Re: The electron shell model is in challenge

From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 08/24/04 

Date: Tue, 24 Aug 2004 11:20:34 +0200

Y.Porat wrote:
> Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cgckm9$4b3$1@news.urz.uni-heidelberg.de>...
>
>>Y.Porat wrote:
>>
>>>Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cg4dap$p63$1@news.urz.uni-heidelberg.de>...
>>>
>>
>>
>>
>>>>>>Who are you to judge this? You don't know the heap of literature
>>>>>>of applying it to the heavy elements. You do not even know a single
>>>>>>one of those thousands of articles!
>>
>>You keep ignoring this argument.
>
> -----------------
> a lot of bla bla bla that has to be snipped:

You mean, a lot of arguments you keep ignoring.

>>>>>and who are you dto judge it while your understanding of physics
>>>>>led you to the FERTZ invention??!!!
>>>>
>>>>*sigh* Did you notice that you are *still* the only one who thinks
>>>>that that shows *any* problems with my physics education?
>>>
>>>--------------
>>>you are lucky enough that not too many people heared about it
>>>or else you would become the joker of the town
>>
>>Unsupported assertion. How do you know?

How do you know?

[snip]

>>>>>we are geting no where with every one in his stand
>>>>
>>>>We could be getting somewhere if you would finally tell me where
>>>>you got that idea from.
>>>
>>>only from myself
>>
>>And what makes you think you are qualified to judge if this idea
>>is right?

Care to answer this?

>>What is your education in electrodynamics? For starters, do you
>>know Maxwell's equations?
>
> -------------
> at leat i am not aparrot like you

The only one here who is a parrot is you, as I explained several
times now in detail.

Who of the two of us simply keeps repeating what he once had been
told in this newsgroup, but never bothered to check this up for
himself? In contrast, who of the two of us has in a lot of cases gone
to the original literature describing the experiments and checked
what had been done and what reasoning was employed there in detail?

Oh, BTW, thanks for admitting that you do *not* even know Maxwell's
equation. That means that you are not qualified in any way to discuss
any topic of electrodynamics, i.e.: your assertion that EMR comes
always from resonance is *totally* baseless.

>>>the fact that there is Au 79 does not prove that
>>>there is no Au 80
>>
>>*sigh* Thousands of calculations were based on the assumption that
>>gold has only 79 electrons. All these calculations yielded results
>>which are consistent with the experimental tests.
>>
>>How do you explain that?
>
> -------------
> bla bla

In other words: you prefer to ignore this, like all inconvenient
things, like reality.

You have closed your eyes, put your fingers into your ears and are
humming "la, la, la, I can't hear you..."

>>>that there are protons inside the nuc that are not
>>>electrically charged
>>
>>That contradicts Rutherford scattering, for starters.
>>
>>If you can explain Rutherford's results ***QUANTITATIVELY***
>>*without* using the assumption that the charge of the gold
>>nucleus is 79+, feel free to do so. Unless you can't do that,
>>this experiment is evidence that the gold nucleus indeed
>>*has* this charge.
>
> -------------
> lets see you first using the Rutherford in caculating
> the 79 charges

I told you that by *assuming* that the charge of the gold nucleus
is 79+, and then calculating what should come out of the scattering,
the results agree very nicely with the experimental results.

If you can explain Rutherford's results ***QUANTITATIVELY***
*without* using the assumption that the charge of the gold
nucleus is 79+, feel free to do so. Unless you can't do that,
this experiment is evidence that the gold nucleus indeed
*has* this charge.

> ---------------
>
>>And, BTW, protons which are not electrically charged would
>>yet again contradict charge conservation.
>
> ---------------
> if you coock youself with your own juice you are right (:-)

Huh?

> you didnt notice that you made here an sssumption based on the same
> ssumption

What assumption would that be?

If you mean the assumption that charge conservation applies to protons,
then a small hint for you: that assumption leads to predictions which
agree with the experimental tests. If you can predict the same,
*quantitatively*, *without* using this assumption, feel free to do
so.

> how do we call it in the logics scince ? i forgot.

As I said several times: you don't understand science.

Science is *based* on using hypotheses and checking them. Among
the hypotheses checked in Rutherford scattering is the assumption
that the charge of the gold nucleus is 79+. The experimental results
agree with the predictions that are based on this assumption. Hence
this experiment is evidence that the gold nucleus indeed has this
charge.

*That* is science.

