the number of photons emitted per second by the laser? (Plank’s constant Light of wave length 6730 Å is emitted by a He-Ne laser with output power of 2.3 mW. Calculate h = 6.63 × [10]^(-34) J)

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May 30, 2016

The intensity or power output of a laser (#P#) is equal to #nhf#, where #n# is the number of photons emitted per second, #h# is Planck's constant and #f# is the frequency.

Therefore, by rearranging:
#n=P/(hf)#

Since we don't have frequency but rather wavelength, we can use #c=f lambda# (rearranged to give #f=c/lambda#) to give a final, more useful equation:
#n=P/(hc/lambda)#

One further rearrangement, though you could use the above form:
#n=(P lambda)/(hc)#

We need to get everything into the correct units before we apply the equation.
The power output is #2.3mW#, this must be in Watts (so #2.3*10^-3# Watts).
The wavelength must be in metres not Ångströms and since #1# Ångström #=10^-10# metres, then our wavelength in metres is #6370 * 10^-10#.
#c# represents the speed of light #3*10^8 ms^-1#
#h# is Planck's constant - #6.63*10^-34 Js#

So finally, #n=(2.3*10^-3*6370*10^-10)/(6.63*10^-34*3*10^8)#
#n=7.37*10^15 s^-1#

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