# What is the difference between the Rydberg constants R_H = "109678.77 cm"^(-1) and R_(H) = 2.18 xx 10^(-18) "J"?

Jan 6, 2017

For a comparison of $\approx 109677$ and $109737.32$ ${\text{cm}}^{- 1}$, see this answer I gave earlier today.

As for ${R}_{H} = {109678.77174307}_{10}$ ${\text{cm}}^{- 1}$ and ${R}_{H} = 2.1787 \times {10}^{- 18} \text{J}$ (Inorganic Chemistry, Miessler et al.), clearly, the difference is in the units. We have reciprocal distance vs. energy.

These do NOT have equivalent meanings, but they DO interconvert!

The only difference in their usage, which is the more interesting part, is the lefthand side of the Rydberg equation for electron relaxation (i.e. $\Delta E < 0$) in the HYDROGEN atom:

$\boldsymbol{\frac{1}{\lambda} = - {R}_{H} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)}$

where ${R}_{H} = {109678.77174307}_{10}$ ${\text{cm}}^{- 1}$, $\lambda > 0$ is the wavelength in $\text{cm}$, and ${n}_{f}$ and ${n}_{i}$ are final and initial energy levels that the electron moved to and from, respectively.

$\boldsymbol{\Delta E = - {R}_{H} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)}$

where ${R}_{H} = 2.1787 \times {10}^{- 18} \text{J}$, the negative of the ground-state energy of the hydrogen atom (whose actual signed value was $- \text{13.6 eV}$).

You can verify that the ${\text{cm}}^{- 1}$ version converts to $\text{13.6 eV}$ here.

If you recall, the energy absorbed by the electron is equal to the energy of the photon emitted during the electronic relaxation.

So, ${E}_{\text{photon}} = | \Delta E | = h \nu = \frac{h c}{\lambda}$.

You can convert one Rydberg equation into the other by dividing by $h c$:

$\Delta E = \frac{h c}{\lambda}$

$\implies \frac{\Delta E}{h c} = \frac{1}{\lambda}$

Therefore, you can convert one Rydberg constant to the other. If we temporarily label the ${\text{cm}}^{- 1}$ version as ${R}_{H}^{\text{*}}$ and the $\text{J}$ version as ${R}_{H}$, then:

${R}_{H}^{\text{*}} = {R}_{H} / \left(h c\right)$

= 2.1787xx10^(-18) cancel"J" xx 1/((6.626xx10^(-34) cancel"J"cdotcancel"s")(2.998xx10^(10) "cm/"cancel"s"))

$= {\text{109676.6996 cm}}^{- 1}$

$\approx {109678.77174307}_{10}$ ${\text{cm}}^{- 1}$