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

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Ozone effectively absorbs UV light in the wavelength range of 2200 to 2900 angstroms, which protects the Earth from harmful radiation. The discussion calculates the frequencies and energies of the most energetic photons within this range, determining that the highest energy photon corresponds to a wavelength of 2200 angstroms, yielding an energy of approximately 9.0E-19 J. The relationship between photon energy and frequency is clarified, emphasizing that energy is inversely proportional to wavelength; shorter wavelengths result in higher energy photons. The calculations confirm the expected outcomes based on established physical principles. Understanding these interactions is crucial for comprehending ozone's protective role in the atmosphere.
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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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