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4hv.org :: Forums :: Electromagnetic Radiation
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Spectral line energy versus anode current?

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Proud Mary
Thu Jul 30 2009, 12:40AM Print
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
Any of the common target elements - W, Cu, Ag, Mo, Ge, Fe, Co etc have a general pattern of low level braking radiation broken up by very often strong emissions on highly defined spectral lines. The example I have given in the graph shows emission from a silver target with increasing anode voltage.

My question is this: if we take the silver spectrum as an example, if I gradually increase the anode voltage until I arrive at one of the strong emission peaks, will there be a corresponding change in Va/Ia equal to the energy radiated away in the spectral lines (minus various losses, I suppose)

Ray Spectrum 399pixels
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klugesmith
Thu Jul 30 2009, 01:53AM
klugesmith Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
My guess is that when you reach the voltage which allows the emission peak,
the extra energy radiated as X-rays will mean slightly less than proportional heating of anode,
rather than a sudden tweak in the I/V curve. Might be hard to measure --
have you ever converted your X-ray intensity measurements into total radiant watts?

By very rough analogy, the temperature of a photovoltaic cell in the sun
goes down when a maximum-power electrical load is connected to its terminals,
compared with an open- or short-circuit.
It's not hard to measure that effect with garden-variety Si cells from Radio Shack.
It's a critical consideration for passive cooling of ultra-efficient multi-junction cells
where tracking mirrors or lenses concentrate light to intensity of 500 suns.
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Proud Mary
Thu Jul 30 2009, 02:24AM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
Well, Kludge, we'll soon find out, eh! I'll put my differential voltmeter across a measuring resistor* in the kathode line and then 'tune' the target voltage through the peaks and see what happens. Experiments wouldn't be much fun if we knew the outcome in advance!

But I do see what you mean about the problem, of nano measurments. Perhaps it would not be possible for the amateur to detect if indeed it does exist. I would expect the X-ray emission of a typical 10W tungsten tube would be of the order of 10mW, but beam angles of 2 and 3 deg. of the type I use by cranking up the focusing grid bias means that the beam has considerable fluence, comparable in some respects to a 10mW laser at close range.

I have a medium-low power tube (30kVp/10W/0.3 mm sq anode spot/ focusing grid/ambient air cooling) with a silver anode, which is the easiest to deal with as detection and measurement of peaks around 21keV is a routine maatter compared with target elements peaking well below 10kV like iron, nickel and cobalt.

If you are right, and there would be no change in Va/Ia but a change in the anode temperature, then a tube with a very low mass target could be created and its thermal radiation measured in search of the peaks. What's the betting that both Boltzmann's and Planck's Constant will be involved 'ere long. (I've always wanted to do something practical that involved both Boltzmann's Constant and Planck's Constant, which this experiment looks to do. Quanta talking to eachother in German accents! What fun! smile

Perhaps I shall discover nothing at all, but who can say? You have to be an optimist to construct wonderful things from designs of your own invention. smile

* But maybe safer from rogue spikes to use some very accurate current mirror, as in "Oh no not my Fluke!"



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