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Author Topic: Is there any voltage in between electron and proton in hydrogen atoms?  (Read 15185 times)

Offline jccc

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Quote from: jccc
Particle decay may caused by electricity leakage whinin atom. Obviously atom is build by electrical potential.
Nope. But what are you talking about specifically? What particles inside the atom are you referring to when you say that they are decaying? Nuclei? Proton? Neutron? Electron?

Quote from: jccc
Seems to complete the puzzle of atom structure, we need a negative sub-charged new particle,
Nope. There are currently no missing pieces of any puzzle dealing with atomic structure.

Quote from: jccc
it fills the gape in between nucleus and electrons in atoms, and it fills the whole space to conduct EM wave.
That's meaningless.

Pete, maybe I am blind into my soul.
 

Offline PmbPhy

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Quote from: jccc
Pete, maybe I am blind into my soul.
That's only because you don't want to give up the idea that nature should be intuitively meaningful. There's no reason to expect that. Our minds were constructed by millions to billions of years of evolution to survive I a macroscopic world. Our minds were never meant to understand the quantum mechanical world. Those of use who accept that easily learn the subject and move on. Those who refuse to accept that fact never understand it and stay in a rut and keep quoting Feynman who said that nobody understands quantum physics. While in a certain sense that has a bit of truth to it, it's often used as an excuse not to learn it.
 

Offline jccc

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Whenever a neutron decays, the proton must attract some negative goo from somewhere, so we are looking for a continuum of negatively charged....particles? Each with infinitesimal charge, pervading the universe.

Do you understand/agree with the above? 
 

Offline PmbPhy

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Quote from: jccc
Whenever a neutron decays, the proton must attract some negative goo from somewhere, ...
What is "goo"? By the way. Neutrons don't decay while they're in the nucleus.

Quote from: jccc
... so we are looking for a continuum of negatively charged....particles?
First you started out with one neutron decaying into one proton which attracts one negatively charged particle such as an electron. Why the continuum of negatively charged particles? And negatively charged particles cannot form a continuum because they're quantized, i.e. come in discrete amounts.

Quote from: jccc
Do you understand/agree with the above?
As usual, you're not making any sense.
 

Offline jccc

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whenever a neutron decays, the proton must attract some negative goo from somewhere, so we are looking for a continuum of negatively charged....particles? each with infinitesimal charge, pervading the universe.

So far, it seems that charge is quantised, but if you can find and demonstrate a continuous subquantum of charge, you may have a point.

Pete, that's from a comment about my atomic structure theory. Read his other comments learned he knows lot more than me. Whole lot more.

 

Offline jccc

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Back to the topic.

If high voltage within atom is real thing, why can't we utilize it?

It is like a water reserver in the air, all we need is to open it to release the water.

 

Offline UltimateTheory

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If high voltage within atom is real thing, why can't we utilize it?

There is no high voltage within atom..

It is like a water reserver in the air, all we need is to open it to release the water.

I told you how to do it - using antimatter/antiparticles.
That's the only way to get the all energy from rest-mass of f.e. Hydrogen.
« Last Edit: 30/07/2014 17:46:33 by UltimateTheory »
 

Offline JP

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Again, please knock off the personal attacks.  I just deleted several posts that went over the line.  If you disagree with each other, please keep it to the science and if you disagree with someone's science, point out the error and back it up with evidence.

-The Mods
 

Offline PmbPhy

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Quote from: UltimateTheory
I told you how to do it - using antimatter/antiparticles.
That's the only way to get the all energy from rest-mass of f.e. Hydrogen.
Note: I believe that when UT wrote get the all energy he reallly meant to write get all the energy. Therefore when I quote him below I'll use the later phrasing.

Using matter/antimatter annihilation to get all the energy from rest mass is a very common misconception. This is all explained in the article Does nature convert mass into energy? by Ralph Baierlein, Am. J. Phys., 75(4), Apr. (2007). See - http://scitation.aip.org/content/aapt/journal/ajp/75/4/10.1119/1.2431183

I put the article on my website at http://home.comcast.net/~peter.m.brown/ref/baierlein.pdf

The abstract reads
Quote
First I provide some history of how the equation E=mc2 arose, establish what “mass” means in the context of this relation, and present some aspects of how the relation can be understood. Then I address the question, Does E=mc2 mean that one can “convert mass into energy” and vice versa?

In it the author writes
Quote
Q. Does the equation E = mc2 mean that one can "convert mass into energy" and vice versa"?
A. Not really, but the issue is complex, and eminent physicists have used that phrase and variants of it.

