Sciencemadness Discussion Board - Why has Mg no flame color ?

Why has Mg no flame color ?

metalresearcher - 20-9-2010 at 03:10

According to this picture (from Wikipedia)

Mg has green spectral lines. But bwhen burning Mg you see a blinding white flame.
Other alkali (earth) metals do have colors: Ca: orange-red, Sr red, Ba green. Why do Mg salts or Mg itself show no flame color ?

hissingnoise - 20-9-2010 at 03:35

The colour fidelity from wiki can't be trusted very far. . .
The two distinct lines are actually more in the blue range of the spectrum.
And the hotter the flame, the bigger is the blue component!

len1 - 20-9-2010 at 04:17

Thats an interesting question.

the colour radiated by a body can be divided into two components: black body radiation and the convolution of the latter by its characterstic properties, such as absorption due to the elctronic properties of its atoms.

The colour of BB radiation chanes from visible red at about 700C, to white at about 1600C, and blue beyond - the exact spectrum is easily calculated from the laws of physics - from memory its ~lambda^3 exp(-hc/kT clambda) with phase space increase (the ultraviolet catastrophy) modified by quantum exponential damping.

The energy radiated also increases very rapidly with temperature

power ~ T^4

So the Mg flame is so hot (> 2000C) that the thermal component (which is independent of material, dwarfs the modulation by the material contribution

As another example the colour of an arc welding light is bright white, and to first approximation independent of the metal you are welding

[Edited on 20-9-2010 by len1]

unionised - 20-9-2010 at 05:28

Most of the radiation you see from a Mg flame is from hot MgO. Like most solids it has, at best, very broad bands or a continuum emission. I think the picture depicts the spectrum of Mg metal, rather than the oxide.

ScienceSquirrel - 20-9-2010 at 05:38

I think you might get a better result with a spectroscope.
You can buy them quite cheaply on eBay etc or from specialist suppliers;

http://gyroscope.com/d.asp?product=SPECTROSCOPESMALL

or you can build your own;

http://homechemistry.blogspot.com/2008/03/more-light-and-che...

More in line with science madness

len1 - 20-9-2010 at 06:01

When one takes an EE or EA spectrum, one always uses a metal salt rather than the metal, for electronic excitation it doesnt matter if you use Mg or MgO

metalresearcher - 20-9-2010 at 06:14

Two comments:

- Even a stick (e.g. Ti or MgO) with MgCl2 solution does not color while one with CaCl2 colors orange, with NaCl bright yellow and with KCl purplish.

- An electric arc does have an element dependant color: welding steel has a blue color doe to many lines of Fe in the blue range, while holding CaO in a carbon arc colors pink. Sodium salts color the arc bright yellow. Melting ... vaporizing Cu metal in the arc colrs the arc greenish blue.

[Edited on 2010-9-20 by metalresearcher]

unionised - 20-9-2010 at 06:40

Quote: Originally posted by len1

When one takes an EE or EA spectrum, one always uses a metal salt rather than the metal, for electronic excitation it doesnt matter if you use Mg or MgO

Do you really think the excited states of Mg are the same as those of MgO? I think you may find the vibrational and rotational structures are different. That's why MgO gives bands but Mg gives lines.

The great merit of, for example, ICP is that it's so hot it atomises everything (nearly). Of course the problem is that it tends to ionise things too. The reason for using really hot flames is only partly because they are brighter. You get less interference from molecular species.

len1 - 20-9-2010 at 07:18

There coat of welding rods contains salts of different compositions to aid ionisation, and there is also the iron core, but to me all these arcs look bright white, the coloration is a second order effect

The AA spectrometer flame atomizes metal salts, so the cation spectrum is independent of the anion. Also the ionisation energy ~10eV is much larger than vibrational energy so the sharp lines broaden - but these are centred in the same place

But if you test a Mg salt in an AA spectrometer you get essentialy no color, so to answer the question again what you see in burning Mg is essentially blackbody radiation

[Edited on 20-9-2010 by len1]

watson.fawkes - 20-9-2010 at 12:43

Quote: Originally posted by unionised

Do you really think the excited states of Mg are the same as those of MgO? I think you may find the vibrational and rotational structures are different. That's why MgO gives bands but Mg gives lines.

The key physical measure here is the partition of energy among the different ways that the system can have energy. In the present case, you have thermal motion (which is principally translational kinetic energy, but also include rotational modes) and electron excitation. Thermal motion, when in contact with a colder environment, loses energy in a continuous spectrum by black body radiation. Electronic excitation, when in contact with a colder environment, loses energy in a discrete spectrum by electron decay between energy levels. (Electronic excitation is modified by Doppler shifts and some others with less effect.) The combination of these two effects account for the total emission from the system in question. Both Mg and MgO contribute to the black body component of the emission in exactly the same way. They contribute in different ways to the electronic emissions.

