How Many Photons Does a 1-kW Radio Transmitter Emit at 880 kHz?

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Homework Help Overview

The discussion revolves around calculating the number of photons emitted by a 1-kW radio transmitter operating at a frequency of 880 kHz. Participants are exploring the relationship between power, energy, and photon emission in the context of electromagnetic radiation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate power to kinetic energy and frequency but expresses uncertainty about their approach. Some participants suggest using the equation for photon energy, questioning the meaning of Planck's constant in this context.

Discussion Status

Participants are actively engaging with the problem, with some providing relevant equations and definitions. There is a mix of interpretations regarding the application of formulas, and while guidance has been offered, no consensus has been reached on the correct method to proceed.

Contextual Notes

There appears to be some confusion regarding the definitions and relationships between power, energy, and frequency, as well as the specific constants involved in the calculations.

GravityGirl
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a 1-kW raido transmitter operates at 880kHz. How many photons per second does it emit?

Power=work/time
Power=change in KE/time

f=1/T
T=1/880kHz

so 1 kw=change in KE/1.136x10^-6

change in KE=.00136 J

i don't know if i am doing this right. please point me in the right direction.
 
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should i use the equation E(light)=hc/wavelength?

what is h in this equation?
 
should i use the equation E(light)=hc/wavelength?

what is h in this equation?
 
The energy of photon is:

[tex]E=h\nu = hc/\lambda[/tex], h is Planck's constant (6.6x10^-34 Js or 4.1x10^-15 eVs)

So everything that you should do is to divide the power of your transmitter (1 kW) with that energy E.
 

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