• Quick note - the problem with Youtube videos not embedding on the forum appears to have been fixed, thanks to ZiprHead. If you do still see problems let me know.

why Nuclear Physics cannot be entirelly correct

Does he realize just how many real physicists he has been talking to here?

Well, to be fair, and as asshattish as this may sound, I cannot believe any one person in here is a physicist. I do believe there are probably one, two....three...four...maybe more (:D) physicists in here. But I wouldn't put any money on any particular person being one. The whole anonymity of the internets, and all....

But I would take everybody seriously, until they have proven themselves to be less than credible. Which is quite easy to pick out rather quickly. But my seriousness I take of everybody, only extends as far as the sources they cite. Give me great answers backed up with great sources, your credibility will rise farther than the "default" credibility I give to any particular screen name. Do the opposite, and my impressions of you suffers.

So no. I don't think he realizes how many he has talked to. I have no idea how many he has conversed with, if any at all. (I couldn't care less, as I skipped over most of the posts that had to do with nuclear physics. I literally felt like I was reading Greek, and it all was going so far over my head, it owuld have made me dizzy with confusion.) I was originally only posting in here for the lulz. Now I am continuing to post in here simply to try to give Pedrone a hand in advising him how to better conduct himself.
 
Abstract
The theory of quantum electrodynamics (QED) predicts that beta decay of the neutron into a proton, electron and antineutrino should be accompanied by a continuous spectrum of soft photons. While this inner bremsstrahlung branch has been previously measured in nuclear beta and electron capture decay, it has never been observed in free neutron decay
http://www.ncbi.nlm.nih.gov/pubmed/17183320
:D:D:D

From said abstract:
Here we observe the radiative decay mode of free neutrons, measuring photons in coincidence with both the emitted electron and proton. We determined a branching ratio of (3.13 +/- 0.34) x 10(-3) (68 per cent level of confidence) in the energy region between 15 and 340 keV, where the uncertainty is dominated by systematic effects. The value is consistent with the predictions of both theoretical approaches; the characteristic energy spectrum of the radiated photons, which differs from the uncorrelated background spectrum, is also consistent with the calculated spectrum.

If I understand the link correctly (though I'd have to read the paper to check) they're actually claiming that they have measured the Brehmsstrahlung in the very experiment that is reported in the paper!
 
:p
Tubbythin,
why did you hide the link you posted with that "Err..." ?
Standard practice. Its easier to look at than some nonsensical URL.

Well, let's look what he have there:
url='http://prola.aps.org/abstract/PR/v93/i3/p518_1'

:D
Wow !!! :D
:D
This is not beta decay of neutron:
Internal bremsstrahlung spectra from S35 and Pm147 have been investigated with the NaI scintillation spectrometer
It is. It may not be of a free neutron but it is beta decay of a neutron.

Tubbythin,
dont you know the difference between the beta decay of a free neutron and the beta decay of nuclei ? (as S35 and Pm147)
:D
Yes.

Tubbythin,
I suspect that you're acting dishonestly on purpose
:mad:
Nope. You said: "Therefore, according to Maxwell theory, the electron would have to emit photons, in any neutron beta decay". In my mind "any neutron beta decay" refers to any weak decay event in which a neutron turns into a proton, including the decay of 35S.
 
Tubbythin,
I seriously suspect you are lying.

The most probable occurrence within the nuclei is the deuteron, as mentioned by Eisberg and Resnick (in the end of the week I'll post what they state in here).
Nope. If that were the case then nuclei with odd numbers of protons and neutrons would be as stable as those with even numbers of protons and neutrons. They're not. In fact I can list all the stable odd-odd nuclei for you: 2H, 6Li, 10B and 14N.
 
If I understand the link correctly (though I'd have to read the paper to check) they're actually claiming that they have measured the Brehmsstrahlung in the very experiment that is reported in the paper!
I agree. The experiment measured the inner Bremsstrahlung from the radiative branch of the free neutron decay.
The "uncorrelated background spectrum" sounds like the Bremsstrahlung from the other sources (e.g. the electrons from all of the decays being decelerated by the protons).
 
Tubbythin,
I seriously suspect you are lying.

