Question

An FM radio station broadcasts at a frequency of \(89.3 \mathrm{MHz}\). The power radiated from the antenna is \(50.0 \mathrm{kW} .\) (a) What is the energy in electron-volts of each photon radiated by the antenna? (b) How many photons per second does the antenna emit?

Short answer

Question: Find the energy of each photon radiated by an antenna with a frequency of 89.3 MHz and calculate the number of photons emitted per second if the power radiated from the antenna is 50.0 kW. Answer: The energy of each photon radiated by the antenna is approximately \(3.69 \times 10^{-7}\, \text{eV}\). The antenna emits approximately \(8.45 \times 10^{28}\) photons per second.

Step-by-step solution

Part (a): Finding the energy of each photon

To find the energy of each photon, we will use Planck's relation which relates the energy and frequency of a photon: \(E = h\nu\), where \(E\) is the energy of the photon, \(h\) is the Planck constant, and \(\nu\) is the frequency. Given frequency, \(\nu = 89.3\,\text{MHz} = 89.3 \times 10^6\, \text{Hz}\) Planck's constant, \(h = 6.626 \times 10^{-34}\, \text{Js}\) Now, let's calculate the energy of a photon: \(E = h\nu\) \(E = (6.626 \times 10^{-34}\, \text{Js}) (89.3 \times 10^6\, \text{Hz})\) \(E = 5.919 \times 10^{-26}\, \text{J}\) To convert this energy into electron-volts (eV), we use the conversion factor: \(1\,\text{eV} = 1.602 \times 10^{-19}\, \text{J}\): \(E_\text{eV} = \frac{5.919 \times 10^{-26}\, \text{J}}{1.602 \times 10^{-19}\, \text{J/eV}}\) \(E_\text{eV} \approx 3.69 \times 10^{-7}\, \text{eV}\) So, the energy of each photon radiated by the antenna is approximately \(3.69 \times 10^{-7}\, \text{eV}\).

Part (b): Finding the number of photons emitted per second

To find the number of photons emitted per second, we need to consider the power radiated from the antenna and the energy of each photon. Given power radiated from the antenna: \(P = 50.0\, \text{kW} = 50.0 \times 10^3\, \text{W}\) We have already calculated the energy of each photon as \(E = 5.919 \times 10^{-26}\, \text{J}\). The number of photons emitted per second can be found by dividing the power by the energy of each photon: \(num\_photons = \frac{P}{E}\) \(num\_photons = \frac{50.0 \times 10^3\, \text{W}}{5.919 \times 10^{-26}\, \text{J}}\) \(num\_photons \approx 8.45 \times 10^{28}\) Thus, the antenna emits approximately \(8.45 \times 10^{28}\) photons per second.