# Wavelength and penetration by EM radiation

I'd like you to clarify the relationship which relates penetration power by EM radiation with its wavelength (or inverse of frequency).

Suppose we conduct an experiment, irradiating a body with different wavelengths but keeping the amount of power per surface unit constant (so I reckon we have to reduce the power by the emitter as we shorten the wavelength, correct me if I'm wrong). Please appreciate that the body is always the same, so we can leave out the material's characteristics.

We should observe that as we move to higher frequencies, the radiation penetrates less and less into the body (again, tell me if I'm wrong).

First question: why EM radiation is less penetrating at higher frequencies?

However, one knows that the most penetrating form of EM are the gamma rays (that is, the radiation with the shortest wavelength), which seems to be in contrast with the above observation.

So I gather that gamma rays are highly penetrative just because of the high energy they carry by the vitue of their high frequency. If we could manage to keep constant the net amount of electromagnetic power per surface unit, gamma rays would be LESS penetrating than, say, visible light.

Am I correct?

• It depends strongly what energy range you are in. Microwave? Visible? Medical x-ray? Something else? – user93146 Jul 3 '18 at 15:39
• The wave types you listed are characterized by wavelengths, more than energy. I'd like to view the problem from a wavelength/frequency POV. A question: if we fix the electromagnetic power per surface unit, does the energy stay fixed as well? – MadHatter Jul 3 '18 at 16:40
• "I reckon we have to reduce the power by the emitter as we shorten the wavelength, correct me if I'm wrong". If you are referring to the fact that $E = h \nu$, so as $\nu$ increases, $E$ increases too: $E = h \nu$ is the energy of a single photon, not the energy of the whole wave. The energy of the wave is proportional to the amplitude of the Electric (or alternatively Magnetic) field: you must only keep this amplitude constant in order to have constant energy: so, you should not reduce the power as frequency increases. – BowPark Jul 3 '18 at 17:07
• Uhm, Understood. It makes sense, too: a vibrating string is intuitively more energetic as it vibrates at high amplitude. But then, why they say that high frequency radiation is more energetic? It just relates to the sum of single photons? Please, eleborate a bit further about energy of the wave vs energy of the photons. Thanks. – MadHatter Jul 3 '18 at 19:44
• @BowPark The amplitudes of the field components of a photon as well as of EM radiation are stretched to infinity according to the theory. The amplitude of the EM radiation is associated with the intensity of the radiation. – HolgerFiedler Jul 3 '18 at 20:26

Visible light has wavelength of 400-700nm.

Gamma rays have a wavelength of 10^-12m.

In classical view, a surface will be reflective if it is smooth enough, and if the wavelength is much bigger then the molecular and atomic structure. In this case, for visible light, the atoms are much smaller then the wavelength, and so for visible light, the surface looks smooth. So it can be modeled classically, and depending on the material, light might be:

1. absorbed,

2. decoherently reflected by many point sources

3. coherently reflected in elastic scattering (mirror)

When you take gamma rays, you have to use the micro level, and QM, because if you try to see at the micro level if the surface is still smooth for wavelengths of 10^-12m, you will see it is not smooth.

So it cannot be modeled classically, and you have to use QM for gamma rays.

The spacing between atoms is a few tenth of a nanometer, so for visible light (400-700nm), it will appear smooth.

But for gamma rays, classically, they will see mostly empty space between the atoms. That is why higher frequencies go into the material, because they see empty space.

So you have to use QM, and Heisenberg uncertainty principle, gamma rays photons will see empty space.

• So, you are basically stating that all the Classical (Maxwell) computations are no more valid for gamma-rays? For example, refer to the skin depth $\delta$ and its expression $\sqrt{2/(\omega \mu_0 \sigma)} \ \mathrm{m}$ in this document: is it not applicable for the frequencies $\omega$s of the gamma-rays? – BowPark Jul 3 '18 at 17:02
• This document is about EM waves inside material. It does not speak about EM waves' (visible or gamma rays) refraction or reflection. It does not speak about when there is a boundary between two medias, like vacuum and a metal. – Árpád Szendrei Jul 3 '18 at 17:10
• I'm beginning to understand. For the sake of simplicity, let's limit ourselves to the cases where the body is smooth with respect to the wavelength. For example, visible light and microwaves. Is it correct to say that macrowaves will penetrate more in depth wrt visible light, and radio waves even futher? If so, why? – MadHatter Jul 3 '18 at 18:12
• Moreover, is it correct to say that once the wavelength begins to be substantially shorter than the average atomic distances, the relationship is inverted? E.g. gamma rays will be more penetrating than X-rays? – MadHatter Jul 3 '18 at 18:14
• @ÁrpádSzendrei Sorry, my comment was not meant to doubt about your statements, but to better understand them. Putting aside for a moment the reflection/refraction issues, assume that a gamma-ray wave already exists inside a material: does it "violate" that skin depth law, because the spacing of the atoms is not "small" when compared to the wavelength? – BowPark Jul 3 '18 at 21:01

Being a comment too short for this topic, I write an answer, with the necessary premise that it is only about this observation (so it could not be a full answer to your post):

I reckon we have to reduce the power by the emitter as we shorten the wavelength, correct me if I'm wrong.

