Question #202b2

1 Answer
Jul 16, 2017

Answer:

I got: #2.5xx10^20# photons each second

Explanation:

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

#E=hf#

where #f# is the frequaency that can be found using:

#c=lambdaf#

with #c# 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^-19J#

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)=2.5xx10^20#

photons per second.