If you can explain Rutherford's results ***QUANTITATIVELY***
*without* using the assumption that the charge of the gold
nucleus is 79+, feel free to do so. Unless you can't do that,
this experiment is evidence that the gold nucleus indeed
*has* this charge.

>>>the electric charge can be only on an edge proton
>>>a proton that is connected to other neutrons
>>>inside the nuc lost its edge orbital - the electron.
>>>that is part of the binding energy
>>>binding energy is matter that got lost to space.
>>
>>A bunch of unsupported assertions, contradicted by the evidence.
>>What's new?
>
> we will see about it in future

Yes, that's one of your standard cop outs when you have no evidence.
And you conveniently ignore the contradicting evidence.

If you now say that the evidence is misinterpreted, please tell
me precisely at which point this misinterpretation occurs, and how
one could interpret the same results in another way. Please pay
attention to the fact that such interpretations should explain the
results *quantitatively*. Mere handwaving won't suffice.

>>>----------
>>>take it or leave it
>>
>>Support it or drop it.
>
> to drop it or mot is my bussiness not yours.

"mot"?

And, BTW, as soon as you publish your stuff, it *is* indeed your
responsibility to support it. That is *science*.

>>>for me it is not experiments- it is the interpretation of those
>>> experiments
>>
>>If you can explain Rutherford's results ***QUANTITATIVELY***
>>*without* using the assumption that the charge of the gold
>>nucleus is 79+, feel free to do so. Unless you can't do that,
>>this experiment is evidence that the gold nucleus indeed
>>*has* this charge.
>>----------------
>
> surprise !!!! i can do :

No, you can't. What you present below is not *quantitative*,
it is a bunch of *qualitative* handwavings.

Rutherford's formula (4-6, 4-8, 4-9, 4-10 - these are all
essentially equivalent formulations) was derived on the assumption
that a nucleus with Z protons in it has a charge of Z*e. The
experiments show that this formula describes the results very nicely. If
you can derive this formula without this assumption, feel free to do so.

> btw i dont know why you call it 'predictions of rutherford
> he di dhis experiment at rhe beginning of the 20th century
> Geiger did this nonsense physically calculation many decadeslater

Absolute utter nonsense. Both the experiments and the theoretical
explanations were done in the years 1911 to 1913. Add "history of
science" to the list of things you know nothing about.

> the mass of a nucleid is 188 times that of an electron

Already totally utterly wrong.

The mass of a single proton is 1853 times the mass of a single electron.
For a neutron, it is about the same number. Gold has an atomic weight
of about 197. Thus a single gold nucleus is about 365 000 times more
massive than an electron.

> 4 of them are about 7000 times more massive

Interesting. 4 times 188 gives 7000?

Either above you meant 1880, or here you mean 700. Well, both are still
wrong.

Or did you mean "nucleon" above instead of "nucleid"? If yes, then
your numbers begin to become sensible - but they are quite irrelevant,
since there is not only one nucleon in the gold nucleus, but about 197
of them!

> ie all your 79 electrons around the nuc are like a fly to an elephant
> so it is not the electrons but the nuc.

Well, no one ever disputed that Rutherford scattering is caused
by the nucleus. Hint: THAT WAS THE CRUCIAL POINT OF THE EXPERIMENT!!!!!

Say, how ignorant *are* you about the history of science? Even a *basic*
science education tells you that the point of Rutherford's experiment
was the discovery of the nucleus!

> now the nuc is very small no matter if charged or not
> acording to the 4 relations at the end of your rutherford quote
> paragraph no 3
> the scattering per atom of foil differs aproximately (aproxymately?)

*sigh* approximately.
If you are not sure of the spelling, why don't you look it up? Your
laziness is *really* incredible! Hey, you only needed to look
at this very paragraph 3 you mentioned here in order to find out
the spelling of that word!!!

> as a square of the atomic weight

Yes, the text really says this.

Hint: the reason for this is that the charge of a nucleus is
approximately proportional to the atomic weight, and the scattering
is proportional to the square of the charge of a nucleus.

If you have got another explanation for why (Ze^2)^2 appears in
Rutherford's formula (4-6, 4-8, 4-9, 4-10), feel free to provide it.

> the elctric force afaik if linearily proportion to the
> electrich intensity so

What on earth is "electric intensity"?????

Do you perhaps mean the strength of the electric field???

> why as a square of the atomic weight ??
> can you expalin that ??

Yes. See above.