The most probable occurrence within the nuclei is the deuteron, as mentioned by Eisberg and Resnick (in the end of the week I'll post what they state in here).
I suspect that Eisberg and Resnick mention "deuteron" in the context of stable nuclei, e.g. the doubly magic 16O.
It is obvious that proton - proton and neutron-neutron "pairings" are possible as the istopes of oxygen show
  1. 14O with 2 excess protons.
  2. 18O with 2 excess neutrons.
But I would be interested in the full quote from Eisberg and Resnick that you promise.
 
Nope. If that were the case then nuclei with odd numbers of protons and neutrons would be as stable as those with even numbers of protons and neutrons. They're not. In fact I can list all the stable odd-odd nuclei for you: 2H, 6Li, 10B and 14N.

He doesn't mean that they are kept in a stable,pedrone.
 
Tubbythin focused your attention to your completely incorrect use of the word trepidation.

You might want to be honorable and just acknowledge it, instead of blaming it on google translate and act like a child: "wheeehhehehee, Tybbythin is mean to me and has it in for me, wheeeeehheheheee".

Oh, by the way, it seems you still haven't answered the credentials question in a respectable manner:

So, Pedrone, there is no shame in acknowledging that you have no relevant formal training.
Perhaps you acquired your ideas through self study, perhaps you'd want to consider yourself an auto didact.

Is that the case Pedrone? Do you not have relevant formal training and do you consider yourself an auto didact?

So Pedrone, what are your credentials.
You know, credenciais.

Quais são as suas credenciais Pedrone?

Please provide a respectable answer!
 
Originally Posted by pedrone
Tubbythin,
I seriously suspect you are lying.

The most probable occurrence within the nuclei is the deuteron, as mentioned by Eisberg and Resnick (in the end of the week I'll post what they state in here).


Nope. If that were the case then nuclei with odd numbers of protons and neutrons would be as stable as those with even numbers of protons and neutrons. They're not. In fact I can list all the stable odd-odd nuclei for you: 2H, 6Li, 10B and 14N.
:confused:
Wow !!
what a stupid argument.
:D

A nucleus A with odd number of protons is not stable as a nucleus B with pair number of protons because A has one unmatched deuteron.

The 20Ca40 has 20 deuterons, and 20 unmatched neutrons.
It is a magic number because of the pair number of protons favors the spin orbit interactions (Mayer-Jensen theory), ie, all the deuterons are matched

The 19K38 is not so stable because the unmatched deuteron does not favor the spin orbit interaction.


Tubbythin,
you have to ask to Mayer and Jensen to return the Nobel Prize they won with the theory of magic numbers they proposed.

I bet you win the Nobel Prize in 2012, Tubbythin, by propsing such new theory of yours which explains the magic numbers.
Your new theory will replace the Mayer-Jensen theory
:D:D:D:D:D:D
 
:confused:
Wow !!
what a stupid argument.
:D

A nucleus A with odd number of protons is not stable as a nucleus B with pair number of protons because A has one unmatched deuteron.

The 20Ca40 has 20 deuterons, and 20 unmatched neutrons.
It is a magic number because of the pair number of protons favors the spin orbit interactions (Mayer-Jensen theory), ie, all the deuterons are matched

The 19K38 is not so stable because the unmatched deuteron does not favor the spin orbit interaction.


Tubbythin,
you have to ask to Mayer and Jensen to return the Nobel Prize they won with the theory of magic numbers they proposed.

I bet you win the Nobel Prize in 2012, Tubbythin, by propsing such new theory of yours which explains the magic numbers.
Your new theory will replace the Mayer-Jensen theory
:D:D:D:D:D:D

Ok. I guess I am finished with you. The lulz are over, and you have failed to heed any of my advice, and continue to wantonly call people names, use inappropriate type formats, and obnoxiously use smilies. I will now proceed to block you.

Good day to you sir! :mad:
 
Ok. I guess I am finished with you. The lulz are over, and you have failed to heed any of my advice, and continue to wantonly call people names, use inappropriate type formats, and obnoxiously use smilies. I will now proceed to block you.

Good day to you sir! :mad:

Now I won't be able to read your responses to Pedrone, I quite enjoyed them.
You are just doing this to be mean to me, it's not fair, wheehehehehe.

Oh, hi Pedrone.
 
:confused:
Wow !!
what a stupid argument.
:D

A nucleus A with odd number of protons is not stable as a nucleus B with pair number of protons because A has one unmatched deuteron.

