Rydberg Formula

In spectroscopy, the Rydberg constant, symbolized as R∞ for heavy atoms or RH for hydrogen, is named after the Swedish physicist Johannes Rydberg. 

A rydberg constant is a fundamental physical constant that relates the electromagnetic spectrum of an atom. 

The constant first emerged as an experimental fitting boundary in the Rydberg formula for the hydrogen spectral lines, however, Niels Bohr later showed that its value could be determined from more fundamental constants through his Bohr model.

On this page, we will discuss the rydberg constant formula along with the Rydberg equation derivation. We will also have a look at the Balmer Rydberg equation.

Rydberg Constant Definition

The Rydberg constant has a value of 1.097 x 107 m-1. This value is valid in the formulae for the wavenumbers of lines in atomic spectra. It is a function of the following attributes:

• The rest mass and charge of the electron,

• The speed of light, and 

• Planck's constant.

Formula of Rydberg Constant

The Rydberg equation for heavy atoms is given as;

R∞ = \[\frac{me\ e^{4}}{8\varepsilon o^{2}h^{3}c}\], this equation is equal to 1097 (21) m-1.

Here,

me = the rest mass of the electron

e  =  the elementary charge, whose value is 1.6 x 10-19 C

ε0 = is the permittivity of free space, whose value is 8.85 x  10-12 Farad/meter

h = is the Planck constant, whose value is 6.626 x 10-34 J.s

c = is the speed of light in a vacuum, whose value is 3 x 108 m/s

Similarly, the Rydberg equation for a hydrogen atom is:

RH = \[\frac{mp}{mp + me}\approx\]1.09687 x 107 m-1

me = is the mass of the electron, and

mp = is the rest mass of the nucleus (a proton) 

Now, we will do the Rydberg constant derivation to verify its value for heavy atoms and the hydrogen atom.

Rydberg Equation Derivation

We know that the Balmer Rydberg equation gives the wavenumbers of the spectral lines of the hydrogen atom, the formula for the wavenumber is given by: 

\[w = \frac{1}{\lambda} = R (\frac{1}{n^{2}}-\frac{1}{q^{2}})\]

Where,

n and q are integers, and 

R is the Rydberg constant. 

Niels Bohr smartly derived the Rydberg formula by invoking quantum. Now, let’s see the Rydberg constant derivation:

Using the Balmer Rydberg Equation:    

\[\frac{1}{\lambda} = R (\frac{1}{2^{2}}-\frac{1}{n^{2}})\]

Also, the energy of a photon is given by:

\[\triangle E = E_{i} - E_{f}\]

Also,

\[\triangle E = hf = h\frac{c}{\lambda}\]

By Bohr Model:

E   =    Ei   -   Ef   =   KE + PE

So, the energy of an atom is:

\[E = \frac{1}{2}mv^{2} + (- \frac{kZe^{2}}{r})\]

\[= \frac{1}{2}mv^{2} - (\frac{kZe^{2}}{r})\]......(1)

Here,

Z = atomic number

r  = radius of the atom

v  = velocity

m = mass of an atom

As per the Newtonian mechanics:

FC   =   Fe

→ \[\frac{mv^{2}}{r} = \frac{kZe^{2}}{r}\]

→ \[mv^{2} = \frac{kZe^{2}}{r}\]  ………(2)

Now, putting the value of eq (2) in (1), we have:

→\[E= \frac{1}{2}mv^{2} - (\frac{kZe^{2}}{r})\]      

 = \[\frac{1}{2}(\frac{kZe^{2}}{r}) - \frac{kZe^{2}}{r}\]

E = - \[\frac{kZe^{2}}{r}\] (is the required energy of an atom)

As per the rotational mechanics, we know that angular momentum, L is given by:

L  =   I\[\omega\]

Here, 

\[\omega = \frac{v}{r}\] angular velocity in rad/s

Or,

\[L = I.\frac{v}{r}\]

Also, the moment of inertia I is given as;

\[I = mr^{2}\]

Now,

\[L =  mr^{2}.\frac{v}{r} = mvr\]

According to Bohr Postulate:

\[L = mvr_{n} = n\frac{h}{2\pi}\], where n = 1, 2, 3,......., n

So, the velocity will be:    

\[v = n\frac{h}{2\pi mrn}\]……(3)                                 

Besides this, we have an expression \[mv^{2} = \frac{kZe^{2}}{r}\] , putting the value of “v” from eq (3):

\[m(n\frac{h}{2\pi mrn})^{2} =\frac{kZe^{2}}{rn}\]

Now, let’s solve the above equation:

→\[\frac{n^{2}h^{2}}{4\pi^{2} mrn^{2}} =\frac{kZe^{2}}{rn}\]

Rearranging the above equation, we get:

→\[r_{n} = \frac{n^{2}h^{2}}{4\pi^{2} kZe^{2}m} = (\frac {h^{2}}{4\pi^{2} ke^{2}m})\frac{n^{2}}{Z}\] ; where n = 1, 2, 3,……,n

The Bohr radius of an atom is:

r1   =   (5.29. 10-11) m

Also,

\[E_{n} = - \frac{kZe^{2}}{r} = (\frac{2\pi^{2} mk^{2}e^{4}}{h^{2}}).\frac{Z^{2}}{n}\]…..(4)

Here, the part of the above equation has a value, which is given by:

\[(\frac{2\pi^{2} mk^{2}e^{4}}{h^{2}}) = 2.18.10^{-18}J\]

So, putting the value in eq (4):

\[E_{n} = (-2.18.10^{-18}J) \frac{Z^{2}}{n}\]

However, in eV, the equation becomes:          

\[E_{n} = (-13.6\ eV) \frac{Z^{2}}{n}\]

Therefore, the energy difference E is:

\[\triangle E\]   =   Ei   -   Ef  

Or, 

\[\triangle E = hf = h\frac{c}{\lambda}\]

\[\frac{1}{\lambda} =(\frac{2\pi^{2} mk^{2}e^{4}}{h^{2}}).Z^{4}\ (\frac{1}{nf^{2}}-\frac{1}{ni^{2}})\]

Replacing the part of the equation with “R,” Rydberg constant (to get the Rydberg formula) in the following manner:

\[\frac{1}{\lambda} =R.(Z^{2})\ (\frac{1}{nf^{2}}-\frac{1}{ni^{2}})\].....(5)

Hence, eq (5) is the required Rydberg equation derivation.

Where,

The value for the Rydberg constant formula, R   =  1097 m-1

Dimensional Formula of Rydberg Constant

After deriving the formula of Rydberg constant, let’s determine the dimensional formula of Rydberg constant:

From the above value of R   =  1097 m-1, the dimensional formula of Rydberg constant is [M0L-1T0].

Conclusion

The Rydberg formula carries a value of 1097 m-1. Here, 1 R =   2.18. 10-18 J or 13.6 eV. The Rydberg value can be calculated for the hydrogen atom and heavy atoms as well.