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40

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. $$ ...


30

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 ...


25

The X-rays come from hot gas orbiting around the black hole in an accretion disk. As the gas orbits, magnetic stresses cause it to lose energy and angular momentum, thus spiralling slowly in towards the black hole. The orbital energy is transformed into thermal energy, heating up the gas to millions of degrees, so it then emits blackbody radiation in the ...


24

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 ...


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 ...


11

Yes, X-ray, UV, and even radio-waves ares made of photons. The differences is the Energy (or equivalently the wavelength). See the picture of the Electromagnetic spectra . The different nomination comes from the time of the discovery. Youre eyes can see the visible part. the radiowaves can be observes with antenna etc... The only differences is the way we ...


11

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 ...


10

My understanding is that any radiation emitted "from a black hole" is actually originating from or just outside the event horizon of the black hole. This is the region beyond which the escape velocity is at or above the speed of light (electromagnetic radiation). Outside the event horizon, escape is still possible. There is another kind of radiation, ...


8

The goal of such a treatment is to induce damages in the cells of the tumor by mean of ionizing radiation. These radiations can be X-rays (photons), electron, proton or things like carbon ions. The problem is: if you try to irradiate a tumor, you first have to go through normal tissues and the risk is to damage them also. Photons will transfer energy ...


8

X-rays are electromagnetic waves, just as light rays are. The difference is in the wavelength (thus frequency and Energy ). So your question has the same answer as "What happens if you shine light on light" or "What happens if you point a light ray at a light ray". Classically, you will see the same effects you see with usual light rays, interference, ...


7

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 ...


5

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 ...


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 believe most of the em-spectrum from a nuke is low energy. So when you take a photo you get a lot of visible light and heat, and only small amounts of high energy radiation on your film. Also the lens might be transparent to visible light, but non transparent to high-energy em-waves. Edit: just found http://www.fas.org/nuke/intro/nuke/thermal.htm It ...


4

It is well-known that x-rays are blocked by metal. [Ref: Kid's Science] Obviously the doctor wants to look at your internal organs, unobscured by a fuzzy outline of your house keys and pendant. So, the sign is requesting that you removing metal items from the external parts of your body, to allow visibility to the internal parts. (MRI is a totally ...


4

Roughly yes. Radiation is broadly divided into two from a safety point of view. Ionising radiation can break chemical bonds and so has an obvious way to cause damage to your body - how much depends on the energy, how much radiation you absorb and where in your body it gets to. Both X-Rays and particles from radioactive material are ionising, as is the ...


4

Black holes are formed from collapsed supermassive stars. If a black hole ends up completely isolated then it will be very black. However, many stars are binaries. Moreover, within dense globular clusters stars can often pass very close to other stars. Just as with stars, when a black hole is near enough to another star it can collect matter from it. Black ...


4

I don't believe X-ray diffraction machines focus the X-rays. Generally you want to put them in along a straight line and detect them undisturbed from the sample. The straight line input beam is just a source an energy selector and a slit (unless the machines have got a lot more complex) The Wolter mirrors used in telescopes are because you only have very ...


4

Exactly how and to what extent diamond shows up in x-rays depends on factors such as type of x-ray apparatus, size of diamond, orientation and so on. Carbon has an atomic mass of 12. That's fairly low. Diamond exhibits a bunch of unique properties such as extreme hardness, high thermal conductivity and chemical inertness. In terms of X-ray windows ...


4

Matter in solid, liquid gas or plasma form emits radiation in a spectrum . To a good approximation this is the black body radiation spectrum: As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. From the plot you can see that radiation into the UV is very small even for the ...


4

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 ...


3

They have various properties that differ, but the differences are quantitative, not qualitative, and there is no sharp boundary. The differences occur because of the difference in frequency. A wave that is a gamma ray in one frame of reference could be an x-ray if observed in a different frame. An example of their different properties is that gamma rays are ...


3

X-rays very much do travel through water. I think your quote may be out of context. For example, being deep in the ocean would protect against X-rays because there is so much water above you. However, concrete or lead are two more common materials which provide more protection against X-rays. Using the link you provided, I generated the following plot of ...


3

If we set the attenuation per unit length of each of the $n$ segments as $\mu_i$, and they have a length $\ell_i$, then the transmitted intensity is $$I_t = I_0 e^{-\mu_1 \ell_1}\cdot e^{-\mu_2 \ell_2} ... = I_0 \prod_{i=1}^{n} e^{-\mu_i \ell_i} = I_0 e^{-\sum{\mu_i \ell_i}}$$ The value of this expression is the same whether you evaluate from $i=1$ to ...


3

I would agree with the general conclusions regarding soft x-rays. I have measured the attenuation of x-rays in every day materials at energies below 20 Kev and to a first approximation can be described by :- E½ = K t^1/3.25 where t is the thickness in cm. E½ the energy at which the intensity is reduction to ½ . K is the coefficient of the material. For ...


3

The angular distribution of the x-rays is basically uniform as you indicated, but there are beam restriction devices in use that aren't usually shown in a diagram, as in the typical diagram below. Here is a more detailed drawing showing the isotropic radiation and a beam restricting device, which is less fancy that it sounds. Perhaps there are other ...


3

Xrays are photons, so as suggested in the comments above, the photoelectric effect could be exploited by using an appropriate material. For example, SAXS (Small Angle XRay scattering) machines use many kinds of Si based detectors.


3

Technology has evolved a bit compared to what was described by Martin Beckett. Many powder diffractometers still use the setup he describes, but mirrors are becoming more and more common at laboratory powder diffractometer sources. However, these mirrors do not aim to focus the beam, they collimate the beam (try to make sure that as much as possible of the ...


3

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 ...


3

Yes it is possible but as BarsMonster points out it isn't like an optical mirror. X-ray reflectors are used in the construction of nuclear weapons and are critical to increasing the yield. How they work is the initial fusion reaction releases high energy radiation, this is then reflected back into the reaction mass increasing the energy levels of the ...



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