This thread is for youtuber’s: TheUFOefect, and freeThisone

But any one is welcome to answer.

Teach me!

My question’s are that I need to know about light and energy.

Question one.

If a led puts out a wave length of 409 nm can I change the wave length by pulsing the led at high speed? (This just a thought I had while researching but cant find an answer…)

Question two.

What wave length of light is needed to “knock off” electrons of these gases?

1. Hydrogen,

2. Oxygen,

3. Argon,

I know there are deferent valence rings and wave lengths so I guess ill make an led array for the deferent gases and valence rings… ???  so I may have 3 deferent wave length led’s for the same gas… but for each valance ring… yeah?

Now Stan was using just red led’s the enough light energy?  Or did he have 6 deferent wave length red led’s?
 
I would like to order some led’s and make this work but I need to get it right!

Please let me know,

Thanks,

~Russ

I found this searching for hydrogen spectrum:

http://en.wikipedia.org/wiki/Hydrogen_spectral_series  

This site looks like we can use UV and IR LED's also.  Something that Stan didn't have access to:
http://spiff.rit.edu/classes/phys314/lectures/spectra/spectra.html

lol But I didn't pay attention in chemistry class, so I need to learn too. :)

Nate

No, you cannot change the wavelength of the light by pusling the light coming out. c=(freq)(wavelenght)

when dealing with the photoeletric effect you need to know the work function of the object. This is a value in electron volts eV....multiply by the charge of an electron to get your value in joules of energy. The equation for the photoelectric effect is
hf=KE-w. w is the work finction. To knock off an electron you need at least enough energy to overcome the work function. EX: if you supply the photon with 2 ev's and knock it into the electron with the objects workfunction of 1 ev the ejected electron will have a KE of 1 ev.

h is planks constant. 6.63x10^-34. using these values you can figure out how much energy is needed and then figuring out the frequency and wavelength needed. It's tricky stuff no doubt.

If you cant figure it out just try to find some of the work function values and frequencies your using and give the numbers to me. Here are some equations rewritten to find certain values. good luck...Jason

Ps. you need to know the cuttoff frequency (fc)and wavelenght.(LAMBDAc)
KEmax=hFc-w -> Hc/LAMBDAc - w -> LAMBDAc=hc/w

Quote from Rwg42985 on June 8th, 2011, 04:41 AM

This thread is for youtuber’s: TheUFOefect, and freeThisone

But any one is welcome to answer.

Teach me!

My question’s are that I need to know about light and energy.

Question one.

If a led puts out a wave length of 409 nm can I change the wave length by pulsing the led at high speed? (This just a thought I had while researching but cant find an answer…)


Question two.

What wave length of light is needed to “knock off” electrons of these gases?

1. Hydrogen,

2. Oxygen,

3. Argon,

I know there are deferent valence rings and wave lengths so I guess ill make an led array for the deferent gases and valence rings… ???  so I may have 3 deferent wave length led’s for the same gas… but for each valance ring… yeah?

Now Stan was using just red led’s the enough light energy?  Or did he have 6 deferent wave length red led’s?
 
I would like to order some led’s and make this work but I need to get it right!

Please let me know,

Thanks,

~Russ

United nuclear carries the best IR LEDS invisible to the eye.
smaller wave is good, any other wave such as a flash from a camera UV would cause a movment perhaps due to the dilectric properties of the gas, but also carries more energy.
I like the steady wave low frequency, and i high pulse of  blue light for a kick.
But the high intensity of a laser would do a better job all around, because you can focus it on a reflector. Ionising radiation, same this is happening at the gap to a degree, because of heating.

It is possible that a laser pulse would heat the gas much quicker, wasn't it heat that caused the electrons to move to the outer rings? wasn't it the heat from a laser that ionized the air around the pulsed laser rocket? Stan had mentioned pushing the intensity up in several levels. that's why he stepped up the current in the conductor, and, or light intensity?  laser pulsed because of heating, but hey IR is a heat source..

The energy in the electron of a hydrogen atom, are conditioned to add charge, or induce charge on tiny dipoles, and electrons. I think Walter lewin said lower wave lenghth, more energy in the photon emf.  All by itself It will make a great fuel.

