How Many Photons Are Produced from a Falling Mass?

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SUMMARY

A 0.39 kg mass falling from a height of 2.9 m can produce a specific number of photons if its gravitational potential energy is fully converted into visible light. The energy of the falling mass is calculated using the formula E=mgh, yielding 11.43 J. The energy of a single photon at a wavelength of 5.5 × 10−7 m is determined using E=hc/λ, resulting in approximately 3.618 × 10−10 J. Dividing the total energy by the energy per photon reveals that approximately 3.16 × 1010 photons are produced.

PREREQUISITES
  • Understanding of gravitational potential energy (E=mgh)
  • Knowledge of Planck’s constant (6.63 × 10−34 J·s)
  • Familiarity with the speed of light (3 × 108 m/s)
  • Basic principles of photon energy calculation (E=hc/λ)
NEXT STEPS
  • Study the derivation of gravitational potential energy formulas
  • Learn about the relationship between energy and wavelength in photons
  • Explore applications of Planck’s constant in quantum mechanics
  • Investigate the implications of energy conversion in physical systems
USEFUL FOR

Students in physics, educators teaching energy concepts, and anyone interested in the relationship between mass, energy, and light production.

NRip897
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Homework Statement


A 0.39 kg mass falls from a height of 2.9 m. If all of the energy of this mass could be converted to visible light of wavelength 5.5 × 10^−7 m, how many photons would be produced? The acceleration of gravity is 9.8 m/s^2 . Planck’s constant is 6.63 × 10^−34 J · s, and the speed of light is 3 × 10^8 m/s.

Homework Equations


E=hc/λ
p=h/λ
x=.5at^2

The Attempt at a Solution


E=hc/λ=(1.99x10^-25)/(5.5x10^-7)=3.618x10^-10J
x=.5at^2 > 2.9=.5(9.8)t^2 > t=.769s
 
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NRip897 said:

Homework Statement


A 0.39 kg mass falls from a height of 2.9 m. If all of the energy of this mass could be converted to visible light of wavelength 5.5 × 10^−7 m, how many photons would be produced? The acceleration of gravity is 9.8 m/s^2 . Planck’s constant is 6.63 × 10^−34 J · s, and the speed of light is 3 × 10^8 m/s.

Homework Equations


E=hc/λ
p=h/λ
x=.5at^2

The Attempt at a Solution


E=hc/λ=(1.99x10^-25)/(5.5x10^-7)=3.618x10^-10J
x=.5at^2 > 2.9=.5(9.8)t^2 > t=.769s

Well, I think you can just calculate the initial energy E=mgh, assuming it was at rest before the fall.
You already have the energy of one photon Ephoton=hc/λ, so the number of photons will be just E/Ephoton
 

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