Author Topic: ionized HHO  (Read 20150 times)

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Two different things?
« Reply #16 on: November 30, 2012, 08:51:42 am »
Dear members,

Please answer my question:
Is the burnrate of hydrogen a constant?
Is the speed quicker if hydrogen is in a higher energy state or is it the same as in lower energy states?

Is the burnrate of ortho hydrogen the same as from para hydrogen?

If anybody has a hint, please please answer.

Steve

http://www.thch.uni-bonn.de/pctc/bargon/PHIP/parahydrogen.html


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Re: ionized HHO
« Reply #17 on: November 30, 2012, 11:39:28 am »
Ok, i did some things wrong in my previous statements.
The amount of EV to bring the hydrogen atom into full ionization state is 13,6V.
This is the state that the electron leaves the last orbit of the proton.
This is the most energetic state of the hydrogen ion.

The trick is to keep it at that level. I think here comes in the electron extraction idea of Meyer.


Steve

ps.

This works with water with no additives, like KOH or whatever you use...




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Re: ionized HHO
« Reply #18 on: November 30, 2012, 11:41:13 am »
Now the oxygen atom.... 8) 8) 8) 8) 8) 8)


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Re: ionized HHO
« Reply #19 on: November 30, 2012, 11:58:20 am »
see topic:

http://www.ionizationx.com/index.php?topic=642.0


stumbled upon this from this site.

"I will deal with your questions one at a time.

1) Oxygen ionizes in the presence of an energy discharge such as a
UltraViolet lamp. Do the ionized oxygen atoms remain ionized as long
as they are in the presence of the energy field?

Once the oxygen atoms are ionized, they enter into a series of very rapid
reactions, including recombinations with electrons to become neutral
species again. They do not "remain ionized" much at all. But there is a
continuing pool of ionized oxygen atoms because new ones are being formed
all the time.

A side issue: you cannot form oxygen ions with a UV lamp. The quartz
windows of the lamp will only transmit light of wavelength 160 nm or more;
oxygen will only ionize efficiently in UV light of wavelength 130 nm or
less. The oxygen reaction that you can drive with a UV lamp involves
neutral atomic oxygen, which can be generated by photodissociation with
light of 210 nm or less. The easier and more usual way to generate oxygen
ions is in an electrical discharge.
---------

2) How soon after being ionized do the oxygen atoms reform into Ozone?

Much less than one second. But the reaction that forms ozone does not
involve oxygen ions; ozone is formed by

O + O2 + inert third molecule ---> O3 + third molecule

Atomic oxygen (O) is readily formed from oxygen ions; molecular oxygen (O2)
is abundantly present in air anyway.
---------

3) If an ionized oxygen atom has a -2 charge, how can these atoms
bond into Ozone when they should be magnetically repulsed by one
another?

Ionized oxygen, as formed in an electrical discharge or by very low
wavelength UV light has a +1 charge. The most readily formed ion is
molecular oxygen missing one electron: O2+ . Atomic oxygen missing one
electron, O+ , also enters into the reaction system, but it is a much
higher energy species. Because of the electrostatic repulsion (not
magnetic) between ions with like charge, the chemical reactions in the
system do not involve two ions reacting together; it is always one ion with
a neutral species.
----------

4) Is there some way to speed up the decomposition of Ozone?

There are many ways to speed up the decomposition of ozone. Ozone (O3) is a
high energy form of oxygen, so it is quite ready to revert to the lower
energy, more stable form of molecular oxygen (O2). There are many ways of
achieving the transformation. Atomic chlorine or bromine -- from molecular
chlorine or bromine and visible light -- are particularly effective as gas
phase catalysts, as is nitric oxide. Nearly all organic substances are
directly oxidized by ozone. Rubber perishes very rapidly into a powdery
heap, for example. Reducing solutions -- ferrous sulfate or potassium
iodide -- will effectively remove ozone from a gas bubbled through them.
"

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Re: ionized HHO
« Reply #20 on: November 30, 2012, 12:04:07 pm »
http://www.knowledgedoor.com/2/elements_handbook/ionization_energies_part_3.html

Hydrogen I (1)    13.598433 eV

Oxygen
I    (1)
 13.61805 eV

II   (2)
 35.1211 eV

III  (3)
 54.9355 eV

IV   (4)
 77.41353 eV

V    (5)
113.8990 eV

VI   (6)
138.1197 eV

VII  (7)
739.29 eV   

VIII (8)
871.4101 eV


Ionizing oxygen needs 871 eV?




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Re: ionized HHO
« Reply #21 on: November 30, 2012, 15:47:29 pm »
Hello Steve

this are the energies for each electron. 871 is only the innermost electron...

Here at the physics department they got a linear accelerator and they mess up with oxygen charge to get it accelerated to 9Mev

They initially give two electrons to the oxygens so the get an initial charge, than they accelerate it to a certain degree where than the oxygen is ionized again but at this time taking out 6 electrons i guess, than it is further accelerated...  if i remember well.

A potential difference of 2000v should be able to ionize providing the mean free path is big and you get some free electrons to start with.. this mean a hot filament and some vacuum.


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Re: ionized HHO
« Reply #22 on: November 30, 2012, 16:00:06 pm »
Actually sub-sequential ionization is possible if the relation between electric field and mean free path is right. This mean that most the molecules in the volume will get ionized at once by a cascading event... however if this happens it should appear to the source as a short circuit, since ionized atoms will hit the cathode and extract electrons from it.

A forum member use to tell this theory in the past, without knowing this high current problem... since he never came across it. 

If you think about this is where a choke is needed in home laser devices, to impede shorting the source... since the coil will fight changes in current.

with water should be the same?
raise the volts up to the point where current flows?

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Re: ionized HHO
« Reply #23 on: December 01, 2012, 07:13:52 am »
http://cas.sdss.org/dr6/en/proj/advanced/spectraltypes/energylevels.asp#graph

    amp*volt/sec = watt/sec = joule

Calculate the photon energy of light (in J) with a wavelength of 254 nm
 
E=hf is the photon energy. We can sub in v/w for f and h is plancks constant which is 6.63x10^-34
 
E = h(v/w) = 6.63x10^-34 x ((3x10^8)/254x10^-9)
 = 7.83x10^-19J
 
« Last Edit: December 01, 2012, 07:29:08 am by newguy »