As an observer i find it amazing that unpolarized light has a radial field in all directions, due to photon radiation. And at 90 degrees to the source it becomes 100 percent polarized to the observer.  
hope that helpes good stuff Russ. be cool.:cool:

I dont think polarized light is gonna affect anything. if you know what polarized light than you know that it means all of the light waves are on the same plane. All you need to worry about is the frequency, the wavelength of the light depends on the frequency and the speed of light. if you pulsing 1 frequency you cant adjusts its wavelength becasue to do that you would have to change the speed of light in the medium...which isnt gonna happen. A higher frequency has a lower wavelength and a lower frequency has a higher wavelength. a higher frequency/lower wavelength photon will have more energy because the frequenvy is higher, and the oppsite for lower frequency/higher wavelength.

If i were you i would look up MIT's videos on the Photoelectric effect or even khanacademy's videos if they have any.

It's such a simple an beautiful idea that einstien came up with.

Quote from TheUFOeffect on June 8th, 2011, 11:09 PM

I dont think polarized light is gonna affect anything. if you know what polarized light than you know that it means all of the light waves are on the same plane. All you need to worry about is the frequency, the wavelength of the light depends on the frequency and the speed of light. if you pulsing 1 frequency you cant adjusts its wavelength becasue to do that you would have to change the speed of light in the medium...which isnt gonna happen. A higher frequency has a lower wavelength and a lower frequency has a higher wavelength. a higher frequency/lower wavelength photon will have more energy because the frequenvy is higher, and the oppsite for lower frequency/higher wavelength.

If i were you i would look up MIT's videos on the Photoelectric effect or even khanacademy's videos if they have any.

It's such a simple an beautiful idea that einstien came up with.

thanks for the information, i did check the photoelectric effect ,either way,its not only about the plane of the light that becomes polarized, it also works on magnetic domaines to line them up in one direction in a lattice medium, creating the lattice itself perhaps. When dealing with the electrons i believe it was intensity of the light, that matters most for the production of free electrons.
there is more information availible, and thanks for pointing it out. click here photo effect

this is what i found, "Exactly. Higher-frequency photons have more energy, so they should make the electrons come flying out faster; thus, switching to light with the same intensity but a higher frequency should increase the maximum kinetic energy of the emitted electrons. If you leave the frequency the same but crank up the intensity, more electrons should come out (because there are more photons to hit them), but they won't come out any faster, because each individual photon still has the same energy"

In that case, the frequency of the light would make a difference in the photoelectric effect.     And if the frequency is low enough, then none of the photons will have enough energy to knock an electron out of an atom. So if you use really low-frequency light, you shouldn't get any electrons, no matter how high the intensity is. Whereas if you use a high frequency, you should still knock out some electrons even if the intensity is very low.

Quote from freethisone on June 9th, 2011, 04:21 AM

Quote from TheUFOeffect on June 8th, 2011, 11:09 PM

I dont think polarized light is gonna affect anything. if you know what polarized light than you know that it means all of the light waves are on the same plane. All you need to worry about is the frequency, the wavelength of the light depends on the frequency and the speed of light. if you pulsing 1 frequency you cant adjusts its wavelength becasue to do that you would have to change the speed of light in the medium...which isnt gonna happen. A higher frequency has a lower wavelength and a lower frequency has a higher wavelength. a higher frequency/lower wavelength photon will have more energy because the frequenvy is higher, and the oppsite for lower frequency/higher wavelength.

If i were you i would look up MIT's videos on the Photoelectric effect or even khanacademy's videos if they have any.

It's such a simple an beautiful idea that einstien came up with.

thanks for the information, i did check the photoelectric effect ,either way,its not only about the plane of the light that becomes polarized, it also works on magnetic domaines to line them up in one direction in a lattice medium, creating the lattice itself perhaps. When dealing with the electrons i believe it was intensity of the light, that matters most for the production of free electrons.
there is more information availible, and thanks for pointing it out. click here photo effect

this is what i found, "Exactly. Higher-frequency photons have more energy, so they should make the electrons come flying out faster; thus, switching to light with the same intensity but a higher frequency should increase the maximum kinetic energy of the emitted electrons. If you leave the frequency the same but crank up the intensity, more electrons should come out (because there are more photons to hit them), but they won't come out any faster, because each individual photon still has the same energy"

In that case, the frequency of the light would make a difference in the photoelectric effect.     And if the frequency is low enough, then none of the photons will have enough energy to knock an electron out of an atom. So if you use really low-frequency light, you shouldn't get any electrons, no matter how high the intensity is. Whereas if you use a high frequency, you should still knock out some electrons even if the intensity is very low.

