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Why does $\ce{Li}$ give a red flame color which has lower photon energy compared to $\ce{Cs}$ which gives a violet flame color according to $E=h\nu$?

This is my attempt: Since lithium has smaller size it has a high ionization energy, thus, if an electron jumps from a higher energy level to the ground state then it releases a large amount of energy compared to cesium which has a larger size and lower ionization energy. Therefore lithium's photon energy should be higher compared to cesium. But it is not. Why?

The problem is the photon energy of flame coloration should gradually decrease down the group of alkali metals but rather it's increasing.

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The colours you get in flame tests / burning metals, comes from the electrons in one shell being 'knocked' (by the heat), into a higher shell; then shortly after the electron 'drops' back (in one or more steps). The electron which dropped down a shell has a lower energy state, so to conserve energy, a photon is emitted. The further it 'drops' the higher energy/frequency/ colour it has. (Lower energy Infrared -> ROYGBV -> UltraViolet -> X-rays Higher Energy)

An element with more protons has a 'deeper' bottom shell. More protons => more attraction to electrons => more energy required to 'knock' up to a higher shell => higher energy/frequency photons.

So in general, a larger element would give higher energy photons off; in a flame test. However it is a little more complex than that.

Because elements have many shells, an electron may jump back in more than one hop; giving off several different colours. For example Calcium has an orange flame colour, but heaver Strontium has a Scarlet red colour. If you look at a Strontium flame through a prism, you get Red, Green, and Violet.

Note: You may get infrared and ultraviolet in the mix too.

The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron. Normally an electron in its outermost (filled) electron shell.
Lithium only has electrons in its bottom 2 shells. Cesium has electrons in its bottom 6 shells. Its outer most one has a single electron and is relatively "further away" from the protons in the nucleus. So it is easier to liberate Cesium's outer most electron than Lithium's outer most. In a flame test, if the flame is hot enough, electrons from deeper shells may be 'knocked' out of place and so you get higher energy photons.

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  • $\begingroup$ I don't mean to hijack the question, but I'm curious about something. If photons are only emitted when the electron drops down a shell, why is there color while they are in the flame? When they are in the flame wouldn't the electron have the energy to remain in the higher shell? @DarcyThomas $\endgroup$ – Noel Jul 27 '16 at 6:33
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    $\begingroup$ @Noel Good comment/question. The metal atoms in the flame will constantly be having their electrons 'knocked up' as the hot atoms bang into each other, but almost immediately after (less than a nano second I think) the electron will 'drop' down; before being bumped into an excited state again shortly after that etc, etc. Note also that the metal atoms emitting photons is one of the sources of light/photons. That gives you narrow specific colours, there are others too. Like hot bits of carbon soot which glow with a range of colour (but looks yellow/orange when all mixed together). $\endgroup$ – DarcyThomas Jul 27 '16 at 8:15
  • $\begingroup$ But the photon released should have energy somehow nearly equal to the ionization energy of the atom. I think this gives more specific information of the energy of photon. And since Lithium has higher I.E.(Ionization Energy) to Cesium, thus Li should have high photon energy in flame coloration. But it's not. Why? $\endgroup$ – Brett Leigh Jul 27 '16 at 16:04
  • $\begingroup$ @YashpalYadav Lithium only has electrons in its bottom 2 shells. Cesium has 6 shells. Its outer most one has one electron and is relatively "further away" from the protons in the nucleus. So it is easier to liberate Cesium's outer most electron than Lithium's outer most. The outer electron shell is the one which normally is important in (basic) chemistry ( not the case with transition metals ). However in a flame test, if the flame is hot enough, electrons from deeper shells may be 'knocked' out of place and so you get higher energy photons. $\endgroup$ – DarcyThomas Jul 27 '16 at 21:11
  • $\begingroup$ Can you explain this "An element with more protons has a 'deeper' bottom shell. More protons => more attraction to electrons => more energy required to 'knock' up to a higher shell => higher energy/frequency photons." further? $\endgroup$ – chemsonraun Oct 8 '17 at 14:06
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To estimate the approximate wavelength of electronic transitions is alkaline metals, the Rydberg formula $(1)$ may be used. Strictly speaking, it is only valid for hydrogen and hydrogen-like atoms ($\ce{He+, Li^2+, \dots}$), however, if we assume the core electrons to be of spherical symmetry we can assume the effective nuclear charge to be $1+$ and the entire core to resemble the proton of hydrogen.

$$\frac{1}{\lambda_\text{obs}} = R\left( \frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\tag{1}\\ \begin{array}{ll} \text{where:}\\ &R = \pu{1.097e-7 m^{-1}}\\ & n_1 = \text{lower shell’s principal quantum number}\\ & n_2 = \text{higher shell’s principal quantum number}\end{array}$$

For lithium, we expect the lowest transition to be $\ce{3\bond{->}2}$ while for caesium we expect it to be $\ce{7\bond{->}6}$. Plugging these values into the formula gives us:

$$\begin{align} \lambda_\ce{3\bond{->}2}(\ce{Li}) &= \pu{656nm}\\ \lambda_\ce{7\bond{->}6}(\ce{Cs}) &= \pu{12369nm}\end{align}$$

Thus, we expect the lowest-energy photon emitted by caesium indeed to have a lower energy than that of lithium. However, this transition is firmly within the infrared wavelength range. It is therefore not observable visually.

What you are seeing in caesium’s case are a number of different transitions that do not correspond to the valence electron being excited by one shell and dropping back down. You can see all, part or a combination of:

  • high-shell core electrons being displaced into the valence shell and subsequent relaxations
  • relaxation of the valence electron from higher outer shells
  • high-shell core electrons being displaced into the valence shell and another, higher-shell electron relaxing

There are a lot of possible transitions covering the infrared range, the visible range and probably even the UV range. Some of them will be intense enough for the colour to be visible in a spectroscope or as part of the flame colour.

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