Photons per second from a radio transmitter

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SUMMARY

The discussion centers on calculating the number of photons emitted per second by a radio transmitter operating at 10 kW and a wavelength of 100 m. The key equations involved are E=hv, where E is energy, h is Planck's constant, and v is frequency. The user correctly identifies that to find the number of photons, one must first calculate the energy per photon and then divide the total power output (in Joules per second) by this energy value. This approach leads to the determination of photons emitted per second.

PREREQUISITES
  • Understanding of wave mechanics principles
  • Familiarity with Planck's constant and its application in photon energy calculations
  • Knowledge of power, energy, and their units (Joules and Watts)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Learn about Planck's constant and its significance in quantum mechanics
  • Study the relationship between wavelength, frequency, and energy of photons
  • Explore the concept of power in physics and its units
  • Practice problems involving photon calculations using different wavelengths and power levels
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Students and educators in physics, particularly those focusing on wave mechanics and quantum physics, as well as anyone interested in understanding the relationship between energy and photons in electromagnetic radiation.

Crazy Tosser
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Homework Statement



A radio transmitter radiates at the power of 10 kW at a wavelength of 100 m. How many photons does it emit per second?

Homework Equations



E=hv, and some other equations of beginnings of wave mechanics.

The Attempt at a Solution



I had no idea where to start.Please do not give me the whole solution, but, rather, a hint (big hints) so that I understand how to do problems like that myself. Oh btw, I am completely beginner in WM, so go easy.

Edit: I think I started to figure out some things.

If we know the wavelength, we can find out the energy of the photon, and then calculate how many photons are there to make up the 10 kW. Is that right? And if it is, how exactly would you know the units of photon's energy?
 
Last edited:
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You have the "Joules/sec" for the transmitter. Also, you're aware of how to calculate "Joules/photon" (Joules per photon).

How would you combine the two quantities quoted in the above paragraph, in order to get something with units of "Photons/sec", i.e photons per second?
 
ohhh =D

yessir clear now
 

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