After applying sunscreen, Cherie lies in the summer sun to get a tan. The ultraviolet light responsible for tanning has a wavelength over 31

After applying sunscreen, Cherie lies in the summer sun to get a tan. The ultraviolet light responsible for tanning has a wavelength over 310. nm, while the burning rays can range down to 280. nm. Which ultraviolet photons emit more energy, those that tan or those that burn?

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  1. Answer:

    The photons that burn, emit more energy.

    Explanation:

    The energy of the ultraviolet photons can be determined by means of the following equation:

    [tex]E = h\nu[/tex]  (1)

    Equation 1 can be rewritten in terms of [tex]\lamba[/tex]

    Since, [tex]c = \nu \lambda[/tex]  

    [tex]\nu =  \frac{c}{\lambda}[/tex]

    [tex]E = \frac{hc}{\lambda}[/tex]  (2)

    Where h is the planck’s constant, c is the speed of light and [tex]\lambda[/tex] is the wavelength.

    Case for the photon of with [tex]\lambda = 310nm[/tex]

    [tex]\lambda = 310nm . \frac{1×10^{-9}m}{1nm}[/tex] ⇒ [tex]3.1×10^{-7}m[/tex]

    [tex]E = \frac{hc}{\lambda}[/tex]      

    [tex]E = \frac{(6.626×10^{-34}J.s)(3×10^{8}m/s)}{3.1×10^{-7}m}[/tex]                                                        

    [tex]E = 6.412×10^{-19}J[/tex]

    Case for the photon of with [tex]\lambda = 280nm[/tex]

    [tex]\lambda = 280nm . \frac{1×10^{-9}m}{1nm}[/tex] ⇒ [tex]2.8×10^{-7}m[/tex]

    [tex]E = \frac{hc}{\lambda}[/tex]      

    [tex]E = \frac{(6.626×10^{-34}J.s)(3×10^{8}m/s)}{2.8×10^{-7}m}[/tex]                                                        

    [tex]E = 7.099×10^{-19}J[/tex]  

    Hence, the photons that burn, emit more energy.

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