enter image description hereIn Bohr's model of an atom, the formula used to find the energy between the 2 orbits and wavelength of emitted photon was valid only for single electron species like hydrogen.In the case of a multi-electron system like in the picture given above will the electron absorb a photon to go from 2s to 2p and also remitt a photon while dexciting from 2p to 2s.There are also elements like sulphur with two excited states thus showing variable covalency but how do the electrons not dexcite from higher energy orbital in a short time but give enough time gap to show two excitation states?Is the dexcitation and remission of photon a phenomenon which can only be seen when an electron goes from one shell to another like from n=1 to n=2 or can it also be seen when electron goes from orbitals and sub shells like 2s to 2p?Since there is an energy difference between the 2s and 2p sub shells there must be remission of photon on excitation but I did not find any online sources to verify this, so I need help.

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    $\begingroup$ You have merged all types of atomic theories. What is meant by two excited states and variable covalency. $\endgroup$
    – ACR
    Commented Apr 1, 2022 at 2:40
  • $\begingroup$ Like how in sulphur an electron is excited from 3p to 3d and 3s to 3p thus showing two excited states $\endgroup$
    – AJknight
    Commented Apr 1, 2022 at 2:46
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    $\begingroup$ Imagine an isolated sulfur atom, it will be one excitation/emission at a time. One cannot have multiple excited states and multiple photon emission at the same time. Search Grotrian diagram of simple atoms first. $\endgroup$
    – ACR
    Commented Apr 1, 2022 at 2:53
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    $\begingroup$ Is there something you are looking for aside from what I answered? Except for the long dead and totally obsolete and incorrect Bohr theory for the hydrogen atom (and atomic ions having only 1 electron), the 2p level is not the same energy as the 2s level, so a photon can be absorbed, if an electron is promoted from the 2s level up to the 2p level. And vice versa: a photon can be emitted if an electron drops down to the 2s level from the 2p level. In my answer, I show that yellow light from excited sodium atoms arises when electrons drop from the two 3p levels to the 3s level. $\endgroup$
    – Ed V
    Commented Apr 2, 2022 at 14:09

1 Answer 1


The OP asked

Is the dexcitation and remission of photon a phenomenon which can only be seen when an electron goes from one shell to another like from n=1 to n=2 or can it also be seen when electron goes from orbitals and sub shells like 2s to 2p?

A simple answer can be provided by examining the 3s and 3p levels of sodium atoms. As is well known, adding a little table salt (sodium chloride) to a flame causes the flame to emit fairly intense yellow light that is characteristic and diagnostic of sodium. This photograph shows the sodium emission from my homemade alcohol burner, with the fuel (70% isopropanol/30% water) ‘salted’ with two pinches of salt.

Sodium yellow flame emission

In the flame, sodium ions get their electrons back and some of the resulting sodium atoms get excited by energetic collisions. This results in some sodium atoms having electrons in their 3p levels, as per the following simplified Grotrian diagram:

Simplified sodium Grotrian diagram Figure source: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.html

An excited sodium atom cannot stay excited indefinitely: it must de-excite and return to the ground state. The next figure shows that sodium de-excites from the 3p levels to the 3s level by emitting the famous yellow sodium D lines:

How the sodium emission D lines arise Figure source: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.html

How do we know that the yellow light is actually two slightly different colors (wavelengths) of light, as in the above figure? The answer is that we perform spectroscopy on the light: we send the light to an instrument that disperses light, similar to how a simple prism or grating disperses sunlight into a rainbow of constituent colors.

To do the necessary spectroscopy, we first need a source of excited sodium atoms. There are numerous ways to get excited sodium atoms and a particularly useful and convenient method uses a sodium hollow cathode lamp (HCL). My photograph below shows a sodium and potassium HCL with neon fill gas. (The red color is intense and is due to the neon, similar to a neon light.) The photograph shows the light being collected by a lens, to focus it on the end of an optical fiber at the left. Note in the first figure that the optical fiber also appears just to the left of the flame. So the optical fiber can be used to collect light from the flame, as in the first figure, or from the HCL, as in the figure below.