Freethisone,

This is my conclusion also as it is what stan states he could control the output... So he adjusted the frequncy of the light... Or led's that's why I asked about the wave length and pulsing...

Ufoefect... What don you think?

Also, still kinda confused as I need to get some led's but don't know what wave length to get... ???? Annu thoughts?

Thanks!

~Russ

If the corona field is to work properly, then there you may have to adjust the gap also. Perhaps this will boost the electron ejections quickly, because he must be using the step up to create higher energy in the electron orbits to weaken them, or to get more energy when they fall back into a nucleus. it would take allot of work to find out the exact frequency, because there is some hills and valley in the way.
 Infra red is a nice steady wave. it makes heat, and its particle counterpart is of interest.. A high intensity  halogen bulb was used in some patents i read..  

Over all it is UV white light that has more energy to knock off electrons percentage wise. it would be OK to use the infra red steady wave as a inductor if that is possible to add charge. hopefully the tiny dipoles will also move because they can no longer absorb energy, but in turn reflects it at twice the pressure. something told me the entire medium will become conducting. (plasma cool) what will cause a change in direction in dipoles, or magnetic domains besides the coil? the medium itself will have 2 emf because some dipoles will be attracted to the magnets, and something may be carried in the dielectric of the gas in the opposed direction. you may find the closer the induction coils are to the magnetic pulsed pump the higher the voltage reading. but if the gas becomes entirely conducting, and moves all together then the output would be the same on all coils regardless of the distance from the pump.

have you ever taken apart a instant camera so you can use the flash bulb inside, and its circuit? Even a flash will work, so your not limited.

Quote from Rwg42985 on June 8th, 2011, 04:41 AM

This thread is for youtuber’s: TheUFOefect, and freeThisone

But any one is welcome to answer.

Teach me!

My question’s are that I need to know about light and energy.

Question one.

If a led puts out a wave length of 409 nm can I change the wave length by pulsing the led at high speed? (This just a thought I had while researching but cant find an answer…)

Question two.

What wave length of light is needed to “knock off” electrons of these gases?

1. Hydrogen,

2. Oxygen,

3. Argon,

I know there are deferent valence rings and wave lengths so I guess ill make an led array for the deferent gases and valence rings… ???  so I may have 3 deferent wave length led’s for the same gas… but for each valance ring… yeah?

Now Stan was using just red led’s the enough light energy?  Or did he have 6 deferent wave length red led’s?
 
I would like to order some led’s and make this work but I need to get it right!

Please let me know,

Thanks,

~Russ

Hi all I know Stans first, second and third patents harvested radiant solar energies.  Those patents (mostly 3rd) specify various wave lengths thru the light guide lenses.  He never specifies how he is going to utilize these wave lengths in the patent?? Perhaps for this?  

From what I can find searching, an atom that absorbs a certain wavelength also emits that same wavelength when it is excited.  So does that mean if we test oxygen with a 558 nm green LED bombardment it should cause a flame to burn green when in contact with the oxygen?  



I'm finding lots of spectrum graphs like that, I just haven't had time to compile them all to figure out which LED's would work best for more than one type of gas.

Nate

Quote from firepinto on June 9th, 2011, 06:03 PM

From what I can find searching, an atom that absorbs a certain wavelength also emits that same wavelength when it is excited.  So does that mean if we test oxygen with a 558 nm green LED bombardment it should cause a flame to burn green when in contact with the oxygen?  



I'm finding lots of spectrum graphs like that, I just haven't had time to compile them all to figure out which LED's would work best for more than one type of gas.

Nate

I have done the the same things... And I want to get some 3 defrent led's so
I can do each gas... So please keep it comming!

I did my research just like you are... Let's see if you came up with the same things...

Make sence to every one else?

~Russ

Maybe it is also doable in reverse than high power could be used.
I did some lighting for bands and festivals in the past, there we use light source with wide spectrum and filter out the freq we don't
want to use which get out as heat in the filter.

For example this filter has the highest transmission of 525 nm;Green filter
All there filters have wavelength graphs and can be combined to get very specific wavelength.

