Question

Although electromagnetic waves can always be represented as either photons or waves, in the radio part of the spectrum we typically do not discuss photons (like we do in the visible) because they are at such a low energy. Nevertheless. they exist. Consider such a photon in a radio wave from an AM station has a 1545 kHz broadcast frequency.

Required:


a. What is the energy, in joules, of the photon?


b. What is the energy, in electron volts. of the photon?

Answer 1

a. E = 1.02*10^-27 J

b. E = 6.39*10^-9eV

Step-by-step explanation:

a. In order to calculate the energy of the radio photon, you use the following formula:

`E=hf` (1)

h: Planck's constant = 6.626*10^-34 Js

f: frequency of the photon = 1545kHz = 1.545*10^6 Hz

Then, by replacing you obtain the energy of the photon:

`E=(6.626*10^(-34)Js)(1.545*10^6s^(-1))=1.02*10^(-27)J`

b. In electron volts, the energy of the photon is:

`E=1.02*10^(-27)J*(6.242*10^(18)eV)/(1J)=6.39*10^(-9)eV`