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

Step-by-step explanation:

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

`E =(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) (m^(2) kg )/(s)` -> Planck's constant

`c = 3*10^(8) (m)/(s)` -> Speed of light

So, replacing in the equation:

`E =( 6.626*10^(-34) (m^(2) kg )/(s)*3*10^(8) (m)/(s))/(514*10^-9 m)`

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

`E = 3.867*10^(-19) (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:

`(1)/(3.867*10^(-19)) = 2.586*10^(18) (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)(photons)/(J) * 1.1 (J)/(s) = 2.844*10^(18) (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) (photons)/(s) = 1.593*10^(17) photons` are released during the surgery