# Would HE light applied to mercury vapor produce electricity?

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.

Drakkith
Staff Emeritus
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.

• russ_watters
mfb
Mentor
If it would be a multiplication, then you would have electricity = (light)/(mercury vapor). But it is not a multiplication.

jbriggs444
Homework Helper
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.

sophiecentaur
Gold Member
2020 Award
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 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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• russ_watters
Drakkith
Staff Emeritus
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.

sophiecentaur
Gold Member
2020 Award
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.

russ_watters
Mentor
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.

Drakkith
Staff Emeritus
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.

• russ_watters
sophiecentaur
Gold Member
2020 Award
So . . . . . about the Physics?

jbriggs444