The spectrum of electron decay depends upon the atomic or molecular orbital structure. The ratio of electron decay emission with respect to thermal emission, though, depends on the relative magnitude of the energies involved. At this point, in order to follow the story completely you need to know something about thermodynamic partition functions, which is rather too advanced for this level of explanation. The upshot, though, is that at "low" temperatures electronic emission will predominate and at "high" temperatures thermal emission will predominate. The general rule of thumb is that at white-hot temperatures, thermal emission will predominate over electronic emission from any ordinary kind of matter.

Lastly, there's the issue about the emission partition between the excited states of Mg and MgO. If you were able to put a spectroscope with a high enough dynamic range up to a magnesium flare, you'd see the excitation lines from both compounds (as well as some from O2 and N2, no doubt). The ratio of the strength of these emissions is, as before, related to the available energy states of these substances. Mg is easier to ionize than MgO, so its set of states grows more rapidly with temperature. (This is the partition function at work again, plus some gross approximations about ionization energy and excited states.) Thus Mg electronic emission will also dominate over MgO emission, at any temperature.

blogfast25 - 20-9-2010 at 12:51

I believe the gadgets from gyroscope.com are nicely built and affordable but unfortunately do not tell you the exact wavelength of any observed lines. For homemade spectroscopes I'd recommend this fellow, I 'built' (cut n' glued, more like) a few of his models:

http://astro.u-strasbg.fr/~koppen/spectro/spectroe.html

Personally I warmly recommend this model:

http://astro.u-strasbg.fr/~koppen/spectro/mk2be.html

With a piece of DVD-R (NOT music compact disc) as diffraction grating, that resolves just about the Na D doublet, less than a 1 nm apart...

Also by that same professional physicist: emission spectra of various elements (at the bottom of the page *.txt files that give the relative intensity of the lines, for each spectral line):

http://astro.u-strasbg.fr/~koppen/discharge/index.html

Relative intensity of the (green) Mg line at 518 nm is about 400. For the Na D lines at 589 nm, relative intensities of the doublet lines are... 40,000 and 80,000! That's why the sodium line is so easy to see but the Mg line isn't And why we have sodium street lighting and not magnesium...

Unless you use high purity chemicals you'll see the Na D doublet in just about anything you 'analyse'.

I've got a blueprint for a home made spectrogoniometer that allows not only to see the lines (using a DVD R grating) but also to measure the angle of diffraction and thus the wavelength (but I've had that blueprint for some months now...), w/o any electronics.

[Edited on 21-9-2010 by blogfast25]

densest - 20-9-2010 at 19:37

If one burned Mg in Cl2 then the discrete spectrum of Mg might show up. In making fireworks, Mg is used to heat the color-emitting species (BaCl(2?), SrCl(2?), and (CuCl)n to higher temperatures than an organic fuel would. Excess Cl is used to create MgCl2 which is vaporized and doesn't contribute much to the spectrum. Other methods of getting rid of MgO are known. Any solids in a colored firework emit black body radiation which is red-orange-yellow-white all of which are undesirable in a colored flame.

blogfast25 - 21-9-2010 at 08:04

Quote: Originally posted by densest

If one burned Mg in Cl2 then the discrete spectrum of Mg might show up. In making fireworks, Mg is used to heat the color-emitting species (BaCl(2?), SrCl(2?), and (CuCl)n to higher temperatures than an organic fuel would. Excess Cl is used to create MgCl2 which is vaporized and doesn't contribute much to the spectrum. Other methods of getting rid of MgO are known. Any solids in a colored firework emit black body radiation which is red-orange-yellow-white all of which are undesirable in a colored flame.

Generating the visible emission spectrum of Mg, visible with a decent spectroscope in good viewing conditions, should not be a problem with a controlled electrical arc.

As regards fireworks, I'm no expert but BaCl2 cannot be reduced by Mg metal because the Heat of Formation of BaCl2 is larger (more negative: ΔHstandard = - 858.56 kJ/mol) than that of MgCl2 (ΔHstandard = - 641.62 kJ/mol) (all values from NIST), the reduction reaction BaCl2 + Mg --> Ba + MgCl2 would be strongly endothermic... I don't know about SrCl2 (similar to Ba, of course) but Mg would definitely reduce CuCl2 (ΔHstandard = - 205.85 kJ/mol) with probably enough enthalpy to spare (ΔHreaction = - 436 kJ/mol, standard conditions) to vapourise the reaction products...

In the case of Ba (and possibly Sr) I believe they use oxidising anions like peroxide, nitrate of chlorate (or even sulfate) and either Mg or Al to generate the needed heat...

