R\left({\frac {1}{m^{2}}}-{\frac {1}{n^{2}}}\right)}$.

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

${\displaystyle R_{\infty }={\frac {m_{e}e^{4}}{(4\pi \epsilon _{0})^{2}\hbar ^{3}4\pi c}}=1.0973731568525(73)\cdot 10^{7}\,\mathrm {m} ^{-1}}$
where
${\displaystyle \hbar \ }$ is the reduced Planck's constant,
${\displaystyle m_{e}\ }$ is the rest mass of the electron,
${\displaystyle e\ }$ is the elementary charge,
${\displaystyle c\ }$ is the speed of light in vacuum, and
${\displaystyle \epsilon _{0}\ }$ is the permittivity of free space.

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

${\displaystyle hcR_{\infty }=13.6056923(12)\,\mathrm {eV} \equiv 1\,\mathrm {Ry} }$

The "infinity" constant appears in the formula:

${\displaystyle R_{M}={\frac {R_{\infty }}{1+{\frac {m_{e}}{M}}}}}$
where
${\displaystyle R_{M}}$ is the Rydberg constant for a certain atom with one electron with the rest mass ${\displaystyle m_{e}\ }$
${\displaystyle M\ }$ is the mass of its atomic nucleus.

Alternate expressions

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

${\displaystyle R_{\infty }={\frac {\alpha ^{2}m_{e}c}{4\pi \hbar }}={\frac {\alpha ^{2}}{2\lambda _{e}}}}$

and

${\displaystyle hcR_{\infty }={\frac {hc\alpha ^{2}}{2\lambda _{e}}}={\frac {hf_{C}\alpha ^{2}}{2}}={\frac {\hbar \omega _{C}}{2}}\alpha ^{2}}$

where

${\displaystyle h\ }$ is Planck's constant,
${\displaystyle c\ }$ is the speed of light in a vacuum,
${\displaystyle \alpha \ }$ is the fine-structure constant,
${\displaystyle \lambda _{e}\ }$ is the Compton wavelength of the electron,
${\displaystyle f_{C}\ }$ is the Compton frequency of the electron,
${\displaystyle \hbar \ }$ is the reduced Planck's constant, and
${\displaystyle \omega _{C}\ }$ is the Compton angular frequency of the electron.

Rydberg Constant for hydrogen

Plugging in the rest mass of an electron and an atomic mass ${\displaystyle M}$ of 1 for hydrogen, we find the Rydberg constant for hydrogen, ${\displaystyle R_{H}}$.

${\displaystyle R_{H}=10967758\pm 1m^{-1}}$

Plugging this constant into the Rydberg formula, we can obtain the emission spectrum of hydrogen.

Derivation of Rydberg Constant

The Rydberg Constant can be derived using Bohr's condition, centripetal acceleration, and Potential Energy of the electron to the nucleus.

Bohr's condition,

${\displaystyle 2\pi r=n\lambda }$

where

${\displaystyle n}$ is some integer
${\displaystyle r}$ is the radius of some atom

Centripetal Acceleration,

${\displaystyle PE={\frac {mv^{2}}{r}}}$

where

${\displaystyle m_{e}\ }$ is the rest mass of the electron,
${\displaystyle v}$ is the electron's velocity

PE of Attraction between Electron and Nucleus

${\displaystyle PE={\frac {e^{2}}{4\pi \epsilon _{o}r^{2}}}}$

where

${\displaystyle e\ }$ is the elementary charge,
${\displaystyle \epsilon _{o}\ }$ is the permittivity of free space.

Firstly we substitute ${\displaystyle \lambda ={\frac {h}{p}}}$ into Bohr's condition, then solve for ${\displaystyle v}$ We obtain ${\displaystyle v^{2}={\frac {n^{2}h^{2}}{4\pi ^{2}r^{2}m^{2}}}}$

We equate centripetal acceleration and attraction between nucleus to obtain ${\displaystyle {\frac {mv^{2}}{r}}={\frac {e^{2}}{4\pi \epsilon _{o}r^{2}}}}$

Substitute ${\displaystyle v^{2}}$ in and solve for ${\displaystyle r}$ we obtain ${\displaystyle r={\frac {n^{2}h^{2}\epsilon _{o}}{\pi me^{2}}}}$

We know that ${\displaystyle E_{total}=-{\frac {e^{2}}{8\pi \epsilon _{o}r}}}$

Substitute ${\displaystyle r}$ into this energy equation and we get ${\displaystyle E_{total}={\frac {-me^{4}}{8\epsilon _{o}^{2}h^{2}}}.{\frac {1}{n^{2}}}}$

Therefore a chance in energy would be ${\displaystyle \Delta E={\frac {me^{4}}{8\epsilon _{o}^{2}h^{2}}}({\frac {1}{n_{initial}^{2}}}-{\frac {1}{n_{final}^{2}}})}$

We simply change the units to wave number (${\displaystyle {\frac {1}{\lambda }}={\frac {E}{hc}}}$) and we get ${\displaystyle \Delta {\tilde {\nu }}={\frac {m_{e}e^{4}}{8\epsilon _{o}^{2}h^{3}c}}({\frac {1}{n_{initial}^{2}}}-{\frac {1}{n_{final}^{2}}})}$ where

${\displaystyle h\ }$ is Planck's constant,
${\displaystyle m_{e}\ }$ is the rest mass of the electron,
${\displaystyle e\ }$ is the elementary charge,
${\displaystyle c\ }$ is the speed of light in vacuum, and
${\displaystyle \epsilon _{o}\ }$ is the permittivity of free space.
${\displaystyle n_{initial}}$ and ${\displaystyle n_{final}}$ being the excited states of the atom

We have therefore find the Rydberg constant for Hydrogen to be

${\displaystyle R_{H}={\frac {me^{4}}{8\epsilon _{o}^{2}h^{3}c}}}$