By now you may have guessed that inside a burning flame there are many processes occurring all at once. When methane burns in oxygen, the flame contains many excited CH molecules. This heteronuclear diatomic molecule vibrates and rotates. In addition, the electron cloud of the molecule has energy eigen values that must be considered. It is the electron transitions that make up the spectrum that we look at. This spectrum is, however, affected by the rotational and vibrational modes of the nuclei.
The spectrum of the flame is characterized by bands. The lines appear to form a "head" and then trail off. It is easy to see that some of the lines inside a band appear evenly spaced in wave number with a slight decrease (or increase) in spacing as we go along. We can then make the guess that there is a simple power series formula to represent the line spacing for a band. We can make up a label, let's call it "m" to count these lines, and then we have:
k = c + d m + e m2 + f m3 + ... (1)
where k is the wavenumber, and c, d, e, f are all constants in our power series. We will ignore the m3 term and any higher orders of this expansion as the remaining part is of primary interest. Keeping up to the m2 term, we note that the line spacing will diverge if e>0 and converge if e<0. With CH e<0 because we definitely have a branch with the line spacing closing in and the lines reverse on themselves, forming a "head". The next thing to notice is that "m" can take different values, i.e. "m" can be positive or negative. In the spectra we note that the lines are divided into three main regions, a "head" a middle "bump" and a tail. We've already said the head was formed because our quadratic term carried a minus sign. We might as well start in the middle "bump" region and let m=0 at the beginning of the middle bump. Apart from the bump's lines, which we'll get to later, we can then label all of the other lines and happily they form a nice little parabola called a Fortrat parabola. We call the m>0 side (the one with the head in our case) the R branch, the m<0 side the P branch, and the central bump we call the Q branch. At the m=0 spot we don't see any of the P or R branch lines (we see Q branch "bump" lines). This void spot on our otherwise flawless parabola is referred to as the null gap, and finding it is the first step. We can then write:
k = k0 + d m + e m2 + ... (2)
where k0 is the wavenumber of the null gap.
The vibrational modes of the nuclei give a split to the energies from only the electrons. If we call Ee the energy eigenvalue of the electronic wave function. Then:
E = Ee + (v +½)ωe (3)
Er = B J(J+1) +D J2(J+1)2 +... (4)
Where J is the rotational quantum number, B and D are constants (we ignore D and other higher order terms). Be refers to the completely vibrationless state of the molecule. It helps to approximate these effects of rotation and vibration quite nicely by defining
Bv = Be - ae(v +½) (5)
(where ae is yet another constant) so that vibrations and rotations are now combined. Be then has the value:
Be = (h/8p2cm) (1/re2) (6)where we average the square of the internuclear distance. Now the lines come from transitions so Bv'and Bv" represent the higher and lower energy states, respectively. For CH the transition is 2S- (L=0) ® 2P (L=1). (While I have made a small leap in the theory, it should be noted that volumes have been written about molecular spectroscopy and the reader is referred to the links for more information.) Thus we have the allowable transitions:
DJ = 0, ±1 (7)
Each of the branches represent different D Js. R® D J =1, m=J+1; P® D J =-1, m=-J; and Q® D J =0, m=J. So at last we can write down the equations of our parabolas.
kR = k0 + (Bv'+ Bv") m + (Bv'- Bv") m2 kP = k0 + (Bv'+ Bv") m + (Bv'- Bv") m2 (8) kQ = k0 + (Bv'- Bv") m + (Bv'- Bv") m2
The CH band we are looking at is a 0,0 transition, that is, v' = v" = 0. Having found Bv' and Bv" we can then find Be' and Be" using the known values of ae for the 2S- and 2P states of CH and (5), and thus we calculate re' and re": the internuclear spacings.The rotational temperature of the flame can be found by comparing intensities of peaks with known J' and Bv' values. We use the formula:
I = c (2J' + 1) exp[-Bv' hc J'(J' +1)/kTrot] (9)where c is just some constant and Trot is the (rotational) temperature in Kelvin. All we need to do is measure several intensities (using a suitable picoammeter) to find Trot.
Flame Spectroscopy is of great interest in science and industry (Flame spectroscopy is used in many industrial jobs and is a terrific resume gem.). We can find the temperature, the constituants and other properties of the flame, controled experiments can be predicted. This leads to an expansion in the tools scientists can make use of to understand secretive mother nature.
This page is intended as a brief introduction and is by no means complete. Many of the links give much beter explanations of the physical principles involved.
Well I hope you enjoyed it, it was my first page ever.
~Austin Calder 3-20-2002