Question
Asked Oct 9, 2019
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1.18 A mixture of argon and mercury vapor used in blue
advertising signs emits light of wavelength 470 nm. Calculate the
energy change resulting from the emission of 1.00 mol of photons
at this wavelength
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1.18 A mixture of argon and mercury vapor used in blue advertising signs emits light of wavelength 470 nm. Calculate the energy change resulting from the emission of 1.00 mol of photons at this wavelength

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Expert Answer

Step 1

The energy ‘E’ which is present in a photon is used to represent the smallest possible ‘packet’ of energy in the electromagnetic wave is directly proportional to the frequency of the wave. Planck’s equation is given as follows:

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E-hv hc E = E is the energy of the photon h is the planck's constant c is the speed of the light is the wavelength of the light v is the frequency of the light

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Step 2

The wavelength is given to be 470 nm =470×10-9 m= 47×10-8 m.

The speed of the light is 3×108 m/s

Planck’s constant = 6.626×10-34 m2kg/s

Substituting all the values in the Planck’s equation as follows:

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E-hv hc E=- 6.626x103 m2kg/sx3x10°m/s E= 47x10 m =0.42293 101J

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Step 3

This is the energy of one photon. The energy of one mol of photon can be calcula...

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E-6.022x103 0.42293 1018J -2.5468x10Jmol

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Physical Chemistry

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