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

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SUMMARY

A 1-kW radio transmitter operating at 880 kHz emits photons based on the energy of each photon calculated using the equation E=hc/λ. Planck's constant (h) is 6.626 x 10^-34 Js. The energy of a photon can be derived from the frequency (ν) using E=hν. To find the number of photons emitted per second, divide the transmitter's power (1 kW or 1000 J/s) by the energy of a single photon. This calculation yields the total photon emission rate for the transmitter.

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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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