# What is eV?

Discussion in 'High Voltage' started by Dr_Doggy, Jun 9, 2015.

1. ### Dr_DoggyWell-Known Member

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i found this:

http://www.unitconversion.org/energy/electron-volts-to-watt-seconds-conversion.html

in this example it says I need 2.5eV to generate visible light photons, or 2480 eV to generate xrays,
http://www.unitconversion.org/energy/electron-volts-to-watt-seconds-conversion.html

if that is the case then this should be true:

2.5ev = 9. *10^-19 watt*seconds
or
2480eV = 3.*10^-16 watt*seconds

My hypothesis is that given that information is it true to say that:

9 *10^-19 watt*seconds = V*I*t
= (2V)I * (t/2)---- so by doubling the volts i can do such work in half time, which makes sense to me

but then also this works too:
=v/2 * I * 2t

so even with VI = 10watts
A) 2.5eV = 10 * t
= 0.25s

2480eV = 10 * t
t = 248s

so even though i am running 10watts to get a visible light photon every .25s i will also be generating an xray "photon" every 248 seconds?

I know im missing a few things here , but what?

2. ### kubeekWell-Known Member

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ehmm, what? Where did the 9*10^-19 go in the last equations?
With 10W of input power, you can theoretically produce a bit over 10^18 visible photons every second, or roughly 10^15 xray photons every second.

As to the last sentence, you don't magically genereate photons of any possible wavelength just because you have enough input power, if that is what you were saying.

And to the title of the thread, eV is a unit of energy, which is the same as 1.60217733e-19 Watt-seconds or 1.60217733e-19 Joules. Also equivalent to increasing electric potential of a single electron by one volt (1.602e-19 Coulombs times 1 V gives 1.602e-19 J).

(edited my exponents)

Last edited: Jun 9, 2015
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3. ### DerStrom8Super ModeratorMost Helpful Member

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Kubeek basically answered your question, but to be more clear, "eV" stands for "electron-volt".

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5. ### Dr_DoggyWell-Known Member

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THanks, I did drop those exponents didn't I, thought those numbers looked small!

Also i realize that its more than magic, but in a example of my vacuum chamber, I was told that xrays are produced when there is sufficient bombardment against a metal, where i am getting x amount of light in a oxygen/nitrogen environment with copper electrodes, is a small percentage of that going to be xray then? (do i need to worry about exposure to myself?)

Also is my thinking right then , can i apply this equation to electrolysis of water, ie if I need 5.7eV then that works out to

5.7 = 550Kilo Joules per mol
so for 1 mol of water i need to dump 550 kj in to it
550kj is 152 watt*hour or 0.152kW*hour or 9120 watt Minute
so running at 9kw i will only produces a mol which is only 18 gram of water broken down per minute?
But my power bar only provides about 1800 watts , so using that it will actually take 5minute for said 18grams,

Also going by this:
http://www.carstuffshow.com/blog/how-to-convert-gasoline-energy-to-kilowatt-hours-kwh/
gasoline = 33.7 kWh per gallon, = which will only get me about 30 miles

so 33.7kWh/.152kWh = 221 * 5min = 1108 minutes running electrolysis at 1800watts ,it would take to fill my car to the equivalent to a gallon of gas which is:
= 18hrs of electrolysis @1800watts= per gallon of gas
or
= 18hrs of electrolysis @1800watts per 30miles
now i see why water cars dont work!

and that is before factoring in efficiency losses

Also if the resistance of Pure water is 15 Mohm.cm then with 1 cm between my electrodes i will need:
1800watts = v^2 /r 33.7 kWh
= root(1800w*10,000,000ohm)
= 134 kv ??

or with 1mm electrode spacing =
= root(1800w * 1,000,000ohm)
=42kv & 43mA of current

am i still on the right track?

6. ### kubeekWell-Known Member

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Sorry no idea, i was trying to keep up but the kWh per gallon totally made me lose the track. How can you people mix units like that?
It seems you are comparing gallons of gasoline to mols of water, but really i am not sure what is going on in that calculation.

As for the electrolysis, I guess you would have to use some ions to make the water more conductive.

7. ### Dr_DoggyWell-Known Member

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good, good , i just want to make sure i was understanding the math ok,
i read on another chat that was the equivalent energy for a gal of gas where they compared the btu in gas as well as btu for 1kWh, really i was just comparing the energy between them. but if only the rest of my babble is good thats all i will worry about, i just thought it was interesting to see what those HHO guys were up against
http://www.carstuffshow.com/blog/how-to-convert-gasoline-energy-to-kilowatt-hours-kwh/

the miles were also just an avg guesstimate since every vehicle has different mileage efficiency

" As to the last sentence, you don't magically genereate photons of any possible wavelength just because you have enough input power, if that is what you were saying. "

I notice in my vacuum when my electrodes start to glow I notice that the outer ring of glow is pinker(and thicker) than the rest, as i get closer to the wire the pink becomes more red-er and eventually a thin line of white along the surface of the copper electrode, I also notice that each color shell always seems to be 1/10th of the previous shells width(or so), it just reminds me of harmonics on musical strings, which made me wonder what other harmonic frequency(ies) is/are hiding in there.

8. ### kubeekWell-Known Member

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Not sure about harmonics, but I think that above a certain voltage you are likely to produce X-rays. What is your anode voltage? From what I read anything above 15kV seems to be producing x-rays and could be dangerous.

A geiger counter or similar device might not be a bad idea.

