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Science Forum Index » Engineering - Lighting Forum » New xenon fluorescent lamps...
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| redbelly... |
 Posted: Tue Jul 08, 2008 1:39 pm |
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| Don Klipstein... |
| Posted: Tue Jul 08, 2008 2:06 pm |
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In <c81bd321-ec4e-4f92-8404-6131b9f8a28f at (no spam) b1g2000hsg.googlegroups.com>,
redbelly wrote:
147 nm. Stokes losses will be much higher than with mercury.
- Don Klipstein (don at (no spam) misty.com) |
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| redbelly... |
| Posted: Wed Jul 09, 2008 2:42 am |
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On Jul 8, 8:06 pm, d... at (no spam) manx.misty.com (Don Klipstein) wrote:
Quote: In <c81bd321-ec4e-4f92-8404-6131b9f8a... at (no spam) b1g2000hsg.googlegroups.com>,
redbelly wrote:
From Matsushita Electric.
http://news.yahoo.com/s/afp/20080708/tc_afp/japanelectricallightmatsu...
"... the new device can make xenon gas emit vacuum-ultraviolet light
causing a phosphor to fluoresce ..."
Anybody know offhand what would be the primary u.v. wavelength in a
xenon discharge?
147 nm. Stokes losses will be much higher than with mercury.
- Don Klipstein (d... at (no spam) misty.com)
Thanks Don, that was exactly the reason I asked.
Regards,
Mark |
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| Victor Roberts... |
| Posted: Wed Jul 09, 2008 10:18 am |
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On Tue, 8 Jul 2008 16:39:10 -0700 (PDT), redbelly
<redbelly98 at (no spam) yahoo.com> wrote:
Work on such lamps was done many rears ago at GE and other
labs. The GE work was published at one of the LS
conferences, perhaps LS:9 or LS:10. As Don has indicated
the longest UV wavelength emitted is rather short. In order
for these lamps to compete with Hg-based fluorescent lamps
you need a photon-splitting phosphor, which has not yet been
developed with required characteristics.
--
Vic Roberts
http://www.RobertsResearchInc.com
To reply via e-mail:
replace xxx with vdr in the Reply to: address
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| Don Klipstein... |
| Posted: Thu Jul 10, 2008 3:55 am |
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In article <6g3c74dfdqu5ftb9c3j8ba24h59mf34q7t at (no spam) 4ax.com>, Joe wrote:
Quote: On Wed, 9 Jul 2008 00:06:27 +0000 (UTC), don at (no spam) manx.misty.com (Don
Klipstein) wrote:
Anybody know offhand what would be the primary u.v. wavelength in a
xenon discharge?
147 nm. Stokes losses will be much higher than with mercury.
- Don Klipstein (don at (no spam) misty.com)
I don't understand the term "Stokes Loss". Is that simply the lessened
efficiency (energy in vs energy out) in the Stokes shift?
Maximum possible efficiency of a usual phosphor (quantum efficiency
limited to 100%) is ratio of input wavelength to output wavelength.
Photon energy is inversely proportional to wavelength. The loss of
photon energy from fluorescence due to output wavelength being longer
than input wavelength is the Stokes loss.
Quote: If so, what is the determining factor that makes Xenon less efficient
than Mercury?
The main UV wavelength of low pressure mercury is 254 nm. For xenon
this is 147 nm.
- Don Klipstein (don at (no spam) misty.com) |
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| Joe... |
| Posted: Thu Jul 10, 2008 8:29 am |
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On Wed, 9 Jul 2008 00:06:27 +0000 (UTC), don at (no spam) manx.misty.com (Don
Klipstein) wrote:
Quote: Anybody know offhand what would be the primary u.v. wavelength in a
xenon discharge?
147 nm. Stokes losses will be much higher than with mercury.
- Don Klipstein (don at (no spam) misty.com)
I don't understand the term "Stokes Loss". Is that simply the lessened
efficiency (energy in vs energy out) in the Stokes shift?
If so, what is the determining factor that makes Xenon less efficient
than Mercury?
Thanks,
Joe |
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| Andrew Usher... |
| Posted: Thu Jul 10, 2008 12:04 pm |
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On Jul 10, 7:55 am, d... at (no spam) manx.misty.com (Don Klipstein) wrote:
Quote: If so, what is the determining factor that makes Xenon less efficient
than Mercury?
