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#### occam

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##### beta decay energy
« on: May 19, 2008, 12:14:38 PM »
I have been looking at beta decay chains, and I have noticed a curious discrepancy between beta minus and beta plus.

For Beta minus, the decay energy (DE) is almost the same as the mass difference(MD) between parent and daughter isotope. (The table is in AMU)
For beta plus there is a difference of about 1022Kev.
This example is Atomic Number 65, but the effect is consistent for every beta decay chain.
Does anyone know why?

DE       MD          DE - MD          DE-MD(keV)
65V   0.017760    0.017760   -0.0000004   -0.33684
65Cr   0.013824    0.013820    0.0000040   3.75028
65Mn   0.010956    0.010960   -0.0000045   -4.17634
65Fe   0.008902    0.008902   -0.0000002   -0.16129
65Co   0.006394    0.006394   0.0000004   0.40559
65Ni   0.002295    0.002295   0.0000000   0.01713
65Cu   STABLE
65Zn   0.000354    0.001452   -0.0010974   -1022.200819
65Ga   0.002396    0.003494   -0.0010974   -1022.228405
65Ge   0.005604    0.006705   -0.0011013   -1025.854851
65As   0.009031    0.010120   -0.0010893   -1014.721215
65Se   0.013999    0.015100       -0.0011010   -1025.562287

#### gippgig

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##### Re: beta decay energy
« Reply #1 on: May 19, 2008, 07:46:15 PM »
When an atom decays by positron emission the atomic number goes down by 1 so an electron will be lost reducing the mass. When an atom decays by electron emission the atomic number goes up by 1 so an electron will be gained increasing the mass. The difference is twice the electron mass which as I recall is 1022 keV.

#### Valdorod

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##### Re: beta decay energy
« Reply #2 on: May 20, 2008, 12:28:20 AM »
When an atom decays by positron emission the atomic number goes down by 1 so an electron will be lost reducing the mass.

Technically a positron is lost and while the mass (amu) is reduced, the mass number does not change.

When an atom decays by electron emission the atomic number goes up by 1 so an electron will be gained increasing the mass.

There is a difference between electron emission and electron capture.  During electron emission an electron is lost not gained and once again the mass number does not change.

Both types of decay are internal conversions in which a neutron is converted into a proton, or a proton is converted into a neutron.  However, while Z changes, the number of neutrons is adjusted accordingly and the mass number stays the same.

In the case where there is neutron excess in the nucleus you have electron emission

0n1 —>  +1p1 + e-   (Total mass number before and after is 1)

In the case where there is an excess of protons in the nucleus you have positron emission

1p1 —> 0n1 + e+

Both processes involve the release of neutrinos, which are necessary because of conservation of angular momentum.

While electrons are easier to detect, positrons (beta +) are a little more difficult.  They almost instantaneously combine with an electron to produce a photon, which has an energy equal to 1.02 MeV

Recall that the mass of a neutron is sightly larger than that of a proton.  Thus, relativistically one can see that the loss of mass from a neutron to a proton is part of the mass and energy of the electron given off.

However, for the conversion from a proton into a neutron there has to be an increase in mass (not for the isotope) in addition to the mass of the positron being emitted, since the mass of the neutron is slighlty larger than that of a proton.

Both processes are nuclear processes which means that they do not occur with the free particles.  The energetics depend on the interactions between all the nucleons.

So, to answer the question the energy available for beta - decay (e - ) is given by:

QB- =  M(A, Z) - M (A, Z + 1)      Thus, the energy available is equal to the difference in the neutral atomic masses.

For the case of beta + decay (e + )

QB+ =  M(A, Z) - M (A, Z + 1) - 2m0      Where m0 is the rest mass of the electron.  Positron emission is energetically forbidden when the mass difference between parent and product neutral atoms is less than 2 m 0 or in energy terms 1.02 MeV.

#### gippgig

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##### Re: beta decay energy
« Reply #3 on: May 20, 2008, 04:50:35 AM »
To clarify my previous comment consider an atom that initially has x protons & x electrons.
If it decays by positron emission the initial product will have x-1 protons & x electrons, so it will be a negatively charged ion. It will therefore lose one of its electrons so the final product will have x-1 protons & x-1 electrons. The net result is that both a positron & an electron are lost so the mass is less by twice the mass of an electron (in addition to the mass difference due to the decay energy).
If it decays by electron emission the initial product will have x+1 protons & x electrons, so it will be a positively charged ion. It will therefore gain an electron from the surroundings so the final product will have x+1 protons & x+1 electrons. 1 electron is lost but 1 electron is then gained so the net result is that the mass stays the same (except for the decay energy).

#### occam

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##### Re: beta decay energy
« Reply #4 on: May 20, 2008, 11:05:28 AM »
OK
I understand the proton-neutron conversion rules,
So the equation Parent > Daughter + e(+/-)  + neutrino(+/-).should only give a mass loss ~ 500keV.
However in mass terms you need Parent – decay energy > Dr + e +n.
For 65V>65Cr the mass difference is 0.177AMU, (16000keV +500keVfor the e+n)
So what happens to the decay energy mass? Does it come out as gamma, or is it absorbed as heat into the general atomic mass? The literature is a bit vague on this.

