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A student uses visible spectrophotometry to determine the concentration of CoCl2(aq) in a sample solution. First the student prepares a set of CoCl2(aq) solutions of known concentration. Then the student uses a spectrophotometer to determine the absorbance of each of the standard solutions at a wavelength of 510nm and constructs a standard curve. Finally, the student determines the absorbance of the sample of unknown concentration. A wavelength of 510nm corresponds to an approximate frequency of 6×1014s−1 . What is the approximate energy of one photon of this light? 9×1047J 9×1047J A 3×1017J 3×1017J B 5×10−7J 5×10−7J C 4×10−19J 4×10−19J D Submit

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Answer:

3.97 x 10^-19 J

Explanation:

The energy E of a photon is given by:

       E = hf where h = Planck's constant = 4.14 × 10−15 eV · s and f = frequency

In this case, f = 6×10^14s−1, hence, the energy E of one photon of the light would be;

E = 4.14 × 10−15 eV · s x 6×10^14s−1

               =  2.484 eV

Note that:

1 eV = 1.60 × 10−19 J

Hence,

2.484 eV = 2.484 x 1.60 × 10−19 J

              = 3.97 x 10^-19 J

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