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

##### 1 Answer

For a comparison of

As for *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.

#bb(1/lambda = -R_H(1/n_f^2 - 1/n_i^2))# where

#R_H = 109678.77174307_(10)# #"cm"^(-1)# ,#lambda > 0# is the wavelength in#"cm"# , and#n_f# and#n_i# are final and initial energy levels that the electron moved to and from, respectively.

#bb(DeltaE = -R_H(1/n_f^2 - 1/n_i^2))# where

#R_(H) = 2.1787xx10^(-18) "J"# , the negative of the ground-state energy of the hydrogen atom (whose actual signed value was#-"13.6 eV"# ).You can verify that the

#"cm"^(-1)# version converts to#"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_"photon" = |DeltaE| = hnu = (hc)/lambda# .

You can convert one Rydberg equation into the other by dividing by

#DeltaE = (hc)/lambda#

#=> (DeltaE)/(hc) = 1/lambda#

Therefore, you can *convert* one Rydberg constant to the other. If we temporarily label the

#R_H^"*" = R_H/(hc)#

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

#= "109676.6996 cm"^(-1)#

#~~ 109678.77174307_(10)# #"cm"^(-1)#