Question

What is the difference between the Rydberg constants
$R_H=109678.77c{m}^{-1}$ and $R_H=2.18\times {10}^{-18}J$ ?

Answer 1

Hint: To compare these two we first need to know the definition of Rydberg constants. These two values do not have equivalent meaning and unit but they are interchangeable from Joules to centimeter inverse or vice versa if one unit is given to us. Both are two different units with two different values.

Complete step by step answer:
Rydberg Constant:
In the science of spectroscopy under physics, this constant is a physical constant that is related to the atomic spectra. It is denoted by the symbol $R_{\mathrm{\infty }}$ that is for heavy atoms but $R_H$ is denoted for hydrogen atoms. This constant is first arising from the Rydberg’s formula as a fitting parameter but later on Neil Bohr has calculated it from the fundamental constant. And this constant is often used and expressed in the form of the Rydberg unit of energy.
The only difference between these two values is in their uses.
Let us see their two different uses for two differ units.
The Rydberg equation for electron relaxation that is change in energy is less than zero for hydrogen atom:

$R_H=109678.77c{m}^{-1}$	$R_H=2.18\times {10}^{-18}J$
This unit can be calculated using the wavelength formula of the hydrogen atom.	This unit can be calculated using the energy formula of the hydrogen atom.
We can find this value with the help of this below formula:${\displaystyle \frac{1}{\lambda }}=-R_H\left({\displaystyle \frac{1}{{n_f}^2}}-{\displaystyle \frac{1}{{n_i}^2}}\right)$	While to find this value we have to use this formula:$\mathrm{\Delta }E=-R_H\left({\displaystyle \frac{1}{{n_f}^2}}-{\displaystyle \frac{1}{{n_i}^2}}\right)$
Here the value of the Rydberg constant is $R_H=109678.77c{m}^{-1}$ and the wavelength is greater than zero and the unit is centimeter.	While here the value of Rydberg constant is $R_H=2.18\times {10}^{-18}J$ energy is the negative of the ground-state energy of the hydrogen atom and the unit is joule.

Note:
Remember that we can convert one unit to another unit by assuming one unit is known to us. Assuming joules term is known to us then we will get,
We know the energy of photon is,
$\mathrm{\Delta }E={\displaystyle \frac{hc}{\lambda }}$
Now on rearranging we will get,
${\displaystyle \frac{\mathrm{\Delta }E}{hc}}={\displaystyle \frac{1}{\lambda }}$
Let ${R_H}^{\ast }$ be the Rydberg constant in centimeter inverse and $R_H$ be in joules.
Putting their respective formula in place of $\mathrm{\Delta }E\text{ and }\lambda$ we will get,
${R_H}^{\ast }={\displaystyle \frac{R_H}{hc}}$
Here h and c are constant values on putting each terms we will get,
${R_H}^{\ast }={\displaystyle \frac{2.18\times {10}^{-18}J}{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{10}cm{s}^{-1}\right)}}$
${R_H}^{\ast }\approx 109678.77c{m}^{-1}$