Question

How many photons per second are emitted by a 50-kW radio transmitter that operates at a frequency of 1200 kHz?

Answer 1

In one second, the transmitter uses the energy of

$$E=Pt.$$

One electron has the following energy:

$$E_e=hf.$$

The number of electrons:

$$N=\frac E{E_e}=\frac{Pt}{hf},\\\space\\ N=\frac{50000·1}{(6.63·10^{-34})·(1200000)}=6.29·10^{31}.$$