In lighting we get the best colors using gas discharge lamps with
the highest color temp. 5000K or higher often with MSD lamps.
(Note: these need high temp glas filter above example is gel type)
This is some info about different lamps acronyms;

Code: [Select]


MSR = Medium Source Rare-Earth
Philips designation for a medium arc length metal halide arc lamp using dysprosium iodide as the halide fill. These are all double-ended discharge tubes mounted in single-ended envelopes and bases.
Power variants available (depending on range) tend to be 125W, 200W, 250W, 400W, 575W, 700W, 1200W, 2500W, 4000W, 6000W, 12kW and 18kW.
The variants above 1200W are rare in theatrical use, mainly being seen on film sets.
Remember the 18kW lamp puts out nearly 1.75 megalumens. This is equivalent to something like 150 500W tungsten halogen lights.

HSR = Osram equivalent part
CSR = GE Equivalent part
/2 variants have higher colour temperature (7200K instead of 5900K).
HR types have hot-restrike capability (usually requires base with better insulation to cope with the >10kV restrike voltages- so 575/HR is only available in G22 base, not GX9.5)
These usually have a lifetime around 750-1000 hours although some low wattage lamps like the MSR 200/HR has a lifetime of only 200h.

CSI
(Compact Source Iodide) A high intensity discharge lamp. Most often used in followspots, because it has a color temperature (approx. 4000K) close to that of the tungsten halogen lamps.

CID
(Compact Iodide Daylight) A high intensity discharge lamp that produces a light similar in color temperature to daylight approx. 5500K). A 1000W CID lamp produces 2.5 times more light than a 2000W tungsten halogen source.

HMI
(Hydrargyrum medium-arc iodide) A mercury-halide gas discharge medium arc-length lamp with a multi-line spectra emission. Hydrargyrum = Mercury (Hg). 6000K. Single or double-ended. Single end has outer bulb. Hot restrike.

HTI: 4300-7500K. Single or double-ended. No outer bulb. Double-ended can have hot restrike.

HMP: A variation of HMI, specifically for OHP's. Features might include dim/boost.
HSD: 150-1200W. 6000-8000K. Single-ended.
HCD: 35/70/150W. 3000-4200K. Sinlge-ended. Includes UV filter in outer bulb.

MSD lamps - slightly different gas fill. (limit to 1200w)
Arcs are slightly longer than the MSR type.
Colour temperature is around 6000K, although 8500K is these as well.
Spectral peaks tend to be much sharper- the overall spectrum appears less balanced than that of the MSR lamp.

The main advantage of the MSD lamp is the significantly longer lifetime at 2000-3000 hours.
MSD lamps are also rather more expensive than the equivalent MSR types.

Electrically the MSR and MSD lamps of a given wattage appear to be interchangeable. Mac500s and 600s support both, for instance.


In scrap these are in beamers and some projection tvs and xenon lamps for cars.
Speaking of xenon lamps I've seen a vid where people flow there hho under vacuum direct over a 5000K xenon bulb and than in car.
(very danger!!)
And claiming it made a huge improvement for the gas, but could not find it again 123.

So the question is; do you need only one wavelength in the light source
or do you need that one wavelength available in the light source ?

If it is the later then I can see why hanging a xenon bulb on some wire in a glass jar with hho vacuum can be a live risking experiment :D

Also when using multiple light sources we get Interference;
http://en.wikipedia.org/wiki/Light_interference
And every other medium light travels through changes light a bit, didn't Stan have the leds front direct in the gas ?
And if using diffent wavelength leds together for tuning, I would not pulse them but dim them by lowering voltage.

Last thought I had, using single wide spectrum light source with a prisma would give a very tunable setup.



my 2ct.

Good thoughts willam!

Now as far as what I'm doing... Looks like 630nm and 660nm it's the closest I can get for what I think I want to do...

So I will be placing these on the HGG with or with out the UV... Haven't decided yet... Really the uv looks good for the argon...

Any hoo thanks for all the input! Stan used red... Red it is...

~Russ

He would have been adjusting the frequency to the right amount of energy to release the electrons of the atoms. I'm not sure if you can get led's to emitt the right wavelength unless theres a company that sells them at an specific wavelength you need.

Quote from TheUFOeffect on June 12th, 2011, 12:53 AM

He would have been adjusting the frequency to the right amount of energy to release the electrons of the atoms. I'm not sure if you can get led's to emitt the right wavelength unless theres a company that sells them at an specific wavelength you need.

