Question #202b2

1 Answer
Jul 16, 2017

I got: 2.5xx10^202.5×1020 photons each second

Explanation:

I am not sure I got the data right but anyway...
You should have a wavelength lambda=0.5xx10^-6mλ=0.5×106m and a power of 100W100W. With these information we can work out the energy of one photon of emitted light as (Einstein's relationship):

E=hfE=hf

where ff is the frequaency that can be found using:

c=lambdafc=λf

with cc the speed of light in vacuum. So we get for one photon:
E=hc/lambda=6.63xx10^-34(3xx10^8)/(0.5xx10^-6)=3.978xx10^-19JE=hcλ=6.63×10343×1080.5×106=3.978×1019J

But the power (energy per second) allows us to find the total number of photons emitted (each one with our calculated energy) as:
n="power"/"energy of one photon"=100/(3.978xx10^-19)=2.5xx10^20n=powerenergy of one photon=1003.978×1019=2.5×1020

photons per second.