Which graph represents the Energy of a photon vs intensity?

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SUMMARY

The discussion centers on the relationship between the intensity of monochromatic blue light and the energy of individual photons. It is established that increasing light intensity does not affect the energy of each photon, which is defined by the equation Energy of a photon = hƒ. Instead, higher intensity results in a greater number of photons, leading to an increased number of ejected electrons. The correct graph representing this relationship is identified as graph E, contrary to the answer key's indication of graph C.

PREREQUISITES
  • Understanding of the photoelectric effect and its implications.
  • Familiarity with the equation KE=hƒ-∅, where ∅ is the work function.
  • Knowledge of the relationship between light intensity and photon quantity.
  • Basic grasp of the concept of photon energy defined by Energy of a photon = hƒ.
NEXT STEPS
  • Study the photoelectric effect in detail, focusing on how intensity affects photon quantity.
  • Explore the implications of the work function in various materials.
  • Learn about the Planck constant (h) and its role in quantum mechanics.
  • Investigate graphical representations of photon energy versus intensity in different contexts.
USEFUL FOR

Students studying physics, educators teaching the photoelectric effect, and anyone interested in the principles of quantum mechanics and light behavior.

YMMMA
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Homework Statement


If X represents the intensity of monochromatic blue light and Y represents the energy of each photon of the light, which one is the correct graph below?

Homework Equations


KE=hƒ-∅, where ∅ is the work function
Energy of a photon= hƒ

The Attempt at a Solution


Increasing the intensity of light does not increase the energy of the photon. However, It does increase the number of photons, which increases the number of ejected electrons (more current). Does that mean the graph is the one with consistent energy in E.
 

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Since "Y represents the energy of each photon of the light," the photoelectrons have nothing to do with the question. The answer is indeed E.
 
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I don’t know why the answer key says it is C, so I had to check on my understanding.
 

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