Well, the led's he was using was ither 630nm or 660nm so would one of these be able to do this? And they make a full spectrum of wave length...

I will have 405nm(uv) 630nm(red) 660nm(red) white  (full spectrum)

So the reds are what i want but the uv and white is for other gases...

What do you think about 330 and 360nm? Will this fit in to your math?

What do you think?

I will have 16-330nm 16-360nm and 4white and 24-405nm supper bright LEDs... So a totle of 60 led's of way brighter than stanly had back then... So it's alot more light then he had at the time...

I can get you specs of the reg led's and white...

Go to http://www.superbrightleds.com/cgi-bin/store/index.cgi?action=DispPage&Page2Disp=%2Fleds.htm and look down the list for these part numbers:

RL5-R8030: 5mm Red LED
RL5-R2415: 5mm Red LED
RL5-W18030: 5mm White LED

It will give you all the info you need to do your calculating?

Please let ne know tour thoughts!

Thanks for the help!

~Russ

No,
1) Oxygen doesnt burn
2) You cant just shoot a certain color light at an object and get its flame to burn that color.
3) to accurately figure out what wavelength you need you need to find the metal or non metal's work function


Just give me the workfunction of the gases and ill try to do the math and ill type it up and show you how i did it.

Ok so i've done some more research and math and stuff and im honestly not sure why he's expecting to knock off electrons with the led's. Were working with gas's of non metals. The photoelectric effect using light of certain frequencies to knock off electrons is used with metals, mainly charged metals with free roaming electrons. Were dealing with actuall electrons that are attatched to the atom still in thier valence shell. In order to do this you need to hit the atoms with energy...thats where the plsma combustion chamber comes in. The energy in the ark needs to be equal or greater than the ionization energy of that atom. The Ionization energy is the amount of energy it takes to remove an electrons from an valence shell.

Neutral hydrogen has 1 electron.
1st Ionization energy of hydrogen is 1312 KJ/mol

Oxygens Ionization energies:
1st Ionization energy of Oxygen is 1313.9 KJ/mol
2nd Ionization energy of Oxygen is 3388.5 KJ/mol
3rd Ionization energy of Oxygen is 5300.5 KJ/mol
4th Ionization energy of Oxygen is 7469.2 KJ/mol
5th Ionization energy of Oxygen is 10989.5 KJ/mol
6th Ionization energy of Oxygen is 13326.5 KJ/mol
7th Ionization energy of OXygen is 71330 KJ/mol
8th Ionization energy of Oxygen is 84078.0 KJ/mol


Argons Ionization Energies:
1st Ionization energy of Argon is 1520.6 KJ/mol
2nd Ionization energy of Argon is 2665.8 KJ/mol
3rd Ionization energy of Argon is 3931   KJ/mol
4th Ionization energy of Argon is 5771   KJ/mol
5th Ionization energy of Argon is 7238 KJ/mol
6th Ionization energy of Argon is 8781   KJ/mol
7th Ionization energy of Argon is 11995 KJ/mol
8th Ionzation energy of Argon is 13842 KJ/mol

Hopre this helps

Quote from TheUFOeffect on June 12th, 2011, 05:52 PM

Ok so i've done some more research and math and stuff and im honestly not sure why he's expecting to knock off electrons with the led's. Were working with gas's of non metals. The photoelectric effect using light of certain frequencies to knock off electrons is used with metals, mainly charged metals with free roaming electrons. Were dealing with actuall electrons that are attatched to the atom still in thier valence shell. In order to do this you need to hit the atoms with energy...thats where the plsma combustion chamber comes in. The energy in the ark needs to be equal or greater than the ionization energy of that atom. The Ionization energy is the amount of energy it takes to remove an electrons from an valence shell.

