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Zinc Emission Spectrum

Scientists can analyze metals using the emission spectrum produced when the sample is introduced into the flame of an emission spectrometer. The flame provides the energy to excite the electrons of the metal atoms to higher energy states. When the electrons return to the ground state, lines of characteristic wavelengths are produced. The lines in the emission spectrum are characteristic of the metal because each atom’s ground-state electron configuration is unique.

The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 8.50×1010atoms of zinc emitting light in the instrument flame at any given instant, what energy (in joules) must the flame continuously supply to achieve this level of emission?

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Posted on October 10, 2013, in Question and tagged , , , , , , , . Bookmark the permalink. 1 Comment.

  1. E = hc / lambda
    [(3.00*10^8)(6.626*10^-34)]/(214*10^-9) = 9.29*10^-19 J
    (9.29*10^-19)/(8.50*10^10) = 7.9*10^-8 J

    🙂

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