What is the Rydberg's formula? Will it be used here?

  • Thread starter prakhargupta3301
  • Start date
  • Tags
    Formula
In summary, the Rydberg formula is used to calculate the energy needed to excite an electron from one energy level to another in an atom. It can be written in various forms, including with or without the constant z^2, and can be extended to other atoms besides hydrogen. The inverse wavelength is sometimes used as an equivalent of frequency in this formula.
  • #1
prakhargupta3301
58
1
I don't know why, but I have a slight ambiguity regrding the Rydberg formula.
In some places it is written as :
1/λ= Rh(1/n12-1/n22)
In some:
E= -Rh(1/n2)
In some:
1/λ= Rh*(z2/n2)

At some:
1/λ= Rh*(z2)(1/n12-1/n2)2

Please tell me. Where these formula are used? Are they even correct?
 
Chemistry news on Phys.org
  • #2
Take a look at wiki page for the formula. It contains enough info.

The original equation is as follows:

$$\frac{1}{\lambda} = R ( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} )$$

by taking ##n_{2} \rightarrow ##, you will get this:

$$\frac{1}{\lambda} = R ( \frac{1}{n_{1}^{2}} )$$

Note: which means, ionizing an atom by fetching an electron from ##n_{1} \rightarrow n_{2}##. Also, I replaced ##n_{1}## by ##n##.

What is the energy needed to excite an electron from ##n_{1}## to ##n_{2}##? You will need to convert the wavelength you obtained to energy. Hence, you need the equation of energy of a photon 'wiki'. Which is ##E=h \nu##, where ##\nu## is the frequency of the photon/light. Now you should work it yourself and find the second equation.

the ##z^{2}## is an extension to the original formula, so that it can be used for other atoms. The formula (the first you wrote) was originally invented for hydrogen atoms only. The one with ##z^{2}## extends the use of the formula for atoms that are similar to Hydrogen.

Please look at the wiki pages.
 
  • Like
Likes prakhargupta3301
  • #3
Disclaimer: I don't know why you have a negative in your second equation. And I don't find your equations consistent with the use of the constant ##R## and ##R^{*}##
 
  • Like
Likes prakhargupta3301
  • #4
The energy is negative because that is the binding energy for the electron with principal quantum number ## n ##. In a transition of energy levels from a higher state ## n_2 ## to a lower state ## n_1 ## , (## n_2>n_1 ##), a photon of energy ## \Delta E=-R(\frac{1}{n_2^2}-\frac{1}{n_1^2})=h \nu ## is emitted in order to conserve energy. There are systems of units (with ## \Delta E=\frac{hc}{\lambda} ##) that have ## h=1 ## and ## c=1 ##, but normally this is not the case. ## \\ ## The simplest derivation of this equation is the Bohr atom model. It does get the right answer for the energies of the states with principal quantum number ## n ##, but lacks some of the detail that is obtained in much more accurate quantum mechanical calculations. See https://en.wikipedia.org/wiki/Bohr_model
 
Last edited:
  • Like
Likes Phylosopher
  • #5
Phylosopher said:
Take a look at wiki page for the formula. It contains enough info.

The original equation is as follows:

$$\frac{1}{\lambda} = R ( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} )$$

by taking ##n_{2} \rightarrow ##, you will get this:

$$\frac{1}{\lambda} = R ( \frac{1}{n_{1}^{2}} )$$

Note: which means, ionizing an atom by fetching an electron from ##n_{1} \rightarrow n_{2}##. Also, I replaced ##n_{1}## by ##n##.

What is the energy needed to excite an electron from ##n_{1}## to ##n_{2}##? You will need to convert the wavelength you obtained to energy. Hence, you need the equation of energy of a photon 'wiki'. Which is ##E=h \nu##, where ##\nu## is the frequency of the photon/light. Now you should work it yourself and find the second equation.

the ##z^{2}## is an extension to the original formula, so that it can be used for other atoms. The formula (the first you wrote) was originally invented for hydrogen atoms only. The one with ##z^{2}## extends the use of the formula for atoms that are similar to Hydrogen.

Please look at the wiki pages.
So the true formula is just where z =1 and hence it is omitted while writing?
 
  • #6
prakhargupta3301 said:
So the true formula is just where z =1 and hence it is omitted while writing?
No. Read the first wiki page. Section 2.
 
  • #7
Isn't the inverse of wavelength just the frequency? Why not just call it the frequency then instead of 1/wavelength? Maybe it's just the expression of the time it takes to make one wavelength? the time of a single packet of electromagnetic energy?
 
  • #8
litup said:
Isn't the inverse of wavelength just the frequency?

No. They are related, but it is not just a simple inverse.
 
  • #9
No, the inverse of wavelength is the number of wavelengths per unit length. Frequency is the number of cycles per second, and is related to the inverse wavelength by
frequency = speed of wave/wavelength
Because it is proportional to frequency, and hence energy, it is sometimes convenient to use inverse wavelength as an equivalent of frequency. For example, in vibrational spectroscopy, because the numbers are a convenient magnitude, we often use the number of waves per cm, which we call the "wavenumber", with units cm-1. Often we loosely call it "frequency", e.g "CO2 absorbs at a frequency of 2350 cm-1".
True frequency (Hz) = speed of light (cm/s) * wavenumber (cm-1)
 

Related to What is the Rydberg's formula? Will it be used here?

What is the Rydberg's formula?

The Rydberg's formula is a mathematical equation that describes the wavelengths of light emitted by hydrogen atoms. It was developed by physicist Johannes Rydberg in the late 19th century.

How is the Rydberg's formula used?

The Rydberg's formula is used to calculate the wavelengths of light emitted by hydrogen atoms. It is also used to determine the energy levels of electrons in hydrogen atoms and to study the spectra of other elements.

Can the Rydberg's formula be used in other elements?

Yes, the Rydberg's formula can be used to study the spectra of other elements besides hydrogen. However, slight modifications may need to be made to account for differences in the atomic structure of each element.

Is the Rydberg's formula still used today?

Yes, the Rydberg's formula is still used in modern physics and is an important tool for understanding the behavior of atoms and molecules. It has also been extended to describe the spectra of more complex atoms and molecules.

Why is the Rydberg's formula important?

The Rydberg's formula is important because it helped to lay the foundation for the development of quantum mechanics and our understanding of atomic structure. It also continues to be a useful tool in many areas of physics, including spectroscopy and astrophysics.

Similar threads

Replies
1
Views
2K
Replies
2
Views
8K
  • Thermodynamics
Replies
7
Views
1K
Replies
15
Views
842
Replies
2
Views
785
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
937
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
2K
Replies
16
Views
3K
Back
Top