Particles Question from physics and maths tutor
A light emitting diode (LED) emits blue light with a wavelength of 440 nm. The rate of photon emission is 3.0 × 10^16 s−1 . Show that the power output of the LED is approximately 0.014 W.
If somebody could provide a worked solution that would be fantastic as the mark scheme is quite vague and I am getting different values for the power output. Thanks!
If somebody could provide a worked solution that would be fantastic as the mark scheme is quite vague and I am getting different values for the power output. Thanks!
Reply 1
E=hf
E=hc/lambda
each photon has this amount of energy which you can calculate
so E ~4.5x10^-19J
and there are 3x10^16 photons released per second
so in one second
there are 3x10^16 photons released each with energy E
so total energy in one second (definition of power) = 3x10^16*E ~ 0.014W
E=hc/lambda
each photon has this amount of energy which you can calculate
so E ~4.5x10^-19J
and there are 3x10^16 photons released per second
so in one second
there are 3x10^16 photons released each with energy E
so total energy in one second (definition of power) = 3x10^16*E ~ 0.014W
Reply 2
Original post
by mosaurlodonE=hf
E=hc/lambda
each photon has this amount of energy which you can calculate
so E ~4.5x10^-19J
and there are 3x10^16 photons released per second
so in one second
there are 3x10^16 photons released each with energy E
so total energy in one second (definition of power) = 3x10^16*E ~ 0.014W
E=hc/lambda
each photon has this amount of energy which you can calculate
so E ~4.5x10^-19J
and there are 3x10^16 photons released per second
so in one second
there are 3x10^16 photons released each with energy E
so total energy in one second (definition of power) = 3x10^16*E ~ 0.014W
very helpful, thanks a lot mate! 😀
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