Sodium and potassium HCL emission

The other end of the optical fiber serves to provide light as the input to a spectrometer or spectrograph. The next figure shows my homemade echelle spectrograph, with the optical fiber at the upper left.

EdV’s echelle spectrograph

Input light from the optical fiber is collimated and sent to the echelle grating. Diffracted light from the echelle grating is then cross-dispersed by the prism and the camera records the resulting spectrum, called an echellogram. The next photograph is a composite of two echellograms I obtained with the HCL setup shown in the previous figure.

Composite echellogram of the HCL light

The pair of sodium D lines are indicated in the echellogram. All the rest is mostly due to neon, with potassium also contributing some emission lines.

Finally, the next composite photograph compares the sodium D lines obtained from echellograms with HCL and flame emission sources. The D lines are narrowest in the HCL and are somewhat broadened in the flame.

Comparison of sodium D line emissions

Added echellograms, to illustrate how they show spectral content.

White LED echellogram: White LED echellogram Shorter wavelengths are at left and, for a given grating order (“arc”) at bottom. This is a continuum spectrum: there are no atomic lines in an LED.

Echellogram of the sun: Solar echellogram showing Fraunhofer lines in absorption, e.g., H alpha (the red Balmer line), sodium D lines, the magnesium triplet, etc. This is a continuum spectrum with missing dark spots due to absorption of specific wavelengths: the Fraunhofer lines.

White compact fluorescent lamp: Compact fluorescent lamp echellogram showing mercury atomic emission lines and the red and green broad bands due to the rare earth-doped phosphors on the lamp’s glass. This is a continuum spectrum with some mercury atom spectral lines.

A longer exposure echellogram of the sodium and potassium HCL emission, showing the very bright (very over-exposed) red neon spectral lines. I suppressed them in the previous composite photo by using a blue filter. This is a discrete line spectrum due to sodium, potassium and neon. Maybe also impurities: the HCL is old. Na and K HCL with Ne fill gas

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    $\begingroup$ Sorry. I have not understood what are the eleven curved red lines, and the same number of double yellow lines. Why eleven, why curved, and why are they separated by the same distance ? I had always believed that in a spectroscope, the light ray is decomposed into one optical spectrum, and only one. It is even worse for the last picture, where the colored curves contains now bright spots. Please explain ! I am lost ! $\endgroup$
    – Maurice
    Commented Apr 1, 2022 at 18:26
  • $\begingroup$ @Maurice In an ordinary spectroscope, things are exactly as you say: a long spectrum of intensity versus wavelength (or wavenumber). But the echelle spectrograph achieves high resolution by using an echelle grating in high orders and then cross-dispersing the otherwise overlapped orders. So it is kind of like having sentences on a page rather than one very long sentence. It would take many comments to explain this in detail, so I will add some echellograms of other light sources, e.g., LEDs, a white compact fluorescent lamp, etc. I will link to a physics SE answer as well. $\endgroup$
    – Ed V
    Commented Apr 1, 2022 at 18:35
  • $\begingroup$ @Maurice Here is a link to the answer by someone who knows what they are talking about: physics.stackexchange.com/a/515156/313612. The “curves” are grating orders. The “spots” are atomic emission lines, though maybe a couple are ion emission lines: I have not done line assignments yet. The non-spots are broad emissions of some sort. The curving is due to the diffracted rays traveling through different thicknesses of the prism (it is 50 mm on edge). One neat thing about this is I can remove the camera, put my right eye as close to the prism as possible and see the D lines! Exquisite! $\endgroup$
    – Ed V
    Commented Apr 1, 2022 at 18:44
  • $\begingroup$ @Maurice Maybe my answer at astronomy.stackexchange.com/a/49116/45954 will help clarify how echelle spectrographs are used for high resolution spectroscopic work. $\endgroup$
    – Ed V
    Commented Apr 21, 2022 at 12:46
  • $\begingroup$ I am sorry ! I do not understand the beginning of the beginning of all you are talking about. I don't know what is an echelle, a broad emission, a curving, a dekker, etc. I guess it has not too much to do with chemistry. And here we are supposed to discuss about chemistry $\endgroup$
    – Maurice
    Commented Apr 24, 2022 at 10:12

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