Photon flux from a light source

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SUMMARY

The photon flux at a distance of 1 km from a light source emitting 50W of radiation in the visible domain, specifically at a wavelength of 6000 Angstroms, is calculated to be 1.20 x 1013 photons/m2s. The energy of one photon is determined using the formula E = hc/λ, resulting in an energy value of 3.32 x 10-19 J. The calculation of the number of photons per second is confirmed to be 1.51 x 1020, with a consensus that this value was initially misrepresented as a typo.

PREREQUISITES
  • Understanding of photon energy calculations using the formula E = hc/λ
  • Knowledge of basic physics concepts related to light and radiation
  • Familiarity with units of measurement in photonics (e.g., photons/m2s)
  • Ability to perform calculations involving constants such as Planck's constant (h) and the speed of light (c)
NEXT STEPS
  • Study the derivation and application of the formula E = hc/λ in various contexts
  • Explore the concept of photon flux and its relevance in different fields such as optics and photonics
  • Learn about the effects of distance on light intensity and photon distribution
  • Investigate the implications of wavelength on photon energy and flux calculations
USEFUL FOR

Students in physics, optical engineers, and anyone interested in the quantitative analysis of light sources and their effects on photon distribution.

spaghetti3451
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Homework Statement



What is the photon flux (photons/ m2 s) at a distance of 1 km from a light source emitting 50W of radiation in the visible domain, with wavelength 6000 Angstroms.

Homework Equations



The Attempt at a Solution



Energy of one photon = hc/λ = 3.32 x 10-19J.
So, number of photons per second = 50/3.32 x 10-19 = 1.51 x 10-20.
So, photon flux = 1.51 x 10-20/4π(1000)2 = 1.20 x 1013.

Is this correct?
 
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1.51*10-20 photons per second? I think that is a typo.

The approach is good, and the numbers look reasonable.
 
mfb said:
1.51*10-[/size][/color]20 photons per second? I think that is a typo.

Yeah! That is a typo!

mfb said:
The approach is good, and the numbers look reasonable.

Thanks!
 

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