9. ### Dr_DoggyWell-Known Member

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with the neon sign transformer i measured approx turn ratio to be 100x , so about 14kv, it was stated in another thread that i need closer to 75kv @ a few ma for anything substantial in xrays, so i will not worry about geiger ,but plugging in equations like i did does say the reaction is still there.

but as i plug in equations , like in first post, there is evidence that high energy particles could be produced with even small power inputs, mind that i am totally on a limb with this, and plus the factor that : putting enough energy in does not mean i will get a certain energy frequency out,
Overall, thanks for the help , my initial curiosity has been alleviated i can now electrocute water! , but there is then something i still dont understand about all this, what part does determine what kind of "possible wavelength" energy i will get at output? ie is there a calculation to describe why is a N2 laser blue, and a CO2 laser is red?

10. ### kubeekWell-Known Member

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You cannot get a photon out with energy greater than what the electron already has. So say with 2kV anode voltage you cannot get photons with higher energy than 2keV. The distribution of those energies however is beyond the realm of my knowledge, but certainly it will depend on lots of paramters with the target material probably being the most important.

11. ### Dr_DoggyWell-Known Member

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I think that may be the part that is confusing, eV is a unit of power, while volts are only a part of the equation , by decreasing the volts to less than the keV value, i can still make up that final number by increasing the overall current?

12. ### kubeekWell-Known Member

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No, eV is a unit of energy, not power. Watt-second is also a unit of energy, equivalent producing one watt for one second (or half a watt for two seconds etc.).

13. ### Dr_DoggyWell-Known Member

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I stand corrected, but at the same time my point remains valid?

14. ### kubeekWell-Known Member

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No, one electron will only produce one photon. So whatver kV the electron is accelerated to, that is what the maximum energy of the emitted photon will be in keV which is equivalent to its wavelength. If you have greater current you will get more photon emissions, but each photon will again only have as much energy as the electron had in the first place. There will be more of them with larger current, but the energy of each one will still be the same.

Last edited: Jun 9, 2015
15. ### Dr_DoggyWell-Known Member

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1 eV = 1.60217657*10-19 J
and my substance is 5.1ev,
so my joules for 1 electron are 8.16*10-19 J
and my "minimum voltage is 5.1v"

5.1eV = 1.60217733 e-19 Watt-seconds ----is min energy i can apply
== 8.16 *10^-19 Joules/electron ---- is the same number converted
=8.16 e-19 watts * s / electron ----- same
8.16 e-19 watts * s = (v * I) * s ----- equivilant
8.16 e-19 watts = 5.1 * I ------5.1 is my min volt
1.6 e-19 = I per electron ------reduction isolates current to produce a photon

also can this mean 5.1v/1.6 e-19 = 31.875 e18 ohms of resistance per photon/electron

are those my values per electron then?(see what im doing here, does this make sense)?

I wonder if that means a pulse period has a minimum time as well?

16. ### kubeekWell-Known Member

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Up to here it still makes kind of sense in what you are trying to find out. Now:
What this actually says, is that 1 electron per second is a current of 1.6e-19 ampere, or in other words when the current is 1A there is 6.24e18 electrons moving past some point in the wire each second. That actually is the definiton of ampere.

This one is a bit dodgy, what that equation actually means is that at 5.1V you need resistance of 31.875 e18 ohms in order to get one electron through each second. Not sure how that helps you with anything.

I think you should rewrite those equations and keep better track of you units, that might get you a better insight on what is going on.

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17. ### nsaspookWell-Known Member

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+1
https://en.wikipedia.org/wiki/Bremsstrahlung#X-ray_tube

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18. ### Dr_DoggyWell-Known Member

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OHYA! i remember hearing that years ago back in school, thanks, i would have never noticed that! In fact i remember back then disregarding that as boring! I think that really brought things to a full circle for me here!

Your rite, that resistance i calculated was actually the resistance to produce 1 photon, or strip 1 energy shell, again just wanted to see if i could do it and still make sense!

will need to read that link a few times to understand it, but now I think i will take precautions, since i may go over that value, my chamber is some form of metal, but the lid is acrylic, not interested in frying everyone above me in the living room, this should do the trick?:
http://www.ebay.ca/itm/New-Smart-Ge...432?pt=LH_DefaultDomain_2&hash=item1c501d2468

maybe i should switch to carbon electrodes as well

is there anything i can read, or a topic i can Google to find this out?

Also this again: I wonder if that means a pulse period has a minimum time as well?
using this:
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/cnvcalc.htm
for example i assume that if my shell is 2.5 ev then that frequency is 6.0450e+8 MHz, then if i could do a pulse shorter than that, say twice as fast would the atom/molecule still absorb that energy sufficiently to do an emission?
OR
I have heard that lasers are extremely inefficient, i wonder if this could be related? IE, the breakdown current lasts alot longer than the wavelength of photon emitted? Maybe that is why so much energy is dumped as heat? especially in the case of a ruby laser where you have a HV trigger pulse, then the capacitors dump current into the xenon gas until they are empty and the "spark gap" opens up again, which is alot longer.?

19. ### kubeekWell-Known Member

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6e8 Mhz is 6e14 Hz, which if I am not mistaken is 600 THz, or roughly 10000 times faster than anything humans use now. Generating a pulse of that length is basically impossible, but even if it were I cannot tell you if it could generate that photon or not. I am not even sure if an electron entering an atomĀ“s shell does have some time over which that event is happening, but I would guess that this is a meaningless question in such a situation. But then again, I am not a particle physicist.

20. ### Dr_DoggyWell-Known Member

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Brilliant! I went over to a chemistry forum, they were quite helpful explaining:

entropy
enthalpy
Gibbs free energy!
& Hemholtz free energy!

just in case anyone wants to know