- Jan 1, 1970
[... very good info on UV]
Thanks for posting this information, Phantom.
I have a question. I notice that silver chloride turns black when
exposed to the light from ordinary office flourescent lights.
This reaction occurs when a UV photon forces a chlorine ion to give
up an electron, which then converts a silver ion to a metal atom.
The metal absorbs visible light and appears black. The reaction is
quite strong with only two overhead lights. Here is a description:
2AgCl + 2UV --> Ag(s) + Cl2(g)
The same reaction occurs outdoors in sunlight. Since the short wave
UV cannot penetrate ordinary glass, I assume the UV in this reaction
is long wave UV.
However, manufacturers of flourescent lights, such as GE, insist
that no UV escapes from their product.
They probably mean the main strong shortwave UV wavelengths of
low pressure mercury vapor.
But obviously a great deal does escape.
The 365-366 nm cluster of mercury spectral lines does go through most
glass well, and is a weak but slightly significant spectral feature of
fluorescent lamps. Exception: Triphosphor lamps (including most compact
fluorescents) of color temperature rating 3500K or higher usually in my
experience have a blue-emitting phosphor ingredient that utilizes that UV
Maybe silver chloride responds to the 404.7 nm violet wavelength of
mercury, or has slight response to the violet-blue 435.8 nm wavelength of
mercury that is strongly present in the light from fluorescent lamps.
indicates silver chloride reacting to visible blue and violet light as
well as to UV.
also indicates ability of blue and violet visible light as well as UV to
cause silver chloride to do its photochemical reaction.
Both of these note Ritter discovering UV via its great ability to cause
the photochemical reaction in silver chloride.
Do you have any idea how the UV gets through the phosphor coating?
A lot of fluorescent lamps have phosphor coatings that do not absorb
- Don Klipstein ([email protected])