Friday, October 10, 2014

How many photons are produced in a laser pulse of 0.188 J at 571 nm?

A photon of frequency f has the energy E = hf, where h is the Planck's constant:


`h = 6.63*10^(-34) J*s` .


The frequency is related to the wavelength as `f = c/lambda` , where c is the speed of light: `c = 3*10^8 m/s^2` . So the photons in the laser pulse with the wavelength of 571 nm have frequency


`f = c/lambda = (3*10^8)/(571*10^(-9)) = 5.25*10^14 s^(-1)` .


Each of these photons then have the energy of `E = hf = 6.63*10^(-34)*5.25*10^14 = 34.8*10^(-20) J = 3.48*10^(-19) J` .


So, if the total energy of the pulse is 0.188 J, this pulse contains the number of photons equal to the total energy divided by the energy of a single photon:


`0.188/(3.48*10^(-19)) = 0.054*10^19 = 54*10^16` photons.


The laser pulse of the given energy and wavelength contains


540,000,000,000,000,000 photons.

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