In some sense, yes.
UV continuum $\rightarrow$ UV line
This is probably not the answer you're looking for, but massive stars (O and B stars) emit a very hard UV spectrum of light (a UV "continuum", i.e. a broad range of wavelengths). Since these stars don't live for long (because they burn their fuel fast), they tend to be located in the gas clouds from which they were born. The UV light ionizes the enshrouding hydrogen, but the protons and electrons quickly recombine. If the electron goes directly to the ground state, another UV photon is emitted, capable of ionizing another hydrogen atom. In most cases, however, the electron "cascades" down multiple levels, emitting photons of different energies, which may excite other atoms.
It turns out that for every 3 ionizing photons emitted by the star, 2 will eventually — after several interactions — result in the photon corresponding to the energy difference between the first excited state and the ground state; the so-called "Lyman $\alpha$" photon, with a wavelength of 1216 Ångström (121.6 nm). Although there is some broadening due to thermal motion of the atoms and, in particular, due to the resonant scattering of Lyman $\alpha$ on neutral hydrogen, the result is that most of the light from these stars is converted into a single, very narrow (of the order of a few Ångström) emission line, i.e. very close to monochromatic light.
These stars are probably rarely surrounded by planets (because the radiation pressure will tend to blow away the particles used to build planets), and even if they were, these processes happen farther away from the stars than the planets would be. But if you observe a young galaxy, whose spectrum is dominated by young stars, the Lyman $\alpha$ emission line is often the only light visible.
UV line $\rightarrow$ visible line
The Lyman $\alpha$ line is still in the UV, and thus invisible to humans. However, since light is redshifted as it travels through the expanding Universe, Lyman $\alpha$-emitting galaxies sufficiently far away will have their emission line carried into the visible range. Since the shortest wavelength we can see is 400 nm, it must be redshifted by a factor $400/121.6 = 3.3$ (i.e. $z=2.3$), corresponding to a distance of 18.6 billion lightyears (but if it's farther away than 25.5 billion lightyears, it will shift into the infrared and be invisible again). Note, though, that this is only in principle; such galaxies are far too dim to be visible to the naked eye. Too see them, you must use a telescope and a camera.