The main UV wavelength of low pressure mercury is 254 nm. For xenon
this is 147 nm.
For optically thin plasmas (which I thought fluorescents were) 185 nm
should be stronger than 254 nm.
Andrew Usher |
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| ... |
| Posted: Thu Jul 10, 2008 6:12 pm |
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On Thu, 10 Jul 2008 15:04:37 -0700 (PDT), Andrew Usher
<k_over_hbarc at (no spam) yahoo.com> wrote:
Quote: For optically thin plasmas (which I thought fluorescents were) 185 nm
should be stronger than 254 nm.
they are thick. 254 to 184 is about 5:1 |
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| Don Klipstein... |
| Posted: Thu Jul 10, 2008 6:16 pm |
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In <67fb7bd5-9840-4224-84a2-9933ba2d970e at (no spam) l42g2000hsc.googlegroups.com>,
Andrew Usher wrote:
Quote: On Jul 10, 7:55 am, d... at (no spam) manx.misty.com (Don Klipstein) wrote:
If so, what is the determining factor that makes Xenon less efficient
than Mercury?
The main UV wavelength of low pressure mercury is 254 nm. For xenon
this is 147 nm.
For optically thin plasmas (which I thought fluorescents were) 185 nm
should be stronger than 254 nm.
At the resonance wavelengths (ones from transitions whose lower level is
the "ground state"), low pressure vapors in practical lamps are normally
"optically thick".
Web search for "imprisonment" along with "mercury vapor". This refers
to 254 nm photons being absorbed and re-emitted many times before
escaping - preferably before the quanta of energy become lost one way or
another.
- Don Klipstein (don at (no spam) misty.com) |
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| Joe... |
| Posted: Fri Jul 11, 2008 8:15 am |
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Now I get it. Thanks Don.
Joe
On Thu, 10 Jul 2008 13:55:55 +0000 (UTC), don at (no spam) manx.misty.com (Don
Klipstein) wrote:
Quote: In article <6g3c74dfdqu5ftb9c3j8ba24h59mf34q7t at (no spam) 4ax.com>, Joe wrote:
On Wed, 9 Jul 2008 00:06:27 +0000 (UTC), don at (no spam) manx.misty.com (Don
Klipstein) wrote:
Anybody know offhand what would be the primary u.v. wavelength in a
xenon discharge?
147 nm. Stokes losses will be much higher than with mercury.
- Don Klipstein (don at (no spam) misty.com)
I don't understand the term "Stokes Loss". Is that simply the lessened
efficiency (energy in vs energy out) in the Stokes shift?
Maximum possible efficiency of a usual phosphor (quantum efficiency
limited to 100%) is ratio of input wavelength to output wavelength.
Photon energy is inversely proportional to wavelength. The loss of
photon energy from fluorescence due to output wavelength being longer
than input wavelength is the Stokes loss.
If so, what is the determining factor that makes Xenon less efficient
than Mercury?
The main UV wavelength of low pressure mercury is 254 nm. For xenon
this is 147 nm.
- Don Klipstein (don at (no spam) misty.com) |
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| Andrew Usher... |
| Posted: Fri Jul 11, 2008 2:42 pm |
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On Jul 10, 10:16 pm, d... at (no spam) manx.misty.com (Don Klipstein) wrote:
Quote: For optically thin plasmas (which I thought fluorescents were) 185 nm
should be stronger than 254 nm.
At the resonance wavelengths (ones from transitions whose lower level is
the "ground state"), low pressure vapors in practical lamps are normally
"optically thick".
OK, I calculated this. I know the fluorescent lamps optimise around 40
C
wall temperature, which gives 0.01 mb Hg. I assumed the average
temperature inside the tube to be 1/30 ev (380 K) and the electron
temperature 1 ev (yes, I know, the electrons are not actually
Maxwellian
in a glow discharge - but fairly close here). I got an optical depth
of
3 at 254 nm and 110 at 185 nm.
I wonder why this is optimal - at lower pressures the ratio 185 nm/254
nm
goes up, increasing Stokes loss even if the phosphors were made
for 185 nm; also the carrier gas starts to carry more of the current,
wasting it. At higher pressures one gets even more self-absorption,
meaning less of the current goes to the UV lines. I would think these
two would balance at lower optical depths than this.
