Question

Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how many photons are released if the laser is used for 0.056 s during the surgery?

Answer 1

Answer: $1.593*10^{17} photons$ released if the laser is used 0.056 s during the surgery

                           

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

                            $E =\frac{hc}{\lambda}$

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

              $h = 6.626*10^{-34} \frac{m^{2} kg }{s}$  -> Planck's constant

              $c = 3*10^{8} \frac{m}{s}$  -> Speed of light

So, replacing in the equation:

                $E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}$

Then, the energy of each released photon by the laser is:

                $E = 3.867*10^{-19} \frac{J}{photons}$

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

               $\frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}$

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

              $2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}$

And by doing a simple rule of three, if $2.844*10^{18} photons$ are released every second, then in 0.056 s:

            $0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons$ are released during the surgery