A diode laser emits at wavelength of 744nm. All of its output energy is absorbed in a detector that measures a total energy of .63J over a period of 47s. How many photons per second are being emitted by the laser. I got 5.0*10^16 photons/second. Back of the book says 5.0*10^13photons/second.
alright one sec.
E=hv... h being planck's constant. v being frequency.
\[\lambda v = c\]
so \[v=c/\lambda \]
\[E=h(c/\lambda )\]
\[E=6.626^{-34}Js*3*10^{8}m/s*\frac{ 1 }{ 744*10^{-9}m }\]
\[2.67*10^{-19}\]
thats what tha equals.
and that's in J/photon
\[\frac{ .63J }{ 47s}*\frac{ photon }{ 2.67*10^{-19} J}\]
\[=5.0*10^{16}\]
photons/second
@Rushwr that's how I went about it i'm assuming that's the process not sure if i plugged in to the calculator properly.
@IrishBoy123 you good with chemistry?
@freckles you good with chem?
no but I can try to look at your problem
try and just plug in the numbers into a calculator and see if you get the correct answer
I have done that and it doesn't calculate to the back of the book \[E=6.626^{-34}Js*3*10^{8}m/s*\frac{ 1 }{ 744*10^{-9}m }\] where does the 6.626^(-34) come from and the 3*10^8 come from ?
6.636*10^-34 is planck's constant.
which is denoted using h.
oh times 10^(-34)
and c is the speed of light.
which is the 3*10^8m/s
I do have one complaint
when you get that first part the units are in joules/photon and then you continue using dimensional analysis to get photon/second
what is it?
\[7.44 \times 10^{-7} m =744 nm\] right?
yes
you have -9
not -7
i account for it by saying 744 though
i kept the decimal where it was given and put 10^-9m
which is what a nanometer equals
oh ok
ok give me a few i'm thinking
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