This is only one article that points this out. Any physics journal article that I've ever read on the subject agrees with the conclusion of this article. And I've read a lot of articles on this subject. Regardless of what I read I know what the author says is true even without reading it. Proof given upon request. The main part of the argument is that energy is conserved so that the amount of energy that you start with is the same energy you end up with.

When people refer to changing mass to energy using electron/positron annihilation what they really mean is that they start out with rest mass and end up with photons which are all kinetic energy and zero rest mass.  The energy of photons is said to be all kinetic energy.

It's also a common misconception to believe that when a proton and antiproton annihilate the end result is all photons/kinetic energy with zero rest mass. That too is not true. In such annihilations you end up with either d22abb5468abb8cc4a37d91cee06c96d.gif or  cb8b96aee3785dd39803dc965b8ed103.gif. But to be precise see http://en.wikipedia.org/wiki/Annihilation#Proton-antiproton_annihilation  which explains
Quote
When a proton encounters its antiparticle (and more generally, if any species of baryon encounters any species of antibaryon), the reaction is not as simple as electron-positron annihilation. Unlike an electron, a proton is a composite particle consisting of three "valence quarks" and an indeterminate number of "sea quarks" bound by gluons. Thus, when a proton encounters an antiproton, one of its constituent valence quarks may annihilate with an antiquark, while the remaining quarks and antiquarks will undergo rearrangement into a number of mesons (mostly pions and kaons), which will fly away from the annihilation point. The newly created mesons are unstable, and will decay in a series of reactions that ultimately produce nothing but gamma rays, electrons, positrons, and neutrinos. This type of reaction will occur between any baryon (particle consisting of three quarks) and any antibaryon (consisting of three antiquarks). Antiprotons can and do annihilate with neutrons, and likewise antineutrons can annihilate with protons, as discussed below.
 

Offline PmbPhy

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Quote from: jccc
If high voltage within atom is real thing, why can't we utilize it?
Didn't you ever calculate it like I showed you? It's even easier to get an estimate. Suppose

r = 5.3x10-11m
k = 8.99x109Nm2/C2
e = 1.6022x-19
V = ke/r

Then upon plugging these in we get to two significant figures

V = 27 Volts

That's hardly high voltage.

Quote from: jccc
It is like a water reserver in the air, all we need is to open it to release the water.
 

Offline jccc

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Quantum law? Well, put a quantum wire cross electron and proton so complete the circle.


What exactly do you think would be the charge carrier in that wire? Macroscopic wires work by transporting electrons. If you somehow had some sort of "quantum wire" connecting an electron to a proton, you still would not get any transfer of charge in either direction because the wire is smaller than an electron. Also the electron and proton don't have any way to donate their charge--their charges are constant and part of what define them as what they are. In the case of an electron the charge is indivisible. I suppose one could argue that the proton's charge could be divided into components based on quarks, but don't expect the positively charged quarks to start leaking out of a proton just because there is a "wire" connecting it to an electron.

The middle shell electron cloud should be the current carrier without wire. Maybe?
 

Offline PmbPhy

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Quote from: jccc
The middle shell electron cloud should be the current carrier without wire. Maybe?
No.
 

Offline UltimateTheory

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It's also a common misconception to believe that when a proton and antiproton annihilate the end result is all photons/kinetic energy with zero rest mass. That too is not true. In such annihilations you end up with either d22abb5468abb8cc4a37d91cee06c96d.gif or  cb8b96aee3785dd39803dc965b8ed103.gif.

Pair of 2 mesons is very very rare. They happens just in percents of annihilations.
And neither pair of pion-, pion+, nor pair of kaon-,kaon+ are stable. So they quickly (within micro seconds usually), decay.

The most common proton-antiproton annihilation branch is to 3 pions 0, AFAIK.

And the most common pion0 decay branch is to 2 gamma photons. This gives us, the most common (the most probable) branch of decay p+p- -> 5fd36ee593105557d14db2a02769f674.gif -> 6 gamma photons.

Which is what I wrote 2 weeks ago to jccc in thread:
http://www.thenakedscientists.com/forum/index.php?topic=26362.msg438099#msg438099
post #140
« Last Edit: 31/07/2014 02:11:22 by UltimateTheory »
 

Offline PmbPhy

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Quote from: UltimateTheory
Pair of 2 mesons is very very rare.
And yet if I didn't mention them I'd have risked being remiss.

Quote from: UltimateTheory
And neither pair of pion-, pion+, nor pair of kaon-,kaon+ are stable.
Why would you even say such a thing when I gave the reference which explains
Quote
The newly created mesons are unstable, and will decay in a series of reactions that ultimately produce nothing but gamma rays, electrons, positrons, and neutrinos.

Quote from: UltimateTheory
The most common proton-antiproton annihilation branch is to 3 pions 0, AFAIK.

