Question

Question 1:
The fraction of thermally excited atoms in
Flame photometry is determined by
equation​

Answer 1

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.

Answer 2

Concept: Atomization and excitation in spectrochemical analysis is accomplished utilising a spread of emission sources. though the temperature of flames and furnaces (2000–4000 K) is scarce to excite several of the weather, the spectra generated from low-energy sources like flames square measure easier than those derived from electrical discharges. Flame emission spectroscopy continues to be ordinarily used to see alkali components (lithium, sodium, and potassium), whose excitation states square measure low enough to be occupied at flame temperatures. Higher temperatures and therefore a lot of emission lines square measure made by higher-energy sources. Arcs and sparks square measure fashioned by applying currents and potentials across conducting electrodes in electrical discharges, and a substantial portion of the sample surface is gaseous within the method.

Plasma sources like ICP, electricity plasma (DCP), and microwave elicited plasma (MIP) reach temperatures of 7000–8000 K, letting a lot of measure. Metal analysis oft employs glow discharge sources, that use high-energy chemical element atoms and ions to excite atoms expelled from the analyte surface.

Solution: The physicist distribution equation is wont to characterise the degree of excitation caused by a thermal supply. The excited fraction is given by

$\frac{N_{1} }{N_{0} } = \frac{g_{1} }{g_{0} } exp \frac{-E}{kT}$

$T$ is the absolute temperature (K),

$k$ is the Boltzmann constant (1.381023 J K1),

$g_{1}$ and $g_{0}$ are quantum statistical weighting factors,

$E$ is the energy difference between the ground and excited states

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