Atomization Process in the Flame

In the atomic absorption techniques the vapour of free, unionized atoms of the analyte element is led to the optical path of the light beam originating from the radiation source. The success of the analysis depends primarily on the production of the uncombined and unionized atoms in the ground state. 

As the sample solution, in which the analyte element is in the form of a dissolved salt (for example, KCl(s) = K (aq) + Cl (aq)), enters the flame at a temperature of 2000 to 3000 K, the atomization process is considered to occur as follows (i) First the solvent rapidly evaporates and a solid aerosol will be formed (K (aq) + Cl (aq) = KCl(s)) (ii) Then the solid particles melt and vaporize (KCl(s) = KCl(g)). The vaporization is fast, provided the melting and boiling points of the analyte compound are well below the temperature of the flame (iii) The vapour consists of separate molecules or molecule aggregates which tend to decompose into individual atoms (KCl(g) = K(g) + Cl(g)) (iv) Individual atoms may absorb energy by collision and become excited or ionized (K(g) K (g) or K(g) 

The partial pressure of potassium atoms under conventional AA measurement conditions (the sample uptake rate 3 mlmin , and the acetylene and 

This equation can be used to calculate the equilibrium constants at different temperatures between 1800 and 3000 K. The degrees of ionization at various partial pressures can then be calculated using the equation  

Degrees of dissociation of potassium chloride can be derived in an analogous way to the ionization of potassium atoms (Equation 56). Curves IV (100 mgl ) and V (1 mgl ) represent dissociation of KCl. 

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