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43

The formula you want is called Planck's Law. Copying Wikipedia: The spectral radiance of a body, $B_{\nu}$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. $$ B_\...


35

The thing that makes a mirror a mirror is a that it has a high reflectivity (and is very smooth of course, but that doesn't enter into this issue), but all optical properties including reflectivity are functions of wavelength. The mirror is not reflective in the x-ray band, so it looks like a layer of glass (moderately dense) and a very thin layer of heavy ...


27

X-ray (and gamma rays) are quite penetrating. They can pass through solid matter with much less attenuation than visible light as an example. But that doesn't mean that the attenuation is zero. Put enough "stuff" in the way, and the energy is eventually scattered or absorbed. In the case of the atmosphere, it's "just" air, but there is ...


26

dmckee points out that an ordinary mirror doesn't reflect X-rays, but if you could find an X-ray mirror and put it in your case it would just appear black. When you look at yourself in the mirror you're seeing light from your skin/clothes that hits the mirror and is reflected back towards your eyes. But airport X-ray machines work by passing X-rays through ...


23

It can be a little confusing because there are two conventions. The modern convention is to distinguish x-rays from gamma rays by how they are produced. X-rays are produced by electron energy transitions, typically inner orbital transitions, whereas gamma rays are produced by electromagnetic transitions in the nucleus. Usually, gamma rays have shorter ...


20

You have to distinguish, which interactions take place, when electromagnetic radiation passes through a solid and interacts with it. There is a nice plot on Wikipedia, showing the dielectric response of solids for different wavelengths/frequencies. Basically, as the frequency gets higher, the wavelength becomes shorter, and the molecules or atoms are no ...


18

As others have noted, an ordinary mirror will not reflect x-rays. X-ray mirrors do exist, but will also probably not do what you want here. It's very difficult to manipulate the optical trajectory of an x-ray. The critical angle of typical metal foils at x-ray wavelengths is a few degrees at most -- that means that the x-ray is only reflected if it hits the ...


17

Bones absorb more xrays than soft tissue because of the Calcium in the bones (and the high density helps too, but the same densities of, say, Carbon or Silicon, wouldn't absorb as many x-rays). The high atomic number of Calcium (20), dramatically increases the photoelectric effect, which is the main mechanism of xray absorption by bones. If you go ...


15

FELs produce a coherent, monochromatic, intense light beam that can be collimated with an iris (basically a hole in a large lead block). An optical cavity can be arranged by putting two mirrors around the undulators, spaced so each pass of the electrons constructively interferes (go here and click Watch a Movie on How HIGS Works to see a movie demonstrating ...


13

Yes, X-ray, UV, and even radio waves are made of photons. The difference is the Energy (or equivalently, the wavelength). See the picture of the Electromagnetic spectrum . The different nomination comes from the time of the discovery. Your eyes can see the visible part. the radio waves can be observed with antennas, etc. The only difference is the way we ...


13

Light is composed out of a large ensemble of photons, and photons are quantum mechanical elementary particles. Matter is composed out of atoms and molecules , which have small dimensions and are in the quantum mechanical range. The quantum mechanical "size of interaction region" is given by the Heisenberg uncertainty relation. Even though a photon is an ...


12

The wavelengths of light emitted can be calculated using planks law and the temperature of the object. For your average 100W incandescent light bulb, the filament is 2823 kelvin according to google. The spectral radiance, $B$, is equal to $$\frac{1.2\cdot10^{52}}{\mathrm{wavelength}^{5}\cdot e^{\frac{1.99\cdot10^{43}}{\mathrm{wavelength}\cdot4\cdot10^{26}}...


12

You are missing a crucial aspect of the dynamics of a Free Electron Laser: microbunching. This comes from the fact that although electrons at different energies share basically the same velocity $c$, they have different oscillation amplitudes in the undulator, therefore they shift longitudinally. Since you mentioned the SASE mechanism let me expand around ...


10

For data transmission through optical fibers, what you have to worry about is finding the correct material for transmitting the light over long distances. Consider the following diagram for the attenuation through an optical fiber (from NASA through wikipedia): You can see that there is a minimum in the losses through the fiber in the IR because both the ...


10

There is a lot of confusion in this question. Let me try to clear up some of it. First, the LCLS II is not yet running. The LCLS II is expected to begin operation in 2022. However, SLAC is doing experiments with the original LCLS (Linac Coherent Light Source). This delivers pulses of X-rays at 120 Hz (120 pulses per second). Each pulse is ten billion ...


