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

prakhargupta3301
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?

Phylosopher
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.

prakhargupta3301
Phylosopher
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^{*}##

prakhargupta3301
Homework Helper
Gold Member
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

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Phylosopher
prakhargupta3301
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?

Phylosopher
So the true formula is just where z =1 and hence it is omitted while writing?
No. Read the first wiki page. Section 2.

litup
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?

Mentor
Isn't the inverse of wavelength just the frequency?

No. They are related, but it is not just a simple inverse.