Answer :
Comparing the energies of the photons emitted by these two radio stations: the photons from the FM station (E_FM) have a higher energy (6.5 x 10^-23 J) compared to those from the AM station (E_AM, 6.7 x 10^-25 J).
To calculate and compare the energy of the photons emitted by the AM and FM radio stations broadcasting at 1010 kHz and 98.3 MHz respectively, we'll use the formula E = hf, where E is the energy of the photon, h is the Planck's constant (6.63 x 10^-34 Js), and f is the frequency of the radio wave.
First, we'll convert the frequencies into Hz:
AM frequency: 1010 kHz = 1,010,000 Hz
FM frequency: 98.3 MHz = 98,300,000 Hz
Next, we'll use the formula to calculate the energy for each station:
E_AM = (6.63 x 10^-34 Js) x (1,010,000 Hz) = 6.7 x 10^-25 J
E_FM = (6.63 x 10^-34 Js) x (98,300,000 Hz) = 6.5 x 10^-23 J
Comparing the energies of the photons emitted by these two radio stations, we can see that the photons from the FM station (E_FM) have a higher energy (6.5 x 10^-23 J) compared to those from the AM station (E_AM, 6.7 x 10^-25 J). This difference in energy can be attributed to the higher frequency of the FM radio station compared to the AM radio station.
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Complete question:
An AM radio station broadcasts at 1010 kHz, and its FM partner broadcasts at 98.3 MHz. Calculate and compare the energy of the photons emitted by these two radio stations.