Andrew Usher |
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| Victor Roberts... |
| Posted: Sat Jul 12, 2008 8:53 am |
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On Thu, 10 Jul 2008 15:04:37 -0700 (PDT), Andrew Usher
<k_over_hbarc at (no spam) yahoo.com> wrote:
Quote: On Jul 10, 7:55 am, d... at (no spam) manx.misty.com (Don Klipstein) wrote:
If so, what is the determining factor that makes Xenon less efficient
than Mercury?
The main UV wavelength of low pressure mercury is 254 nm. For xenon
this is 147 nm.
For optically thin plasmas (which I thought fluorescents were) 185 nm
should be stronger than 254 nm.
Andrew Usher
Even in optically thin plasmas I would not expect the 185
line to be stronger than the 254 line since it comes from a
much higher energy level.
--
Vic Roberts
http://www.RobertsResearchInc.com
To reply via e-mail:
replace xxx with vdr in the Reply to: address
or use e-mail address listed at the Web site.
This information is provided for educational purposes only.
It may not be used in any publication or posted on any Web
site without written permission. |
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| Andrew Usher... |
| Posted: Sat Jul 12, 2008 3:54 pm |
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On Jul 12, 7:53 am, Victor Roberts <x... at (no spam) lighting-research.com> wrote:
Quote: For optically thin plasmas (which I thought fluorescents were) 185 nm
should be stronger than 254 nm.
Even in optically thin plasmas I would not expect the 185
line to be stronger than the 254 line since it comes from a
much higher energy level.
The transition probability is 90 times higher for 185 nm,
because the 254 nm line is intersystem (LS-forbidden).
So for 254 nm to be inherently stronger the electron
temperature would have to be less than about 0.40 ev (4,650 K).
This is false in all discharges.
Andrew Usher |
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| Don Klipstein... |
| Posted: Sun Jul 13, 2008 6:16 am |
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In <e4accf55-148f-4098-bfd3-a8bd74978567 at (no spam) e39g2000hsf.googlegroups.com>,
Andrew Usher wrote:
Quote: On Jul 12, 7:53 am, Victor Roberts <x... at (no spam) lighting-research.com> wrote:
For optically thin plasmas (which I thought fluorescents were) 185 nm
should be stronger than 254 nm.
Even in optically thin plasmas I would not expect the 185
line to be stronger than the 254 line since it comes from a
much higher energy level.
The transition probability is 90 times higher for 185 nm,
because the 254 nm line is intersystem (LS-forbidden).
So for 254 nm to be inherently stronger the electron
temperature would have to be less than about 0.40 ev (4,650 K).
This is false in all discharges.
I think you should find a mercury atom energy level diagram with
wavelength notations, such as:
http://www.phys.tue.nl/FLTPD/invited/kindel.pdf (Figure 6).
The 185 nm line comes from 6-1 P(1) to ground state (a 6-S level).
The 254 nm line comes from 6-3 P(1) to ground state (a 6-S level).
The 6-3 P(0) and (2) levels have transition to the ground state
forbidden.
- Don Klipstein (don at (no spam) misty.com) |
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| Andrew Usher... |
| Posted: Sun Jul 13, 2008 2:38 pm |
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On Jul 13, 10:16 am, d... at (no spam) manx.misty.com (Don Klipstein) wrote:
Yes, I have seen such many times. I am correct.
Quote: The 185 nm line comes from 6-1 P(1) to ground state (a 6-S level).
The 254 nm line comes from 6-3 P(1) to ground state (a 6-S level).
The ground state is 1S1 (a singlet). Therefore the 254 nm line
involves a change of multiplicity and is called an intersystem
or intercombination transition.
Look at the transition probablities here:
http://physics.nist.gov/PhysRefData/Handbook/Tables/mercurytable3.htm
The 185 nm line says 7.46 and the 254 nm line 0.080, that's
about 90 times.
Quote: The 6-3 P(0) and (2) levels have transition to the ground state
forbidden.
Which is why those are not observed. The 3P2 is allowed by
quadrupole radiation and probably can be seen in special
conditions; the 3P0 is forbidden by any radiation (0->0 rule).
Andrew Usher |
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