How Does Ozone Absorb UV Light's Most Energetic Photons?

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SUMMARY

Ozone effectively absorbs ultraviolet (UV) light with wavelengths ranging from 2200 to 2900 angstroms, providing essential protection against harmful UV radiation. The energy of the most energetic photons in this range can be calculated using the formula E = h*v, where h is Planck's constant. Calculations reveal that the energy of photons at 2200 angstroms is approximately 9.0E-19 Joules, while at 2900 angstroms, it is about 6.9E-19 Joules. This demonstrates that shorter wavelengths correspond to higher photon energy, confirming the inverse relationship between wavelength and energy.

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  • Understanding of Planck's constant (h = 6.626 x 10^-34 J*s)
  • Familiarity with the relationship between wavelength and frequency (c = λ*v)
  • Basic knowledge of photon energy calculations
  • Concept of ultraviolet (UV) radiation and its effects
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Soaring Crane
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Ozone absorbs light having wavelengths of 2200 to 2900 angstroms protecting us from UV radiation. What are the frequencies and energy of the most energetic of these photons?

This is my workings:

E = h*v
lambda*v = c

2200 A *(10^-10 m / 1 A) = 2.2E-7 m
2900 A *(10^-10 m / 1 A) = 2.9E-7 m

v = (3.0 * 10^8 m/s)/lambda

v = (3.0 * 10^8 m/s)/2.2E-7 m = 1.4E15 s^-1
v = (3.0 * 10^8 m/s)/2.9E-7 m = 1.0E15 s^-1

E = (6.626*10^-34 J*s)*1.4E15 s^-1 = 9.0E-19 J Most energetic?
E = (6.626*10^-34 J*s)*1.0E15 s^-1 = 6.9E-19 J

Is this what is being asked?

Thanks.
 
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Energy of a photon is proportional to its frequency \nu, with Planck's constant being the proportionality constant, i.e. Ephoton= h \nu, and the frequency is inversely proportional to wavelength, \nu =c/\lambda,

so E = h c/\lambda,

So the energy of a photon is inversely proportional to wavelength, i.e. the shorter the wavelength, the greater the energy (or higher the frequency).
 

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