No special reason for asking. I’m just curious.
Sure, if it has a filter in front of it.
Also, Lasers do this.
You’re right about lasers, of course. But if a filter bis blocking the visible portion of the spectrum, then obviously the object is radiating visible light.
What I’m thinking of is a non-laser example. Non-coherent light, spreading out in all directions as from a light bulb, rather than the tightly focused beam of a laser.
The Lyman Series of emission lines of Hydrogen is in the UV. This is when an excited electron moves from a higher energy state down to it’s ground state in one shot. I don’t know if that’s physically possible, but if it is, then it would emit strictly in the UV. If any pause at the next higher energy state, then there would be the visible light from the Balmer Series.
Well, you said “object.”
Does that preclude some type of apparatus?
A UV LED is also a possibility, although not as omni-directional as you might want.
You can’t do it with just a blackbody, if that’s what you’re looking for. A sufficiently-hot blackbody will emit a lot more UV than it will visible… but it’ll also emit more visible than it would at a lower temperature.
To my mind yes, but I admit I didn’t phrase the question with a great deal of rigor. So I’ll clarifiy: Can an object omnidirectionally radiate ultraviolet light without also radiating visible light, without the use of an intervening filter?
An ultraviolet LED.
ETA: ok, I see somebody already said that.
You can use photonic structures to tune emission spectra away from blackbody, but I don’t see that eliminating any swaths of the spectrum, e.g. visible.
I am going to say no, with the exception of lasers and LEDs as others have already mentioned. I cannot for the life of me think of any purely UV source.
I work extensively with UV sources, mostly hollow cathode lamps (windowlesss and windowed), electron beam lamps, deuterium lamps, and penray lamps. I also have LEDs that emit at 405 and 385 nm (hardly UV to be honest) and a laser than if you treat it just right that can get down to 220 nm or so (it is tunable), but in the main I am dependent for most of my work on the aforementioned lamps.
What I do with these sources is calibrate instruments that I design and build that image objects in the UV. When I design these instruments I am always cognizant of the visible tail emitted by the objects of interest; typically orders of magnitude brighter that what I am actually designing and building the instrument to study. Luckily, most detectors I use are “solar blind”, meaning their QE (i.e. responsiveness) with visible and IR light is orders of magnitude below that at the UV wavelengths.
Put together a sample of your favorite radioactive element and a light element such as helium, lithium, or beryllium. When the radioactive atoms decay, some of the emitted particles will strike the electrons orbiting the nuclei of the light elements, knocking them up into a higher level. For some of these atoms, the other electrons orbiting the atom will then fall back down into the now-vacant energy level, emitting a photon of light. For most elements, these emitted photons would be in the X-ray region of the spectrum (they’re known as characteristic X-rays), but for the lighter elements, the photons emitted by this process should be in the UV range instead.
All bodies radiate black body radiation, which covers the entire spectrum. That is, all bodies radiate at all frequencies. The better question is how much radiation you’re willing to tolerate in your band of interest.
I don’t know of any way to excite just the Lyman emission. Deuterium arc lamps produce a lot of Lyman emission, but also a lot of UV continuum and visible light.
I work with UV instruments, and haven’t come across any UV sources that don’t need a filter and/or monochromator to suppress visible light. Except LEDs and lasers as already mentioned.
Of course. You just can’t get it by simply heating an object, since heating necessarily produces blackbody radiation. A UV e.g. nitrogen or excimer laser is a perfectly reasonable example. If you want it to be omnidirectional, that’s just a question of how you design your laser cavity and/or output coupling optics.
Hydorogen ions, and hydrogen molecules, are individual seperate objects, and emit on specific frequencies.
“Black Bodies” are bodies of ions/atoms/molecules, and emit in a broad-band fashion.
Only “Black Bodies” emit black body radiation. For all other objects the radiation is a characteristic of the emissivity.
In a “Black body” situation, the material absorbs most strongly at the same frequencies it emits most strongly. This filters the emission peaks down to the “black body” radiation curve.
In situations where the material is not re-absorbing all the energy it emits, you don’t get “black body” radiation.
Have I got this right?
In my mind, “ultraviolet” is just another color, albeit one that our eyes aren’t calibrated for. Aside from its wavelength, UV is not intrinsically different than any other color. The OP might as well have asked, “Can an object radiate green without also radiating other visible colors?” or "“Can an object radiate red without also radiating other visible colors?”
What do y’all think? Am I correct? And if I am indeed correct, I’ll ask a specific question: Does an X-ray machine (either modern or old-fashioned) radiate any colors outside of the x-ray band?
Omit the phosphorus coating of a garden variety fluorescent bulb, and it will emit (mostly) UV.
This type of clear lamp is very dim to the human eye, yet it generates a tremendous amount of UV, to the point of being unhealthy. Hence their use for example, in water treatment facilities to kill pathogens in the process flow.
It’s actually easier than that, even, because UV is a much wider band than all of visible light combined.
Correct. Though UV is a much wider range of wavelengths than “green.”
Most X-ray generators are electron impact sources (aka X-ray tubes). An electron beam is accelerated by a strong electrical field, and impacts onto a target. The interaction between the electron beam and the target generates X-rays (Bremsstrahlung radiation). I don’t think there is a “modern vs. old-fashioned” here, as far as I know. And yes, this type of X-ray generator emits visible light, which is usually blocked by a thin metal window (usually beryllium). (Except for laboratory X-ray generators for generating extremely low-energy X-rays which would be blocked by the beryllium window - I’ve worked with those, and they require very thin metallic filters to block visible light.)
As MikeS mentioned, another way to generate X-rays is to use a radioactive isotope. It’s only used when you need a simple and very weak source of X-rays - e.g. testing and calibrating X-ray detectors. These do not emit any visible light.
So, imagine an object starting at room temperature and getting steadily warmer. To our eyes, it’s black, then dull red, then red, then orange, then yellow, then white. If it keeps warming up, will the emission peak move to the ultraviolet? When this happens, will the visible emissions continue to increase as well?