An FM radio transmitter has a power output of  120 kW and operates at a frequency of 87.5 MHz. How many photons per second does the transmitter emit?

To solve this problem, we need to use the formula for the energy of a photon, which is E = hf, where h is Planck's constant and f is the frequency. We also need to know that power is energy per unit time, so we can write P = E/t.

Step 1: Convert the power output to watts. 1 kW = 1000 W, so 120 kW = 120,000 W.

Step 2: Convert the frequency to Hz. 1 MHz = 1,000,000 Hz, so 87.5 MHz = 87,500,000 Hz.

Step 3: Substitute the values into the formula for the energy of a photon. Using the value for Planck's constant h = 6.626 x 10^-34 Js, we get E = (6.626 x 10^-34 Js)(87,500,000 1/s) = 5.798 x 10^-26 J.

Step 4: Substitute the values into the formula for power. We get P = E/t, so t = E/P = (5.798 x 10^-26 J)/(120,000 W) = 4.832 x 10^-31 s.

Step 5: To find the number of photons per second, we divide the power by the energy per photon. This gives us (120,000 J/s)/(5.798 x 10^-26 J/photon) = 2.07 x 10^31 photons/s.

So, the transmitter emits approximately 2.07 x 10^31 photons per second.