Neutral hydrogen has 1 electron.
1st Ionization energy of hydrogen is 1312 KJ/mol

Oxygens Ionization energies:
1st Ionization energy of Oxygen is 1313.9 KJ/mol
2nd Ionization energy of Oxygen is 3388.5 KJ/mol
3rd Ionization energy of Oxygen is 5300.5 KJ/mol
4th Ionization energy of Oxygen is 7469.2 KJ/mol
5th Ionization energy of Oxygen is 10989.5 KJ/mol
6th Ionization energy of Oxygen is 13326.5 KJ/mol
7th Ionization energy of OXygen is 71330 KJ/mol
8th Ionization energy of Oxygen is 84078.0 KJ/mol


Argons Ionization Energies:
1st Ionization energy of Argon is 1520.6 KJ/mol
2nd Ionization energy of Argon is 2665.8 KJ/mol
3rd Ionization energy of Argon is 3931   KJ/mol
4th Ionization energy of Argon is 5771   KJ/mol
5th Ionization energy of Argon is 7238 KJ/mol
6th Ionization energy of Argon is 8781   KJ/mol
7th Ionization energy of Argon is 11995 KJ/mol
8th Ionzation energy of Argon is 13842 KJ/mol

Hopre this helps

I think the LEDs are half of the solution that Stan had in mind.  The high voltage field between the tubes puts the atoms in resonance.  Possibly stretching the electrons orbit to the point were a photon of the correct wavelength could be absorbed easier.  Then the electron would be forced to move up a level because the space was filled by the photon.  

The energy that is calculated is also the amount of energy released when the electron drops back down a level?


Russ:
The white LEDs will cover a wider spectrum of light, but the wattage  is divided across the spectrum.  A LED with the closest wavelength for a gas will have more wattage applied to the wavelength absorbed by that particular gas.  But maybe modern super bright LEDs will make that less of a problem.

I think the only way to know that this works is test burns of treated gas in a chamber with a slide piston and markings of distance.  Then a burn rate could be found, and weather it is explosion or implosion.  

Nate

Electrons dont move up an energy level because the photon take its place momentarily and pushes the electron up. The electron is given just the right amount of energy to bump up an energy level. The electron will move up because it took in the right amount of eV's adn then back down and will release a photon of the energy it absorbed

Ok, here are some links with some wavelengths.
hydrogen
Strong Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/hydrogentable2.htm
Persistant Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/hydrogentable3.htm
Oxygen
Strong Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/oxygentable2.htm
Persistant Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/oxygentable3.htm
Argon
Strong Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/argontable2.htm
Persistant Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/argontable3.htm

Quote from TheUFOeffect on June 12th, 2011, 10:17 PM

Electrons dont move up an energy level because the photon take its place momentarily and pushes the electron up. The electron is given just the right amount of energy to bump up an energy level. The electron will move up because it took in the right amount of eV's adn then back down and will release a photon of the energy it absorbed


Ok, here are some links with some wavelengths.
hydrogen
Strong Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/hydrogentable2.htm
Persistant Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/hydrogentable3.htm
Oxygen
Strong Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/oxygentable2.htm
Persistant Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/oxygentable3.htm
Argon
Strong Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/argontable2.htm
Persistant Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/argontable3.htm

Here on NASA site it looks like it states 487nm is needed for ionizing the hydrogen atom for removal of the electron to enhance its energy state?? I am no physicist so.... The link below offers the light spectrum at 487nm.

http://imagine.gsfc.nasa.gov/docs/teachers/lessons/xray_spectra/worksheet-energy-sol.html

Solution for Calculate the Energy! Student Worksheet
Neils Bohr numbered the energy levels (n) of hydrogen, with level 1 (n=1) being the ground state, level 2 being the first excited state, and so on. Remember that there is a maximum energy that each electron can have and still be part of its atom. Beyond that energy, the electron is no longer bound to the nucleus of the atom and it is considered to be ionized. In that case n approaches infinity.
The equation for determining the energy of any state (the nth) is as follows:

 
Because the energy is so small, the energy is measured in electron-volts, designated by "eV". 1 eV = 1.6 x 10-19 J.
Answer the following questions:

1. Using the above expression, calculate the energy of the first excited state. Your answer will be negative. This signifies that the electron is bound to the atom (as opposed to being a free electron).


For the first excited state, n=2. Using this in the above equation gives E = -3.40 eV
2 . Use the above expression to find the energy of the photon released when an electron around a hydrogen atom moves from the 4th to the 2nd level.


The energy of the photon is found by computing the difference
in the energies of the fourth (n=4) and second (n=2) levels
E = -13.6/42 - (-13.6/22)
E = -0.85 + 3.40
E = 2.55 eV


3. Now use the above expression to find the energy of the photon released when a free electron is captured to the 2nd level.


We represent a free electron by assigning it an infinite n. Hence, its energy is zero.
The energy of the photon emitted by a free electron captured to the n=2 level is thus
E = 0 - (-13.6/22) = 3.4 eV
4. Use the relationship between a photon's energy and its wavelength to calculate the wavelength of the photon emitted in question 2.