The 20Ca40 has 20 deuterons, and 20 unmatched neutrons.
It is a magic number because of the pair number of protons favors the spin orbit interactions (Mayer-Jensen theory), ie, all the deuterons are matched

The 19K38 is not so stable because the unmatched deuteron does not favor the spin orbit interaction.


Tubbythin,
you have to ask to Mayer and Jensen to return the Nobel Prize they won with the theory of magic numbers they proposed.

I bet you win the Nobel Prize in 2012, Tubbythin, by propsing such new theory of yours which explains the magic numbers.
Your new theory will replace the Mayer-Jensen theory
:D:D:D:D:D:D

On ignore you go until you tell me what your qualifications are.
 
:confused:
Wow !!
what a stupid argument.
:D

A nucleus A with odd number of protons is not stable as a nucleus B with pair number of protons because A has one unmatched deuteron.

The 20Ca40 has 20 deuterons, and 20 unmatched neutrons.
Pardon? It has 20 deuterons and 20 unmatched neutrons? That makes 60 nucleons in calcium-40.

It is a magic number because of the pair number of protons favors the spin orbit interactions (Mayer-Jensen theory), ie, all the deuterons are matched
No, it is a magic nucleus because it has 20 protons and 20 neutrons and 20 is a magic number. Explcitly, there is a large gap between the 1 d3/2 and 1f7/2 orbitals. Excited states can only be made by exciting protons, neutrons or pairs thereof across this large energy gap. Notice this only occurs because of the exclusion principle, an inherently quantum effect.

The 19K38 is not so stable because the unmatched deuteron does not favor the spin orbit interaction.
38K is not so stable because excited states can be formed without exciting nucleons across the shell gap.

Tubbythin,
you have to ask to Mayer and Jensen to return the Nobel Prize they won with the theory of magic numbers they proposed.
No I don't, as I just explained it to you. Incidentally, it would be much use trying to ask them.

I bet you win the Nobel Prize in 2012, Tubbythin, by propsing such new theory of yours which explains the magic numbers.
Your new theory will replace the Mayer-Jensen theory
:D:D:D:D:D:D
It isn't a new theory.
 
Wow !!
what a stupid argument.

A nucleus A with odd number of protons is not stable as a nucleus B with pair number of protons because A has one unmatched deuteron.

...magic numbers...
Wow !!
what a stupid argument.

A nucleus A with odd number of protons is not stable as a nucleus B with pair number of protons because A has one unmatched nucleon.

Magic numbers happen to be even. But even unmagic nuclei are stable, e.g. 18O has 2 excess ("paired") neutrons.
 
I suspect that Eisberg and Resnick mention "deuteron" in the context of stable nuclei, e.g. the doubly magic 16O.
It is obvious that proton - proton and neutron-neutron "pairings" are possible as the istopes of oxygen show
  1. 14O with 2 excess protons.
  2. 18O with 2 excess neutrons.
But I would be interested in the full quote from Eisberg and Resnick that you promise.

To be honest, I have no idea what he is talking about...
Eisberg and Resnick - Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles, 2nd Edition, p541:
There is much evidence indicating that the total angular momentum of the protons in a subshell couple together in pairs, with the total angular momentum of each pair of protons equal to zero, and that the same thing happens for pairs of neutrons in a subshell.
 
Wow !!
what a stupid argument.

A nucleus A with odd number of protons is not stable as a nucleus B with pair number of protons because A has one unmatched nucleon.

Magic numbers happen to be even. But even unmagic nuclei are stable, e.g. 18O has 2 excess ("paired") neutrons.

And 18F is not, despite have an excess neutron-proton pair compared to that excess neutron pair.
 
To be honest, I have no idea what he is talking about...
Eisberg and Resnick - Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles, 2nd Edition, p541:
Since you have the book, it makes pedrone's task about citations even easier. While he is working from a Portuguese version of the book, the page numbers should be roughly the same.

pedrone: State the page numbers where Eisberg and Resnick state that the "most probable occurrence within the nuclei is the deuteron" to support your assertion that

For instance, let's consider the oxygen nucleus 8O16.
The physicists discovered that into the nuclei the protons and neutrons are linked together (the nuclei are filled by deuterons).
 

Back
Top Bottom