Chemistry, asked by omkargurav251198, 5 months ago

GURAV OMKAR HEMANT
Question 1 :
The fraction of thermally excited atoms in
Flame photometry is determined by
equation.
Boltzmann distribution​

Answers

Answered by sweetyk9040
0

Answer:

Flame Emission Spectrophotometry

Flame emission spectrophotometry is based on the characteristic emission of light by atoms of many metallic elements when given sufficient energy, such as that supplied by a hot flame. The wavelength to be used for the measurement of an element depends on the selection of a line of sufficient intensity to provide adequate sensitivity and freedom from other interfering lines at or near the selected wavelength. For example, lithium produces a red, sodium a yellow, potassium a violet, rubidium a red, and magnesium a blue color in a flame. These colors are characteristic of the metal atoms that are present as cations in solution. Under constant and controlled conditions, the light intensity of the characteristic wavelength produced by each of the atoms is directly proportional to the number of atoms that are emitting energy, which in turn is directly proportional to the concentration of the substance of interest in the sample. Although this technique once was used for the analysis of sodium, potassium, and lithium in body fluids, it has been replaced largely by electrochemical techniques

Answered by aniket6669
0

Answer:

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