#### Valdorod

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##### Re: beta decay energy
« Reply #5 on: May 20, 2008, 01:10:32 PM »
During beta (+/-) decay the daughter nucleus is so much heavier than the other particles that it can carry relatively large amounts of the momentum and still have a negligible abount of energy.  So that we can specify that the energy available for the processes is related to the kinetic energy of the particles being emitted.

QB(+/-) = TB(+/-) + TV + MV

Where
TB(+/-) is the kinetic energy of the beta particle
TV is the kinetic energy of the neutrino
and MV is the rest mass of the neutrino, however, the mass of the neutrino is taken to be zero and thus the rest of the energy is carried as kinetic energy.

This is not the whole story however, The above equation assumes that all the available energy is released in the beta decay process.  Often this is not the case because beta ray transitions usually go to an excited state of the daughter nucleus rather than to the ground state (yes nuclei have ground and excited states just like electrons do).  The excited sate then decays by gamma-ray emission (or in some cases internal conversion) to the ground state.  This gamma ray is usually released following the beta transition.

As a side note note that the rest mass of the electron is about 500kev/c2.

Valdo

#### occam

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##### Re: beta decay energy
« Reply #6 on: May 20, 2008, 04:52:42 PM »
Hi Valdo.
I gather that you agree the mass loss is not emitted.
The kinetic energy of an e particle at 0.5c is only 60keV, so the process must have something to do with exchange of momentum, and it is a process since the transitions involve not only mass, but time.
The W boson theory does not appeal to me. The concept of a mass greater than an Iron atom appearing from nowhere to facilitate a transition between tritium and 3He, and then disappearing again, is more alchemy than science.
You may have noticed my parallel post about crystal structures. I am toying with the thought that since the parent has a greater mass than the Daughter, and the crystal structures are different, inducing stresses, then the nuclear forces would transfer momentum to a surrounding body of atoms. But to avoid inducing a transition in the “receiving” atoms, a minimum number (W mass?) of atoms is required to absorb the excess momentum (gain mass) and thus facilitate the transition?

I have put up another post regarding spectral effects.
Graham

#### Valdorod

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##### Re: beta decay energy
« Reply #7 on: May 20, 2008, 10:28:32 PM »
I agree with you that the mass loss is not emitted.

While I am only in my thirties, The little that I know about modern physics and quantum mechanics are from classes in which we used reprints of old books.  One of the texts, is Modern Physics for Engineers by Otto Oldenberg, a book written in 1966.  We obtained a small reprint in the early nineties, because my professor studied under him.  The other text is Introduction to Quantum Mechanics by Linus Pauling and E. Bright Wilson.  A book written in 1935.  Once again I own a reprint.  For the level that I needed to know the physics made sense without the need of all these new particles.

Just as well beta decay made sense to me without the need for W bosons, up quarks and down quarks (which I think it is supposed to be one of the down quarks that interacts in beta decay)  I am clueless when it comes to the "weak force" and the standard model of particle physics.

I agree that a particle with a mass of over 100 protons is hard to conceive as coming out of nowwhere as well as the explanation composite bosons exhibit bosonic behavior at large distances distances and behave like normal particles at small distances.  How can an alpha particle behave like a group of fermions at high energies, when it is considered to be a boson?

On a different note there are a couple of books out there by an Australian physicist named Wallace Thornhill, one of them called The Electric Universe.  I have only read a couple of excerpts, however, to someone like me, someone with very little unerstanding of string thoery, the theories put forth by Thornill make as much sense, if not more than accepting carrier particles of the fundamental forces.

Valdo

#### occam

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##### Re: beta decay energy
« Reply #8 on: May 21, 2008, 05:42:44 PM »
Quote
Its in
http://www.chemicalforums.com/index.php?topic=26629.0

I copied all the isotope tables from Wikipedia, and periodictable.com into one file so I could analyse the data. Its 2.8MB so too big to attach, but you can have a copy if you ask.
•   There are 194 fully stable isotopes and there is only one value of “Z” for a stable atomic number “A”.
•   There are no stable isotopes above 208Pb. For any given Atomic number, most decays are either beta minus, resulting in an increase in the Z number, or beta plus (or electron capture) resulting in a decrease in the Z number both converging on the unique Z number for the stable A number.
•   2beta minus, and 2beta plus decays are from semi-stable long half life Isotopes, which always result in the stable Z of that atomic number. (There is always another single beta decay in the Z+/- 1 position.)
•   All other decays are either to isotopes in another “A” number beta chain or directly to other stable isotopes.

Which gives rise to some questions

•   beta minus and beta plus decay chains are not symmetrical. Question  - electrons and positrons are supposed to be identical with opposite charge and spin. The data suggests there is a subtle difference?
•   Nuclei have two classes: Fermions with fractional spins, and Bosons with 0 or whole number spins. The pattern of decay energies is different for Fermions and Bosons, but analysis shows that the Bosons have two classes, Atomic numbers divisible by 4, and other even numbers
•   The “curve of stability” is actually three curves, one for Fermions, one for Boson numbers divisible by 4, and one for other even numbers. Thus the question about geometry, and spectrum.
•   Generally the further away from the curve of stability, the shorter the half-life, but there isn’t a consistent ratio between mass difference and half-life.

There may be more questions, but I haven’t finished stirring the data.
Graham