IrC - 21-9-2010 at 15:35

You could shift the angle on an etalon while measuring intensity. A graph of light intensity VS frequency could be made. If for no other reason, to prove len1 has been right on in the reasons for the color. I believe in any case it is simply black body temperature (partially masking the lines also being produced).

peach - 22-9-2010 at 07:27

Physics labs tend to use things like discharge tubes that vapourise the element and then pass a current through the gas. The burning elemental metal will be emitting black body light (continuous spectrum based on heat) as well as the spectral lines.

Burning lithium, I can see a reddish glow, but the dominant component is a blinding bright white. And the red tint usually only appears once the burning is coming away from it's peak intensity.

Another example, the sun is basically a massive hydrogen bomb, burning away, yet you don't see the hydrogen spectral lines if you look at the sun with your eyes, you see white.

Or, you see yellow / orange. Yet the major component of sunlight at sealevel is blue to green. There's too much black body noise masking the specific lines and it's entirely overloading your eyes at the same time. Stare at it for a few seconds and it will semi-permanently mess up the receptors; afterglow / colour smearing. Stare a bit longer it'll bleach them. Which is why the physics boys use electrical excitation.

Interesting sun based facts;

If you watch the sun as it sets, it will sometimes go green for a second due to the way the light is being filtered by the atmosphere.

Plants reflect green light, one of the major components of sunlight, absorbing blue and red only (sometimes orange) in surprisingly tight bands. Our eyes are most sensitive to green light (I can take one guess why that's so; nom nom nom).

There's also a very cleaver and funny optical trick a guy did a long while ago (I've been trying to find his name and the experiment again, so let me know if you know) where looking at a single colour (a flame?) with a bit of paper sticking out between your eyes will cause one eye to see magenta and the other to see blue; due to the way your brain interprets the signals.

Quote:

- An electric arc does have an element dependant color: welding steel has a blue color doe to many lines of Fe in the blue range, while holding CaO in a carbon arc colors pink. Sodium salts color the arc bright yellow. Melting ... vaporizing Cu metal in the arc colrs the arc greenish blue.

And it's blue with nothing there because of all the nitrogen around. Nitrogen is used in UV gas lasers.

Here's one from the absolute gold standard page on DIY lasers, Samual M. Goldwasser's (Sam's Laser FAQ). It's basically a PCB with a break etched in it and the nitrogen in a plastic cavity over that. The gain is so high mirrors aren't even needed.

There is a very visual form of proof for what you're saying. If someone interested does a quick search of youtube, you'll discover countless guys making Tesla coils or Jacobs ladders. They usually emit blue arcs. But the JL's sometimes don't start reliably or burn smoothly. So they'll drip some table salt solution over them. Hey prestro, it works, with a yellow arc, the sodium lines (just like the orange light bulbs in the street lamps outside).

Vapourise it, shoot electrons through it and you'll likely see the lines clearly. If there's a few of them, you'll see a homogenization of them with your naked eyes. Argon lasers look teal / aqua blue, but they contain numerous lines and can produce bluey rainbows of those when the beam is sent through a prism (as in the photo at the bottom).

LEDs and Lasers are maybe a good comparison. A gas laser in particular works by electrically exciting a gas into emission. And the light that comes out of both has a tiny line width. A few billionths of a metre. There's no (or very little) black body noise related to heat.

That's why lasers and LEDs produce such vivid, pure colours that catch your eyes. They don't occur in such pure, intense forms naturally, so you're brain will subconsciously recognize it as out of place and go 'OOooooooooo!"

A copper laser heats a salt of it to a vapour state and then passes current through the gas;

My personal favourite, argon (the thing by his feet is the cooler / supply for the beast, as these ones tend to get hot, a lot hot);

[Edited on 22-9-2010 by peach]

Rogeryermaw - 24-9-2010 at 17:07

i have actually seen magnesium produce green fire before. it was in metal shop when i was in school (million years ago) i was running foundry and someone thought it would be wise to melt a piston from a jaguar. turns out they alloy their aluminum pistons with magnesium...turn your back for a split second and all hell breaks loose! you never saw a class empty out that fast! but i distinctly remember the last thing i looked at was the foundry furnace and seeing a green-white pillar of fire about 3-4 feet tall shooting out of the furnace.

peach - 24-9-2010 at 18:58

Ferrari do it as well. The engine blocks are magnesium. There's a video on youtube somewhere of them casting and machining the blocks.

I'd rather have a motorbike, but the mold is impressively complex and precise; it's full of cores.

Contrabasso - 25-9-2010 at 03:13

Spectral emissions from a pure specimen - a sample in a vacuum tube will be from the sample only. Emissions from a flame environment will have components from ALL the species present and in a flame even the most transitory metastable compounds will exist and have spectra!