Would HE light applied to mercury vapor produce electricity?

  • #1
I'm likely going to have a ton of questions in the following weeks about a wide range of particle physics etc.
If (electricity X mercury vapor) = light, then multiplication being transitive, (light X mercury vapor) = electricity.
Can someone inform? I've found only mercury vapor as a metallic gas, and I'm specifically looking to create a model using sustained HE light as a basis for high voltage output.
 

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  • #2
Drakkith
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If (electricity X mercury vapor) = light, then multiplication being transitive, (light X mercury vapor) = electricity.

Electricity X mercury vapor is not multiplication, so that particular rule does not apply. Note that running an electric current through nearly any material can generate light by virtue of simply heating the material up. The reverse (light X some material = electricity) isn't true except in very specific situations, such as solar panels.
 
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  • #3
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If it would be a multiplication, then you would have electricity = (light)/(mercury vapor). But it is not a multiplication.
 
  • #4
jbriggs444
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If (electricity X mercury vapor) = light, then multiplication being transitive, (light X mercury vapor) = electricity.
In addition to being irrelevant, multiplication is not "transitive". Even if you did the algebra correctly, you would have ##electricity=\frac{light}{mercury\ vapor}##

Transitive: If a=b and b=c then a=c. Equality is transitive.
Commutative: ##a \times b = b \times a##. Multiplication is commutative.
Associative: ##a \times (b \times c) = (a \times b) \times c##. Multiplication is associative.
 
  • #5
sophiecentaur
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We should give the OP a break here. He is "guilty" of using inappropriate terminology, agreed, but his Physics is not misplaced. He just needs a 'symbol' which means "combined with in some way". I don't know of one :smile: so he should have just used a verbal description.
If you pass a current though Mercury Vapour it will ionise some molecules. When the electrons recombine there will be Photons emitted. The direction of the current to excite this is defined by an external power supply and the ionisation is sustained by the fast electrons that are supplied at the electrodes.
However, although a beam of incident high energy EM (photons) will ionise some of the mercury atoms, the ions will have no preferred direction to go so you can't expect the system to produce an 'electric current' (i.e. to be a generator). The ions will just recombine and emit a range of EM frequencies.
But you could imagine a detector of X Rays, which consisted of a container with mercury vapour in it and a low voltage across it; not high enough to produce ionisation so it would be 'off'. When sufficient flux of X Rays hits the tube, enough ions will be produced for a measurable current to flow. It would strike an arc which could go on for ever unless the supply voltage is removed and the ions would recombine.
A Geiger Muller tube works in this way but doesn't use Mercury Vapour, which needs to be heated to keep it from condensing. Low pressure Argon gas is commonly used. Also,the design of a GM tube allows very low energy particles through, to ionise the enclosed gas. But it's the same principle.
PS You need some form of 'diode' or one-way device for EM to generate a current in one direction. PV cells and phototransistors do this but not (afaik) ever with mercury.
 
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Drakkith
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We should give the OP a break here. He is "guilty" of using inappropriate terminology, agreed, but his Physics is not misplaced.

It may be if it's based on trying to apply mathematical rules to non-mathematical situations.
 
  • #7
sophiecentaur
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It may be if it's based on trying to apply mathematical rules to non-mathematical situations.
I didn't have any problem with understanding his question though and I think he was only using the Maths notation in order to 'fit in' with the way many threads converse.
It would be nice to hear from him again on this thread.
 
  • #8
russ_watters
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I didn't have any problem with understanding his question....
Nor did I.
It may be if it's based on trying to apply mathematical rules to non-mathematical situations.
Not if it is a literary usage, like a metaphor.
 
  • #9
Drakkith
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I didn't have any problem with understanding his question though and I think he was only using the Maths notation in order to 'fit in' with the way many threads converse.

Not if it is a literary usage, like a metaphor.

That certainly might be the case, but I have no way of knowing. The OP might literally think you can apply math rules directly to non-math situations.

It would be nice to hear from him again on this thread.

Agreed. Clarification would be helpful.
 
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  • #10
sophiecentaur
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So . . . . . about the Physics?
 
  • #11
jbriggs444
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So . . . . . about the Physics?
Well, to point out the obvious... The first law of thermodynamics (energy conservation) is OK with the reverse operation. The second law of thermodynamics says that it will not be a 100% efficient reversal.

However, observation shows very little evidence of Mercury vapor lamps being used in the daytime to pump electrical energy into the mains and quite a lot of evidence of other techniques being used to convert sunlight to electrical energy. This suggests that as a matter of engineering practice, a Mercury vapor lamp is not the best technology for the purpose.
 
  • #12
Yes, math is my one true heartbreak. I love science, astronomy, and a host of both artistic and cerebral interests, but math.....alas.
Yes, I was using the transient property of math, and applying it to a physics problem. And to me, it seems like they're one and the same. Seems not.
 

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