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2007-01-29 00:49:08 · 3 answers · asked by red_0rient 1 in Science & Mathematics Chemistry

3 answers

Outermost electrons of an atom when reach in excited state,they become unstable and try to come back to the ground state in order to become stable.Thus , they releases energy in the form of spectra and spectral lines of different series are obtained.
This is known as atomic emission spectra.
BYE...........

2007-01-29 01:20:15 · answer #1 · answered by john d 1 · 0 0

Atomic emission spectroscopy (AES or OES) uses quantitative measurement of the optical emission from excited atoms to determine analyte concentration. Analyte atoms in solution are aspirated into the excitation region where they are desolvated, vaporized, and atomized by a flame, discharge, or plasma. These high-temperature atomization sources provide sufficient energy to promote the atoms into high energy levels. The atoms decay back to lower levels by emitting light. Since the transitions are between distinct atomic energy levels, the emission lines in the spectra are narrow. The spectra of multi-elemental samples can be very congested, and spectral separation of nearby atomic transitions requires a high-resolution spectrometer. Since all atoms in a sample are excited simultaneously, they can be detected simultaneously, and is the major advantage of AES compared to atomic-absorption (AA) spectroscopy.

2007-01-29 00:54:44 · answer #2 · answered by ? 2 · 1 0

calculate the energy corresponding to 1 photon of 214 nm E=hc/wavelength= 6.626*10^-34*3*10^8/214*10^-9=9.289*10^-... J and for 8.5*10^10 atoms the energy = 9.289*10^-19*8.5*10^10=7.895*10^-8J

2016-03-29 07:54:21 · answer #3 · answered by Anonymous · 0 0

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