> so let me bring an alternative explanation
> as you know the devil is in the interpretations!!
> it is not just merly an electric problem
> ios is as well a mechanical problem!!

Then why does Rutherford's formula, which was derived by assuming
that the scattering occurs entirely by electrostatic forces between
a nucleus with charge Z*e and an electron of charge -e, fit so
nicely to the observations?

> you neglected conpletely the real situation ina methal latice:
> it is not a 'rock solid little sphere that repels the colliding
> alpha as if it was a rock solid body
>
> it is an elastic situation in which *the impact and momentum has to be
> taken into acount!!**

No, that was neglected at all by Rutherford. Recoil effects were
studied and found to be insignificant.

> the more correct model is of a solid sphere (or whateber it is not relavant
> the exact shape at this stage)
> thst body is suspending with 'springs' to its neighbours !!!

And that was indeed studied, and what was found out is that the
coupling to the neighbours by these "strings" is totally insignificant
for the outcome of the experiment.

Hint: Physicists are not totally dumb. They thought about this obvious
problem, and solved it, more than 90 years ago already. They don't need
an engineer to tell them that something which is so obvious.

> if that is the case *the mass of that body is palying a major roll
> in the results:
> if it is a small nuc colided by an alpha
> *it retreats back and sideways* during the collition
> therefore less chaces for a full recoil of the coliding alpha:
> if it is a more massive nuc at the foil
> more chances for back retreat of the alpha

Nice. And now please tell me why the result depends on the *square*
of the mass.

> we have to use here a model of a mass suspended on springs
> and use the momentum laws.

Well, please do.

> that is only the beginning of my interpretation that i drawed
> out of my sleeve by say an hour thinking.

In other words: so far, you have got only qualitative hand waving,
but can't explain even the dependence on the *square* of the
mass *quantitatively*.

So when you above said that you can reproduce the results
*quantitatively*, you lied. Yet again.

> to do it numerically needs much more time
> but this is the beginning of the revolusion in interpretation......

That's a totally unsupported assertion. Why do you boast already now,
although you have not succeeded even in showing only *one* thing
*quantitatively*?

[snip]

>>>you have to wory first for (your' model.
>>>to be completed while obviously it is not.
>>
>> > ant serious scintist wil admitt it,
>>
>>*No* physical theory is *ever* completed! What on earth do you
>>mean?
>
> -----------
> so just lye (and lie) down on your olympic laurals!!

Answer the question, moron.

[snip]

>>>but does it cover all the particles ans subparticles there
>>
>>Huh? Rutherford scattering depends merely on the total charge of
>>the nucleus. The details of how the nucleus is structured are irrelevant
>>for Rutherford scattering, since those can only be seen for higher
>>energies of the projectiles. Try reading up on the Rosenbluth formula...
>>-----------------
>
> see above

What you wrote above had nothing to do with this.

Have you read up on the Rosenbluth formula?

>>>and 2 what are the error margines?
>>
>>See the link.
>
> i cant see it

References to the literature are given. Look it up.

> i can see for instance that in figure 4-12
> the results fior gold and silver are too close!!
> there is even one point (the one just befor the top one
> that neerly coincide for gold and silver!!

Yes. So what??? They also coincede nicely with the predicted lines.

If the assumption that the charge of the nucleus is Z*e is wrong,
then how can it be than even *one* of the points lies on the predicted
line? In reality, at least 90% of the points lie on the predicted
lines!

> --------------
> my above imterpretationexplaines as well
> the influence of thickness at the adjacent figure.

How?

>>Because scattering experiments provide evidence that
>>the charge of the nucleus is indeed the same as the number of protons
>>in it.
>>
>>Spectra also provide such evidence, but it is more indirect.
>
> more indirect but much more acurate !!!

Yes. So what?

All spectra calculations are based on the assumption that the
charge of the nucleus is Z*e, and that there are Z electrons in the
atoms. All such calculations give results which agree nicely with
the observations. How do you explain that? Oh, I know, you prefer
to ignore this, right?

>>>>>>A heavy nucleus has no electrons in it.
>>>>>
>>>>>so go learn something new
>>>>>even neutrons contain electrons.
>>>>
>>>>That's a totally unsupported assertion which contradicts vast heaps
>>>>of experimental data.
>>>
>>>--------------
>>>vat heaps is your interpretation.
>>
>>No. Try looking at Hofstadter's data and tell me how on earth they
>>are compatible with electrons inside the neutrons.
>
> ----------------
> i have no access to it

Then you are not in the position to judge if "vast heaps" is my
interpretation or not.

