An AM radio transmitter radiates 5.00 times 10 to the third watts of electromagnetic radiation at a frequency of 7.60 times 10 to the third hertz. How many photons per second does the transmitter emit?
We’re told the transmitter gives off 5.00 times 10 to the third watts of power; we’ll call this value 𝑃. We’re also told the radiation has a frequency of 7.60 times 10 to the third hertz; we’ll call this value 𝑓. We want to solve for the number of photons per second emitted by the transmitter, what we’ll call capital 𝑁.
In this example then, we have a radio transmitter giving off radiation with a power 𝑃 at a frequency 𝑓. To start on our solution, we can recall the relationship for the energy of a photon. The energy 𝐸 of an individual photon is equal to Planck’s constant ℎ times the frequency of that photon 𝑓.
In this exercise, we’ll assume that ℎ is equal to exactly 6.626 times 10 to the negative 34th joule seconds. So the energy of a single photon emitted by this radio transmitter equals ℎ times the frequency 𝑓. If we also recall that power 𝑃 is defined as energy per unit time, we can also write that 𝑃, the power radiated by the transmitter, equals 𝐸, the energy of a single photon, times 𝑁, the number of photons per second, which also equals ℎ times 𝑓 times 𝑁 per one second.
Since 𝑃 equals ℎ𝑓𝑁 in one second, we can rearrange and say that 𝑁 equals the power times the time interval, one second, divided by Planck’s constant times 𝑓. We’re now ready to plug in for 𝑃, ℎ, and 𝑓. With those values entered in, when we calculate this fraction, we find it’s equal to 9.93 times 10 to the 32 power. That’s the number of photons emitted by this transmitter every second.