In electromagnetic waves, what exactly do changes in frequency and amplitude do? [duplicate]

Question

I recently learned about how lower frequency electromagnetic radiation is less harmful compared to photons above the ultraviolet type C frequency which have ionizing radiation because they contain lots of energy due to their high frequency.

I'm curious now about what amplitude and frequency change interactions between matter and photons. I'm guessing that amplitude changes the amount of heat transferred from photons to matter, and frequency amplifies this effect until it reaches the UV threshold where it starts to knock off electrons from atoms, causing even more damage.

I'm probably wrong about all of this but it made sense to me because the same intensity of visible light should heat something up more than the same intensity of radio waves(?).

Please correct me if I'm wrong about any of this, I'm still in high school and I only know surface-level information about this.

Answer 1

In quantum mechanics, we learn that the energy E of a single photon with frequency f is E=hf, where h is planck's constant (if you are unfamiliar with h, you can think of it simply as the proportionality factor for the frequency and energy of a photon). However, in classical mechanics, we learn that the energy is proportional to the square of the amplitude of the light wave- i.e., E∝$A^2$. So what gives a more accurate depiction of reality for ionizing radiation, the quantum interpretation or the classical interpretation? Well, if an electromagnetic wave is ionizing, it means that a single photon interacting with a single electron has sufficient energy to detach the electron from the nucleus, meaning that E=hf is the important equation here. Since only one photon can interact with the electron at a time, bombarding an electron with many low energy photons will never lead to ionization.

However, if instead you were concerned with with the energy donated to a macroscopic object in the form of heat by electromagnetic radiation- e.g., a hot pocket in a microwave, the intensity of the wave (roughly, the number of photons hitting a unit area) matters just as much as the energy per photon. In this case, twice as many photons per unit area equals twice as much energy per unit area, and since quantum mechanics tells us that the square of the amplitude $A^2$ is equal to the number of photons in the wave, this leads to something analogous to the classical result, E∝$A^2$.

The physical interpretation of light is not as faithful to quantum mechanics as it could be, but I hope it is sufficient to aide your growing understanding of light phenomena.