Question

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?

Answer 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:

$E = h\nu$  (1)

Equation 1 can be rewritten in terms of $\lamba$

Since, $c = \nu \lambda$  

$\nu = \frac{c}{\lambda}$

$E = \frac{hc}{\lambda}$  (2)

Where h is the planck's constant, c is the speed of light and $\lambda$ is the wavelength.

Case for the photon of with $\lambda = 310nm$

$\lambda = 310nm . \frac{1x10^{-9}m}{1nm}$ ⇒ $3.1x10^{-7}m$

$E = \frac{hc}{\lambda}$      

$E = \frac{(6.626x10^{-34}J.s)(3x10^{8}m/s)}{3.1x10^{-7}m}$                                                        

$E = 6.412x10^{-19}J$

Case for the photon of with $\lambda = 280nm$

$\lambda = 280nm . \frac{1x10^{-9}m}{1nm}$ ⇒ $2.8x10^{-7}m$

$E = \frac{hc}{\lambda}$      

$E = \frac{(6.626x10^{-34}J.s)(3x10^{8}m/s)}{2.8x10^{-7}m}$                                                        

$E = 7.099x10^{-19}J$  

Hence, the photons that burn, emit more energy.