Question
Asked Nov 26, 2019
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Assuming the wavelength for fluorescence is observed at 475 nm and the wavelength of the phosphorescence process occurs at 523 nm, find the energy released from 12.35 g of Calcium atoms. (Assume that there is 1 excited electron per calcium atom and every one of them relaxes via the processes above.) (Fluorescence occurs 2.45 times more frequently than phosphorescence) 

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Expert Answer

Step 1

The number of moles of calcium atoms present (n) in 12.35 g of the sample is calculated using equation (1) in which m is the given mass and M is the molar mass of calcium atom.

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т n=- . (1) 12.35 g 40.078 g/mol = 0.3081 mol

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Step 2

One mole of particle contains an Avogadro number (NA) of particles. One mole of calcium atom is equivalent to the Avogadro number (6.022×1023 atoms). The number of atoms (N) in 0.3081 mol of calcium atom is calculated as follows:

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6.022 x 1023 atoms 1 mol N 0.3081 mol =1.855x1023 atoms

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Step 3

As one electron is excited per atom, the total number of electrons in the excited state is equal to the number of calcium atoms. The number of electrons in the excited state is 1.855×1023. Consider NP be the number of electrons that decays through phosphorescence. As fluorescence occurs 2.45 times more frequently than phosphorescence, the number of ...

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Np 2.45p N 3.45Np 1.855x 1023 Np 5.377x1022

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