Re: UV nail lamps for EPROM


"Mike Monett" <No@xxxxxxxxx> wrote in message
news:Xns9934839B33F4BNoemailadr@xxxxxxxxxxxxxxxx
"Roger Hamlett" <rogerspamignored@xxxxxxxxxxxxxxxxxxx> wrote:

> GE, publish the spectrum of their lamps, and show a small peak at
> 404.656nm, and a second somewhat larger one at 435.833nm, together
> with a general very low level of radiation beyond this. The 'black
> light' type lamps, typically produce perhaps 10* the peak
> intensity of either of these lines, at the shorter wavelength of
> about 370nm. GE, say that their lights should be used with
> 'CovRguard' fittings, or with UV sleeves, to completely eliminate
> UV (I don't know who said the bulbs themselves produce 'no' UV, if
> you talk to their commercial division, and specify that there must
> be no radiation above 400nm, in the illuminated area, they specify
> sleeves to be used. ). They say that the total UV (beyond 400nm),
> is around 1/500th the level from noonday Sun, but not that their
> is 'none'.

> You'd need to work out whether it might be the two visible lines
> at the top of the spectrum, that are producing the effect, or
> 'UV'. I'd suspect it might be the visible lines, rather than 'UV'.

> Best Wishes

Hi Roger,

Thanks for your very interesting post. Can you give the url for the
info on the GE spectrum?
Paper version sent to me at work, when I raised this question with regards
to a commercial application which would be affected by light in the very
near UV.
I'd think they would send another copy, or may have it somewhere on their
website as well.
Just looked. They have a much lower resolution copy, showing the same
peaks, but so poorly resolved, that it is hard to tell the exact
frequencies. The spectra are at:
http://www.gelighting.com/na/business_lighting/education_resources/learn_about_light/distribution_curves.htm
The curve I had, was for the 'WWX' lamp, with the paper copy ending at
370nm, where some significant output is still shown. The computer version
seems to lose the bottom percent or so.

Untill now, my sources for the reaction of light on silver chloride
claimed that UV light was needed to knock an electron loose from the
chlorine.

However, I just discovered the following entry that shows the
reaction is stronger at UV, but it still occurs even with red light:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
After hearing about William Herschel's discovery of infrared light
in 1800, Johann Ritter decided to see if he could detect light
beyond the other end of the spectrum - past violet. In 1801, he was
experimenting with silver chloride, which turned black when exposed
to light. He had heard that blue light caused a greater reaction in
silver chloride than red light did and decided to conduct an
experiment to see if this was indeed true. Ritter directed sunlight
through a glass prism to create a spectrum and then placed silver
chloride in each color. He found that the silver chloride
increasingly darkened from the red to the violet part of the
spectrum as predicted. Ritter then decided to place silver chloride
in the area just beyond the violet end of the spectrum in a region
where no sunlight was visible, and was amazed to see an even more
intense reaction there. This experiment showed for the first time
that an invisible form of light existed beyond the violet end of the
visible spectrum. This is now know as the ultraviolet part of the
electromagnetic spectrum.


http://coolcosmos.ipac.caltech.edu//cosmic_classroom/classroom_activities
/ritter_example.html

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I am very pleased to find this out. I was about to invest some time
on an instrumentation project that would have not worked the way I
planned:)

Regards,

Mike Monett
So it is responding to the visible, but more strongly to UV. A
photo-chemist would probably be able to tell you the minimum energy photon
required to trigger it.

Best Wishes



.

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