Calculating number of Photons absorbed by a surface

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SUMMARY

The discussion centers on calculating the number of photons absorbed by a surface when light of frequency 4.41 x 1015 Hz impinges on it, resulting in an energy gain of 1.13 x 10-16 J. The energy of each photon is determined using the formula E = hf, where h is Planck's constant (6.626 x 10-34 J·s). By dividing the total energy absorbed by the energy per photon, the number of photons absorbed can be accurately calculated.

PREREQUISITES
  • Understanding of the equation E = hf for photon energy calculation
  • Familiarity with Planck's constant (6.626 x 10-34 J·s)
  • Basic knowledge of frequency and energy relationships in physics
  • Ability to perform simple algebraic calculations
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  • Calculate the energy of a photon using E = hf for different frequencies
  • Explore the implications of photon absorption in various materials
  • Investigate the concept of quantum efficiency in photonic applications
  • Learn about the role of photons in photoelectric effect experiments
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Students and professionals in physics, particularly those studying quantum mechanics, photonics, or materials science, will benefit from this discussion.

SirNevermind1
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If light with a frequency of 4.41*10^15 Hz falls on a surface that gains 1.13*10^-16 J of energy, how many photons has the surface absorbed?



Probably simple, I've been using E = hf and such, but I got no idea from there



Help meeee please
 
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since you have the frequency of the light, you can calculate the energy of each photon with E = hf. with the energy of each photon and the total energy the surface has absorbed, you can calculate the number of photons by dividing the total energy by the energy of an individual photon.
 
Excellent, thankyou very much
 

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