You missed the point. You implied that they all end up as photons when in fact
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The newly created mesons are unstable, and will decay in a series of reactions that ultimately produce nothing but gamma rays, electrons, positrons, and neutrinos.
So the series doesn't always end up with photons but with electrons, positrons and neutrinos.
 

Offline UltimateTheory

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So the series doesn't always end up with photons but with electrons, positrons and neutrinos.

True.
But it is in minority of cases.
I don't have to explain the all decay modes every time.
Enough is explaining the main branch, and give reference link for details, for rare exceptions.
(especially explaining something to non-physicist that won't understand quarter of it)

Post #140 in that thread was to jccc. So the same here. No need to repeat writing the same multiple times to the same person.

You should instead concentrate on answering to jccc, as he wants to get 938.272 MeV energy from proton, without any annihilation..
 

Offline PmbPhy

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Quote from: UltimateTheory
You should instead concentrate on answering to jccc, as he wants to get 938.272 MeV energy from proton, without any annihilation..
If that's what he indicated or implied that he wanted I would. But that's not what he's looking for. Just ask him and he'll tell you. Right, Joe?

In any case the assertion That's the only way to get the all energy from rest-mass of f.e. Hydrogen is questionable since the energy is already there to begin with. There's no "getting" energy from anywhere. Unless you really meant that its the only way to change the form of energy, i.e. from rest energy to kinetic energy? Is that it?
« Last Edit: 31/07/2014 15:05:00 by PmbPhy »
 

Offline PmbPhy

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Quote from: UltimateTheory
True.
But it is in minority of cases.
Then you should have said "most" rather than "all."

Quote from: UltimateTheory
I don't have to explain the all decay modes every time.
You should. Never imply something happens all the time if it doesn't. The author of the text I'm proof reading made this same mistake until I pointed it out to him. When he realized that he made that mistake he changed it quickly.

re - Enough is explaining the main branch, and give reference link for details, for rare exceptions. - If that's the case then please show me where you're getting your data from. I want to see this for myself.
« Last Edit: 31/07/2014 15:32:08 by PmbPhy »
 

Offline UltimateTheory

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If that's what he indicated or implied that he wanted I would. But that's not what he's looking for. Just ask him and he'll tell you. Right, Joe?

See post #139 by jccc in thread http://www.thenakedscientists.com/forum/index.php?topic=26362.msg438098#msg438098
 

Offline PmbPhy

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If that's what he indicated or implied that he wanted I would. But that's not what he's looking for. Just ask him and he'll tell you. Right, Joe?

See post #139 by jccc in thread http://www.thenakedscientists.com/forum/index.php?topic=26362.msg438098#msg438098
Okay. Although why you took so long to answer is odd. No wonder I missed it.

That doesn't answer my other question, where did you get your data from? I want to confirm that what you claim is true.
 

Offline jccc

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May I collect a little transpassing fee? Pretty broke lately.
 

Offline PmbPhy

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May I collect a little transpassing fee? Pretty broke lately.
What exactly is transpassing?
 

Offline jccc

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May I collect a little transpassing fee? Pretty broke lately.
What exactly is transpassing?

Every time I am nearing broke, making lot typos. Never been there Pete?
 

Offline jccc

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Quote from: jccc
If high voltage within atom is real thing, why can't we utilize it?
Didn't you ever calculate it like I showed you? It's even easier to get an estimate. Suppose

r = 5.3x10-11m
k = 8.99x109Nm2/C2
e = 1.6022x-19
V = ke/r

Then upon plugging these in we get to two significant figures

V = 27 Volts

That's hardly high voltage.

Quote from: jccc
It is like a water reserver in the air, all we need is to open it to release the water.

Thank you Pete, hope you like my Jokes.

V=Ke/r, so V and r dependent? Why is voltage between two points in a conductor independent with distance?

Also 27 volt is a lot compare with a car battery, can we use liquid hydrogen put in two metal plates to make a 27 V battery? Electron move to anode plate, proton move to the cathode plate. 
 

Offline jccc

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If pete was right on the 27 volt within H atom, why can we use the power?

Isn't he always right?
 

Offline evan_au

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why can't we use the 27 volts within H atom as a battery?

You must have another participant in the reaction which applies the 13V needed to extract an electron from the Hydrogen 1s shell (13eV of energy).

When the electron falls back into the ground state, it will then release this 13V (sometimes as a 13eV ultraviolet photon).

So it's not very useful to have a battery which consumes 13V to produce 13V. Given inefficiencies of typical chemical reactions, you are better off with a brick.

The reason it does not produce 27V is that the electron can't fall all the way into the nucleus, due to the quantum nature of electrons and atoms.
 

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