After applying sunscreen, Cherie lies in the summer sun to get a tan. Theultraviolet light responsible for tanning has a wavelength over 310. nm, whilethe burning rays can range down to 280. nm. Which ultraviolet photons emitmore energy, those that tan or those that burn? How much more?
Answers
Answer:
The photons that burn, emit more energy.
Explanation:
The energy of the ultraviolet photons can be determined by means of the following equation:
E = h\nuE=hν (1)
Equation 1 can be rewritten in terms of \lamba\lamba
Since, c = \nu \lambdac=νλ
\nu = \frac{c}{\lambda}ν= λc
E = \frac{hc}{\lambda}E=λhc (2)
Where h is the planck's constant, c is the speed of light and \lambdaλ is the wavelength.
Case for the photon of with \lambda = 310nmλ=310nm
\lambda = 310nm . \frac{1x10^{-9}m}{1nm}λ=310nm.1nm1x10−9m ⇒ 3.1x10^{-7}m3.1x10−7m
E = \frac{hc}{\lambda}E=λhc
E = \frac{(6.626x10^{-34}J.s)(3x10^{8}m/s)}{3.1x10^{-7}m}E=3.1x10−7m(6.626x10−34J.s)(3x108m/s)
E = 6.412x10^{-19}JE=6.412x10−19J
Case for the photon of with \lambda = 280nmλ=280nm
\lambda = 280nm . \frac{1x10^{-9}m}{1nm}λ=280nm.1nm1x10−9m ⇒ 2.8x10^{-7}m2.8x10−7m
E = \frac{hc}{\lambda}E=λhc
E = \frac{(6.626x10^{-34}J.s)(3x10^{8}m/s)}{2.8x10^{-7}m}E=2.8x10−7m(6.626x10−34J.s)(3x108m/s)
E = 7.099x10^{-19}JE=7.099x10−19J
Hence, the photons that burn, emit more energy.