BTW, Hofstadter's results are available in every university library.
Do you want to tell me you don't have access to any university library?

>>Merely *asserting* that the interpretation is wrong is not enough.
>>*Prove* this.

I notice that you do not bother to do this.

>>Also, this contradicts Heisenberg's uncertainty principle. But
>>unfortunately, you don't understand it...
>
> -----------
> bls bla bla

Yes, that characterizes your ramblings very nicely.

>>>acording to my interpretation there is a vast heap of evidence
>>>that the nuc contain huge mumbers of electrons.
>>
>>Well, present a few pieces of that evidence, please-

I notice that you do not bother to do this.

>>>btw during a nuclear hydrogen bomb blast
>>>is there Em radiation???!!!
>>
>>I don't know for sure, but I would guess yes.
>
> so where from are the electrons there??

Huh??? Which electrons??? You talked about electromagnetic
radiation above!

If your question was supposed to mean if there are *electrons*
radiated during a hydrogen bomb blast, then the answer would be
yes. Where they come from? From the hydrogen atoms, obviously!

>>>i think i read it in past
>>>and i saw there nothing of a prove for 79 electrons per
>>>Au ie no one more no one less!!
>>
>>Err, for the 50th time: experimental science is about evidence,
>>not about prove.
>>
>>By assuming that the nucleus of gold has a charge of 79+, the
>>experimental results can be explained quite nicely. If you can
>>explain the results *quantitatively* *without* that assumption,
>>feel free to do so. If you can't do that, the experiment is
>>evidence that the nucleus indeed *has* that charge.
>
> --------------
> and it hase 80 as well can the rutherford scattering
> prove tha thter is no Au 80 ?
> feel free to show it.

All data obtained on Gold atoms so far *always* showed that
the gold nucleus has a charge of 79. There never was a *single*
experiment which showed another charge.

And now you will again come with your totally braindead argument
"they did not look for it specifically!"...

[snip]

>>>you have to know only the exact nomber of charges of the proton
>>
>>And this is obtained using Rutherford scattering, among other
>>things. Result: the nucleus of gold has a charge of 79+. If you can
>>explain the results *quantitatively* *without* that assumption,
>>feel free to do so. If you can't do that, the experiment is
>>evidence that the nucleus indeed *has* that charge.
>
> -----------
> not at all

Yes, indeed.

> you can finf the mass of gold that is chared only Au +1 !!!
> much simpler than do it with Au +79
> only in your dreawms they did it with au 79

I have no clue at all what you are talking about here. In Rutherford
scattering, one uses a gold foil, not gold ions, neither Au 1+ nor
Au 79+.

[snip]

>>>*you test* it must not be neither its maximum charge
>>>noe its minimum charge!!!
>>
>>Rutherford scattering shows that the charge of the gold nucleus
>>is precisely 79+.
>
> nothing like that it is not acurate to that extent

It is.

If you think otherwise, *prove* this assertion.

> and i brought an alternative explanation to it
> mass momentum and elesticity calculations.

That alternative calculation is mere handwaving so far. You
did not show anything *quantitatively* there.

Do you understand what the word "quantitatively" means? Apparently
not.

>>>yet we are looking for the maximum number of electrons per Au
>>>remember?
>>
>>Since the charge of an electron is -1, the number of electrons
>>has to be equal to the charge of the nucleus.
>
> ------------
> again cooking youyself in your own juice

Pot. Kettle. Black.

>>>it is relavant because
>>>the number of protons andneutrons might change even for the same nuc.
>>
>>When beta decay occurs, the nucleus *changes*. It is *not* the same
>>one anymore.
>
> but not al;ways you can folow the quick changes

Yes, obviously not.

> and my argument was that we cannot know all what is inside the nuc

For Rutherford scattering, what is inside of the nucleus is entirely
irrelevant. The only relevant thing is the total charge of the nucleus.
And Rutherford scattering shows nicely that the total charge of a gold
nucleus is 79+.

> (all did you got what does it mean all??)
> only arrogant people pretend to know all!! guess whom do i mean.

Since I do not pretend to know all, I don't know whom you mean.
Perhaps you? *You* are the one who keeps claiming that he knows where
precisely in the nuclei the protons and neutrons are located, remember?