A neat distinction must be made between

• A Quantum Mechanics topic: photons. Each of them represents a quantum of energy, $E = h \nu$ Joules, if the Electro-Magnetic field is a sinusoidal wave with frequency $\nu$. The observed Electro-Magnetic field is a continuous flow of these quanta. In other words, when an Electro-Magnetic wave is observed, it is like all its photons are all glued to each other.
• A Maxwell's classical theory topic: energy. The "global" energy carried by an Electro-Magnetic field is usually denoted as $u = \frac{1}{2} \epsilon | \mathbf{E} |^2$ (Joules per unit volume). Let an Electro-Magnetic wave go through a certain volume $V$ of the three-dimensional space: if you could absorb all the energy delivered by the Electro-Magnetic wave into the volume $V$, you would absorb $u \cdot V$ Joules of energy.

Photons are about how the Electro-Magnetic energy is delivered: it is delivered by separated "particles" (fragments) called photons, each of them carrying a quantity $h \nu$ of energy. When you absorb energy from the Electro-Magnetic field, you can absorb only an integer multiple of $h \nu$ Joules per time, because you can gather only a discrete number of photons.

When instead you evaluate the field amplitude $| \mathbf{E} |$, you are referring to the whole set of the photons you are receiving, with its global properties. You are not interested in the granularity of this energy (the minimum "quantum" of energy you can absorb), but in the behaviour of the whole wave represented by the field. This wave is capable to fill a volume $V$ with a global amount of energy $u \cdot V$ Joules (you may know that this quantity $u \cdot V$ is being carried there by fragments of $h \nu$ Joules, but you are not interested in this, now).

The global energy $u = \frac{1}{2} \epsilon | \mathbf{E} |^2$ is about the number of photons being delivered, $E = h \nu$ is about the energy amount carried by a single photon. If you want an amount of $A$ Joules of energy, you can obtain it with a large number of low-energy (so, low frequency) photons, or with a small number of high-energy (high frequency) photons. But the final amount of energy will always be $A$ Joules.

So, yes, an Electro-Magnetic field with a big $| \mathbf{E} |$ is like a "string" which is having a big vibration: it can deliver a big amount of energy. When you evaluate the global energy of the wave, you do not wonder how big are its fragments: you only care about their final sum.

A high frequency vibration is more energetic only in the sense that its energy fragments, the photons, are bigger.

Suppose that you can generate two fields:

• a sinusoidal field at frequency $\nu_1$;
• a sinusoidal field at frequency $\nu_2 \gg \nu_1$.

You want to fill with them the volume $V$ with a global energy

$$u_{\mathrm{desired}} \cdot V$$

This is achieved only if both fields, regardless of the frequency and so regardless of how energetic are their photons, have a squared amplitude

$$| \mathbf{E} |^2 = \frac{2 u_{\mathrm{desired}}}{\epsilon}$$

Sorry if it was long, I hope anyway it was a little useful.

Edit 1: Your questions are all very good and absolutely licit, but I think this discussion would make more sense with a basic knowledge of Electromagnetism. You can begin with Coulomb's law. While these concepts will not cancel your doubts, they will make you more able to deal with them, also as regards the linked document.

The photons from the second field [at frequency] $v_2$ are, however, "bigger" as you said. Will $v_2$ carry more energy?

No, because the global energy inside the volume $V$ is $u_{\mathrm{desired}} \cdot V$. The number of photons inside the volume $V$ is such that the sum of their single energies is $u_{\mathrm{desired}} \cdot V$: this is regardless of their frequency and can happen both for the field at frequency $\nu_1$ and the field at frequency $\nu_2$. What does change is instead the number of photons. The field at frequency $\nu_1$ will need to send inside the volume $V$ more photons than the field at frequency $\nu_2$ to reach the same amount of energy $u_{\mathrm{desired}} \cdot V$, because each single photons at frequency $\nu_1$ carry less energy than the photons at frequency $\nu_2$. But the global energy in both cases is $u_{\mathrm{desired}} \cdot V$. And this is an expected result, not weird in this case.

Edit 2 and disclaimer: these concepts are extremely more complex than this trivial, naive, elementary description. The language here used is only inteded to ease the comprehension: it is neither rigorous nor exhaustive and does not claim to be a Quantum Mechanics representation of this problem. This answer only aims at using a suitable and qualitative language for the OP.

Edit 3: This answer and the original question have been downvoted. This is when

the post contains wrong information, is poorly researched, or fails to communicate information

Together with the OP, we are doing our best to deal with a legitimate and meaningful question. If some hint about what we can further improve were given, it could be useful.

• It was useful indeed. But what is epsilon?? Moreover, consider the two fields above, and say they got the same amplitude. The photons from the second field v_2 are, however, "bigger" as you said. Will v_2 carry more energy? It seems not, from what you wrote above, but I find it strange. – MadHatter Jul 3 '18 at 22:37
• @MadHatter I edited the answer trying to write about it. – BowPark Jul 4 '18 at 8:05
• Thank you, I'm digging inside the linked document right now. You edits were useful, as well. As for the downvote, it would be helpful if the downvoter could add a comment for explaining why he downvoted both the question and the answer. – MadHatter Jul 4 '18 at 14:13

Gamma rays are far more penetrating than alpha and beta rays. However they do not penetrate very deeply in materials compared to optical rays. That is because they will inelastically scatter off electrons. They do penetrate more deeply into metals than optical rays, which are stopped by the plasma oscillation phenomenon. Gamma rays lose half their power in lead after just one cm or less. Of course for a very intensive gamma source a thick layer of lead is needed to reduce the intensity to the very low safe level.