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

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A 1-kW radio transmitter operating at 880 kHz emits photons based on its power output and the energy of each photon. The energy of a photon can be calculated using the equation E = hν or E = hc/λ, where h is Planck's constant (approximately 6.626 x 10^-34 Js). To find the number of photons emitted per second, divide the transmitter's power (1 kW) by the energy of a single photon. The discussion emphasizes the need for clarity on the calculations and the correct application of the relevant equations. Understanding these principles is crucial for accurately determining photon emission rates.
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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:

E=h\nu = hc/\lambda, 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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