>>>is it possible ?.......
>>
>>Yes, indeed it is possible to solve 100 equations with 100 unknowns.
>>But that is entirely irrelevant here.
>>
>>
>>
>>>if not see my book!
>>>how does even a computer solves 100 equations with 100 unknowns?
>>
>>Depends on the equations. If they are all linear equations (and in
>>the problem above, they indeed are, as far as I can see, simple
>>matrix inversion can be used.
>>
>>Didn't you even know such totally basic stuff???
>
> you undestood nothing about how i was building my model!!!

That remark has nothing to do with what I said above. This is a rather
lame attempt to cover the fact that you did not even know that
solving 100 equations with 100 variables is no big problem with computers.

> you are a zero pioneering scintist.

Oh, that explains why I was able to publish articles, right?

>>>so ???
>>>they were not torn but vibrate and resonate!!
>>>whats th eproblem??
>>
>>Even if it were right that the radiation comes from electron-electron
>>collisions (which it isn't)

You keep ignoring that.

>>how would that produce a *continuous*
>>spectrum? Have you ever heard the terms "resonance frequency",
>>"Breit-Wigner resonance" and "Lorentz spectrum"?
>
> ---------------
> there are thwere many electrons with different self frequency
> and they vibrate even for a shrt timeand resonate ans stop
> vibrating

Well, please explain how "many different electrons with different
self frequencies" manage to make the typical bremsstrahlung spectrum.

Since one does not see any resonance spectra in bremsstrahlung
(notice that I am *not* speaking about the *characteric* spectra
here!), you have no evidence for your assertion that bremsstrahlung is
caused by resonance.

>>(Hint, where I am aiming at is: a resonance spectrum looks *totally*
>>different from the bremsstrahlung spectrum)
>
> -----------
> how do you know how in a n element no 79 electrons
> each one of them behaves and all of them with cross
> mutual interaction effects??

How do you know that resonating all these electrons would result
in the bremsstrahlung spectrum?

Hint: if one does the calculation of how the spectrum looks like
*without* assuming that there is any resonance, the results of the
calculations agree very nicely with the observations.

[snip]

>>>>>it vibrates the iers are frickshoned
>>>>>ion a railroad while you stopp suddenly on the railes
>>>>>you can see fire under it
>>>>>it is fierce feiction and vibration
>>>>
>>>>Well, what is the analogue to the railroad in bremsstrahlung?

Care to answer that?

>>>>>even the sound of break is a rsult of resonationg the air.
>>>>
>>>>Absolute utter nonsense.
>
> --------------
> for a parrot

I.e. for you.

>>Or tell me what the resonance frequency of the air in this example is.
>>And why the resulting sound spectrum is not a resonance spectrum.
>
> we are speaking about a methaphore

What is metaphoric about the claim "the sound of break is a rsult [sic]
of resonationg [sic] the air"?

> yet is it not clear to you that any sound is created by resonance??

No. In contrast, most sounds are *not* created by resonance.
I.e. the sound created when I hit the keys of my keyboard has nothing
at all to do with resonance.

>>Huh? Which irrelevant issue?
>>
>>It was *you* who brought up the point of deviations between the
>>predictions of your model and experimental values, not me.
>
> ---------------
> i never mentioned my model *before you* croock

Problems with your memory again?

Look at your post from the 13th of august. You wrote there:
"an imbecil like you hjas my book for more than one and a half
year and yet keeps asking idiotic questions about it
(including lies about 25 percent error)"

I then pointed out that this is not a lie, that there are indeed
examples where your model deviates by 25% from observations.

> and it is you who entered my thread
> i never interfere in youtr threads

That could be due to the simple reason that I have opened close
to none threads on my own so far, whereas you keep opening new
threads with nonsensical, unsupported assertions, don't you think?

Anyway, I plan to open a thread today. Feel free to interfere there.
I won't mind, since I don't have paranoia like you.

> you have 2 motivations to interfere in my threads

Only one. Correcting your nonsense.

> 1 revance of a personal enemy

Nonsense yet again.

> 2 after all you finf it some how interesting

"amusing" and "annoying" would be better words.

> btw from time to time i find our disputes constructive and even useful

Well, it's really sad that you nevertheless learned so little.

> in spight the very unsympathic general atmosphere!!

Pot. Kettle. Black.

*Who* keeps writing posts filled with insults? Granted, I also use
an insult now and then - but the most hate in these threads comes
quite obviously from *your* side.

[snip]

Bye,
Bjoern

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