Number of emitted photons per second

Click For Summary
SUMMARY

The discussion focuses on calculating the number of emitted photons per second from a 300mW LED emitting light at an average wavelength of 500nm with an efficiency of 66%. The correct formula for energy per photon is established as E=hf, where h is Planck's constant and f is the frequency derived from the speed of light divided by the wavelength. The final calculation confirms that the number of emitted photons is approximately 5x1017 per second.

PREREQUISITES
  • Understanding of Planck's constant (h) and its application in photon energy calculations
  • Knowledge of the relationship between wavelength, frequency, and the speed of light
  • Familiarity with efficiency calculations in energy conversion
  • Basic proficiency in algebra for manipulating equations
NEXT STEPS
  • Study the derivation and application of Planck's constant in quantum mechanics
  • Learn about the relationship between wavelength and frequency in electromagnetic waves
  • Explore efficiency metrics in LED technology and their impact on light output
  • Investigate advanced photon counting techniques and their applications in photonics
USEFUL FOR

Students in physics or engineering, optical engineers, and anyone interested in the principles of light emission and photon calculations.

Coldslooks
Messages
2
Reaction score
0

Homework Statement


The average wavelength that LED emits is 500nm. The electrical power of the LED is 300mW and the efficiency of turning energy into light is 66%. Number of emitted photons in unit time is?

Homework Equations


##E_f=hcf##
##\frac{Pt}{E_f}=number of photons##

The Attempt at a Solution


After calculating the energy of one photon i am to find the overall energy that is transformed into light. The energy is ##300x10^{-3}J## but how am i to include the efficiency? Just multiply the Energy with 0.66 and use that in the equation? The answer is suppose to yield ##5x10^{17} per second##
 
Physics news on Phys.org
Coldslooks said:

Homework Equations


##E_f=hcf##
First, this must make no sense. The correct one should be ##E=hf.##
Since you can get the energy of a photon, then given the amount of light you can calculate the number of the photons. Like what you said, the energy transferred to light per second is ##300mJ\cdot 66\%.##
 
  • Like
Likes   Reactions: Coldslooks
tommyxu3 said:
First, this must make no sense. The correct one should be ##E=hf.##
Since you can get the energy of a photon, then given the amount of light you can calculate the number of the photons. Like what you said, the energy transferred to light per second is ##300mJ\cdot 66\%.##
Yeah, your right, i haven't seen that I've put ##c## there, thanks.
Well then, ##\frac{E_t}{E_f}=\frac{198*10^{-3}J}{hf}##
##f=\frac{c}{w}=6*10^{14}## yes the rest is clear thanks :)
 

Similar threads

Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
Replies
7
Views
4K
Replies
5
Views
3K
Replies
2
Views
10K
  • · Replies 11 ·
Replies
11
Views
17K