From the Calculation Investigation, we learned that energy and wavelength are related through E = h c / l.
We can solve this for the wavelength, l = h c / E. where h = 6.626 x 10-34J-s, and c = 3 x 108 m/s.
We convert our energy E= 2.55 ev into Joules using 1 eV = 1.6x10-19 J. This gives an energy of E = 4.08 x 10-19 J.
We then find a wavelength of
l = ((6.626 x 10-34) x (3 x 108)) / (4.08 x 10-19)
l = 4.87 x 10-7 m.
OR, using 1 nm = 1 x 10-9 m,

l = 487 nm.

Scott

We's learned all of this in school; unforunitaly you can really only use the bohr arom for Hydrogen and if i remember right other atoms with only 1 electron( He+1, Li +2, Be +3)

Quote from HHOstar on June 13th, 2011, 07:59 PM

Quote from TheUFOeffect on June 12th, 2011, 10:17 PM

Electrons dont move up an energy level because the photon take its place momentarily and pushes the electron up. The electron is given just the right amount of energy to bump up an energy level. The electron will move up because it took in the right amount of eV's adn then back down and will release a photon of the energy it absorbed


Ok, here are some links with some wavelengths.
hydrogen
Strong Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/hydrogentable2.htm
Persistant Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/hydrogentable3.htm
Oxygen
Strong Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/oxygentable2.htm
Persistant Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/oxygentable3.htm
Argon
Strong Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/argontable2.htm
Persistant Lines: http://physics.nist.gov/PhysRefData/Handbook/Tables/argontable3.htm

Here on NASA site it looks like it states 487nm is needed for ionizing the hydrogen atom for removal of the electron to enhance its energy state?? I am no physicist so.... The link below offers the light spectrum at 487nm.

http://imagine.gsfc.nasa.gov/docs/teachers/lessons/xray_spectra/worksheet-energy-sol.html

Solution for Calculate the Energy! Student Worksheet
Neils Bohr numbered the energy levels (n) of hydrogen, with level 1 (n=1) being the ground state, level 2 being the first excited state, and so on. Remember that there is a maximum energy that each electron can have and still be part of its atom. Beyond that energy, the electron is no longer bound to the nucleus of the atom and it is considered to be ionized. In that case n approaches infinity.
The equation for determining the energy of any state (the nth) is as follows:

 
Because the energy is so small, the energy is measured in electron-volts, designated by "eV". 1 eV = 1.6 x 10-19 J.
Answer the following questions:

1. Using the above expression, calculate the energy of the first excited state. Your answer will be negative. This signifies that the electron is bound to the atom (as opposed to being a free electron).


For the first excited state, n=2. Using this in the above equation gives E = -3.40 eV
2 . Use the above expression to find the energy of the photon released when an electron around a hydrogen atom moves from the 4th to the 2nd level.


The energy of the photon is found by computing the difference
in the energies of the fourth (n=4) and second (n=2) levels
E = -13.6/42 - (-13.6/22)
E = -0.85 + 3.40
E = 2.55 eV


3. Now use the above expression to find the energy of the photon released when a free electron is captured to the 2nd level.


We represent a free electron by assigning it an infinite n. Hence, its energy is zero.
The energy of the photon emitted by a free electron captured to the n=2 level is thus
E = 0 - (-13.6/22) = 3.4 eV
4. Use the relationship between a photon's energy and its wavelength to calculate the wavelength of the photon emitted in question 2.


From the Calculation Investigation, we learned that energy and wavelength are related through E = h c / l.
We can solve this for the wavelength, l = h c / E. where h = 6.626 x 10-34J-s, and c = 3 x 108 m/s.
We convert our energy E= 2.55 ev into Joules using 1 eV = 1.6x10-19 J. This gives an energy of E = 4.08 x 10-19 J.
We then find a wavelength of
l = ((6.626 x 10-34) x (3 x 108)) / (4.08 x 10-19)
l = 4.87 x 10-7 m.
OR, using 1 nm = 1 x 10-9 m,

l = 487 nm.

Scott

Also on the spectrum there are also what looks like about 435nm and 658nm witch if I understand is the other valance rings… so not only 487nm but there are also other wavelengths for other valance rings???? Yes????

