Answer:
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}](https://tex.z-dn.net/?f=E%20%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
Where
is the wavelength,
is the Planck's constant and
is the speed of light
-> Planck's constant
-> 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}](https://tex.z-dn.net/?f=E%20%3D%5Cfrac%7B%206.626%2A10%5E%7B-34%7D%20%5Cfrac%7Bm%5E%7B2%7D%20kg%20%7D%7Bs%7D%2A3%2A10%5E%7B8%7D%20%5Cfrac%7Bm%7D%7Bs%7D%7D%7B514%2A10%5E-9%20m%7D)
Then, the energy of each released photon by the laser is:
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:
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}](https://tex.z-dn.net/?f=2.586%2A10%5E%7B18%7D%5Cfrac%7Bphotons%7D%7BJ%7D%20%2A%201.1%20%5Cfrac%7BJ%7D%7Bs%7D%20%3D%202.844%2A10%5E%7B18%7D%20%5Cfrac%7Bphotons%7D%7Bs%7D)
And by doing a simple rule of three, if
are released every second, then in 0.056 s:
are released during the surgery