9

A Faraday cage need not be a continuous conductor — you can make a reasonable Faraday cage out of chicken wire. The rule of thumb is that if the gaps in the conductor are small compared to the wavelength of the electromagnetic wave, the wave "won't notice" and the conductor will appear continuous; if the gaps are bigger than the wavelength, parts of the ...


9

Unfortunately X-ray and gamma mirrors are impossible to build the way you think - mainly because there is much less interaction with the matter comparing to UV - it will go through all materials commonly used for making mirrors. Even for EUV light (wavelength of 13.5nm) building effective mirrors is a royal pain. As wavelength of X-Rays is very small (down ...


8

Since the lattice spacing is about eight angstroms, the issue isn't any sort of unusual lattice spacing. Instead, the issue is that bones are thick. [Another answer points out that the crystals are also pretty small, but I don't think that invalidates the discussion below.] [By NASA/JPL-Caltech/Ames - http://photojournal.jpl.nasa.gov/jpeg/PIA16217.jpg, ...


8

This experiment has sort-of been done. As far as I know, no X-ray satellite has captured an image of the Earth eclipsing the Sun, but HINODE observed an eclipse of the Sun by the Moon at X-ray wavelengths from Earth orbit. The presence of an atmosphere around the Earth would change things only slightly. The atmosphere is opaque to X-rays, so the occulting ...


8

To extend very slightly upon BowlofRed's answer: X-ray photons interact with electrons. More electrons in an atom means more interactions, and more interactions means less penetration. In general terms, this means that the heavier the atom, the more electrons it has and the harder it is for an x-ray photon to pass through a solid chunk of it without getting ...


8

Is water transparent? It seems so in a glass, but a kilometer of water is nearly opaque. The deep ocean is dark. Air at atmospheric pressure is similarly opaque to x-rays on a kilometer scale. It's even more opaque for long-wavelength "soft" x-rays, where you may be troubled by its opacity on a lab bench. But for short-wavelength "hard" x-...


7

There are many ways to detect X-rays. I will list just a few that are (or have been) used in medical imaging. Essentially there are several strategies, but it always involves stopping the radiation and using the energy released to effect a change - chemical or electrical - that can be detected. Photographic film. Exposure to Xrays has a similar effect to ...


7

I think that this is an interesting question and I will try to give a very rough answer (warning: it might include some approximations). Instead of calculating the optical thickness of the D-T plasma, we can try to simply estimate the mean free path $\lambda$ of a neutron in the plasma. It can be estimated as (see for example here for a nice explanation) $$\...


6

If you use "population inversion" as an essential part of the definition of what a laser is, then you're right it's not a laser. But that doesn't deny that the properties of light can be just light from any "laser." So, this is a bit of semantics issue, and less about the physics. Some other "non-laser" coherent light sources: Optical Parametric ...


6

I think the way to read your "hash" ("#") is as "number." What you've plotted here is a differential spectrum: the number of x-ray events observed in each energy window from $E_i$ to $E_i + \Delta E$. The most common way to construct such a spectrum is to build a histogram. But in a histogram, the number of events in each bin depends on the width of the ...


6

The continuous X-ray spectrum comes from Bremsstrahlung radiation, which is the radiation emitted whenever an electric charge is accelerated or decelerated. In this case electrons striking the metal are decelerated by collisions with metal atoms and emit EM radiation as a result. The spectrum is continuous because the electrons experience a range of ...


5

It depends what you consider "radiation". Ultrasound doesn't involve radition. MRI involves radiowaves, which are electro-magnetic radiation (but not ionizing radiation). Proton Imaging and Neutron Imaging do not directly involve electro-magnetic radiation, but the term ionizing radiation is usually defined to include particles that cause ionization ...


5

I can't find details on the pulse energy and duration of the LCLS, but it's entirely plausible the power could be greater than the national grid of a large country. The power is energy divided by time, and if the pulse length is very short then even a modest pulse energy produces an astronomically high power. For example the laser at the National Ignition ...


5

X-rays work by being absorbed in denser material, producing the shadow that is the traditional X-ray image. Medical X-rays are designed to top out looking at calcium-40 ($Z$=20). On an object the size of the Earth, the X-ray shadow isn't going to differ much from the optical shadow (except of course, for the atmosphere). If materials are too dense, or high-$...


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