91Ó°ÊÓ

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 \(3.69 \times 10^{-7}\, \text{eV}\).

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.