And the 658nm is in the red section…  


Also like firepinto was thinking, Stan was using the photon energy to weaken the bonds and then use less electoral energy to the same amount of work… ??? and was extracting electron until there was at least less than 4 electrons left.

This let the gas to become unstable and when burned the water could not form H2o.
and this meant there was a grater energy release than if burned with out being processed...
???

Thanks!

~Russ

good work on UFOeffect lots of info.
 
what about the corona effect? 40,000kv was said by Stan, as It is the Tesla effect. lol Ionizing radiation. cool 03..
its almost like a duel spark gap, one at the base and one for the corona effect. do you have more info on this? will a spark pass between the gap, or will there be no spark?

Stan said he stretch the HHO, and hit it with light at descriete frequency,  were these indeed laser diodes?
lets find the resnant frequency of the corona cavity. Stepped up in Harmonic levels.

Quote from freethisone on June 16th, 2011, 07:59 PM

good work on UFOeffect lots of info.
 
what about the corona effect? 40,000kv was said by Stan, as It is the Tesla effect. lol Ionizing radiation. cool 03..
its almost like a duel spark gap, one at the base and one for the corona effect. do you have more info on this? will a spark pass between the gap, or will there be no spark?

Stan said he stretch the HHO, and hit it with light at descriete frequency,  were these indeed laser diodes?
lets find the resnant frequency of the corona cavity. Stepped up in Harmonic levels.

I ran out of time to go over this in my last video, but the were diodes Led's not lasers. ~Russ

Quote from Rwg42985 on June 9th, 2011, 05:16 AM

Quote from freethisone on June 9th, 2011, 04:21 AM

Quote from TheUFOeffect on June 8th, 2011, 11:09 PM

I dont think polarized light is gonna affect anything. if you know what polarized light than you know that it means all of the light waves are on the same plane. All you need to worry about is the frequency, the wavelength of the light depends on the frequency and the speed of light. if you pulsing 1 frequency you cant adjusts its wavelength becasue to do that you would have to change the speed of light in the medium...which isnt gonna happen. A higher frequency has a lower wavelength and a lower frequency has a higher wavelength. a higher frequency/lower wavelength photon will have more energy because the frequenvy is higher, and the oppsite for lower frequency/higher wavelength.

If i were you i would look up MIT's videos on the Photoelectric effect or even khanacademy's videos if they have any.

It's such a simple an beautiful idea that einstien came up with.

thanks for the information, i did check the photoelectric effect ,either way,its not only about the plane of the light that becomes polarized, it also works on magnetic domaines to line them up in one direction in a lattice medium, creating the lattice itself perhaps. When dealing with the electrons i believe it was intensity of the light, that matters most for the production of free electrons.
there is more information availible, and thanks for pointing it out. click here photo effect

this is what i found, "Exactly. Higher-frequency photons have more energy, so they should make the electrons come flying out faster; thus, switching to light with the same intensity but a higher frequency should increase the maximum kinetic energy of the emitted electrons. If you leave the frequency the same but crank up the intensity, more electrons should come out (because there are more photons to hit them), but they won't come out any faster, because each individual photon still has the same energy"

In that case, the frequency of the light would make a difference in the photoelectric effect.     And if the frequency is low enough, then none of the photons will have enough energy to knock an electron out of an atom. So if you use really low-frequency light, you shouldn't get any electrons, no matter how high the intensity is. Whereas if you use a high frequency, you should still knock out some electrons even if the intensity is very low.

Freethisone,

This is my conclusion also as it is what stan states he could control the output... So he adjusted the frequncy of the light... Or led's that's why I asked about the wave length and pulsing...

Ufoefect... What don you think?

Also, still kinda confused as I need to get some led's but don't know what wave length to get... ???? Annu thoughts?

Thanks!

~Russ

Russ and all, the part of the light spectrum you want to use is dependant on the make up of the mag-gas you are using. Different atoms have different photon emission spectra . Hydrogen has a completely different emission spectra than argon, and depending on what you are doping the gas with, like carbon, gold, magnesium, calcium or what have you - the nanometer rating of the LED's or laser's will be different.
The photon's must be resonant with the atoms or they wont be absorbed.
To instantly ionize hydrogen - 91nm works best. The thing is - no one makes a 91nm laser or LED.
:blush:
If you have a few thousand bucks, I'll make you a 91nm laser.
:D