Understanding Beta Particles and Their Role in Radioactivity | Diagram Included

[Physicsissuef]

Hi! I read in my textbook, that when there is beta radioactivity, the beta particles are "carrying" the kinetic energy from the nuclei, and they are taking the most of the kinetic energy of the nuclei, so the electrons have less energy than them...
Here is diagram. I can't understand something about the diagram. If I put straight horizontal line among one point, there will be two points $$M_1$$ and $$M_2$$, for different energy levels of the electrons (W is the kinetic energy of the electrons). Why it is like that?

http://img151.imageshack.us/img151/7273/picture001copy1jg2.jpg

http://img329.imageshack.us/img329/3677/picture001copyuc2.jpg

Please help! Thank you.

 

[jtbell]

Those images are too small for me to read. I think you linked to thumbnail images, not to the full-size images.

 

[Astronuc]

Try

http://img151.imageshack.us/img151/7273/picture001copy1jg2.jpg

http://img329.imageshack.us/img329/3677/picture001copyuc2.jpg


They are energy distributions of beta particles. The peak is about W_max/3.


The beta particle is very light compared to nuclei. M[nucleus] ~ A*1836*m_e.

If one looks at the momentum equation and energy equation, then one will see that the KE is apportioned by the momentum. If there are two particles MV = mv, and v = M/mV, for equal but opposite momentum vectors.

 

[Physicsissuef]

and why there is same energy level of the electrons for some number of beta particles, like on the picture 2? jtbell sorry for the pictures, I will correct now.

 

[Astronuc]

If one looks at a Poisson or Gaussian distribution, one will find two values of the population function f(x) with the same value but for two different x. All it means is that there is equal probability at those two different values of x, which in the case of the beta decay is the energy of the beta particle. Don't forget that there is a third, nearly massless particle, the neutrino (or anti-neutrino in the case of beta decay) which also takes some momentum and energy.

 

[Physicsissuef]

The neutrino is the cause that there are two equal points for x, if the function is f(x)?

 

[neu]

The neutrino is the cause that there are two equal points for x, if the function is f(x)?

No.

The diagram you posted is a frequency distribution. f(x) = the number of particles with value x.

As Astronuc said, the fact that there are 2 values of x which give the same f(x) simply shows that the same number of particles are produced with x1 and x2.

 

[Physicsissuef]

how is possible that with (lets say 100 beta particles), there will be 2 kinetic energy levels for the electrons. What is that cause? Why the energy of the electrons is not constant?

 

[Astronuc]

how is possible that with (lets say 100 beta particles), there will be 2 kinetic energy levels for the electrons. What is that cause? Why the energy of the electrons is not constant?

Because,

n -> p(in nucleus) + e^- + $\overline{\nu_e}$, and the beta and antineutrino can each take any direction and share energy and momentum.

 

[Physicsissuef]

Somebody told me that because of different kinetic energy distribution, the energy of electrons is not constant. Is this correct?
I have few questions more:

How the neutrino is "taking" the most of the kinetic energy of the nuclei, when it is electro neutral?

When there is $$\beta^-^1$$ decay (lets say of the C atom):

$$\stackrel{14}{6}C \rightarrow \stackrel{14}{7}N + \stackrel{0}{-1}e + \overline{\nu_e}$$

We decay C atom with 6 protons, and receive 7 protons. How come that?

Then, how is possible the period of half decay?

 

[Astronuc]

Somebody told me that because of different kinetic energy distribution, the energy of electrons is not constant. Is this correct?

This is not quite correct. If one were to measures a population of a specific nuclide, one would find a spectrum of energy of the beta particle and anti-neutrino. The energy distributions of the beta and anti-neutrino are complementary since there is a fixed energy available - i.e. the mass-energy difference between the radionuclide (parent) and the subsequent nuclide (product of decay).

I have few questions more:

How the neutrino is "taking" the most of the kinetic energy of the nuclei, when it is electro neutral?

When there is $$\beta^-^1$$ decay (lets say of the C atom):

$$\stackrel{14}{6}C \rightarrow \stackrel{14}{7}N + \stackrel{0}{-1}e + \overline{\nu_e}$$

We decay C atom with 6 protons, and receive 7 protons. How come that?

Then, how is possible the period of half decay?

The anti-neutrino has energy and momentum, just as photons (also neutral) have energy and momentum.

The neutron (n) decays into a proton, beta and anti-neutrino, so if ¹⁴C decays by beta emission, then 6p,8n becomes 7p,7n, because 1 n -> 1 p. The number of nucleons is preserved but Z increases by 1, and N (number of neutrons) decreases by 1.

What is meant by "period of half-decay"? Is one referring to half-life?

 

[jtbell]

One of the neutrons in the C atom converts to a proton via the weak interaction:

$$n \rightarrow p + e^{-} + {\overline \nu}_e$$

More fundamentally, one of the d quarks in the proton converts to a u quark by emitting a virtual W boson:

$$d \rightarrow u + W^{-}$$

and then the W boson converts to an electron and antineutrino pair:

$$W^{-} \rightarrow e^{-} + {\overline \nu}_e$$

Both of these are fundamental weak-interaction processes. To anticipate your next "why?" question , they are this way because quarks have local SU(2) x U(1) gauge symmetry. We don't know why they have this particular symmetry. I suppose that's one of the things that string theory is trying to address.

 

[Physicsissuef]

so because of the weak ineraction with the nucleus, they carry the kinetic energy of it, right?
By period of half-decay, I meant [itex]T_1_/_2[/tex]

 

[Astronuc]

With regard to beta decay and the weak process, please see - http://hyperphysics.phy-astr.gsu.edu/hbase/particles/proton.html#c3

It's not that the weak process causes kinetic energy, but that particles (electron and neutrino) have energy and momentum.


The half-life is derived from the 'statistical' treatment of radioactive decay. Basically, for each single nucleus of a given radionuclide, there is no way to determine when it will decay. It might decay in 1 second, or 10 years, or one century or a millenium from now.

However, if we have a collection (population) of the same radionuclide, one can measure the decays and one will find 'on average' that half of the radioactive nuclei will decay in some amount of time T. Then between T and 2T, half of the remaining nuclei (or 1/4 of the original) will decay. For each period of T, 'approximately' one-half of the remaining nuclei will decay, until eventually the last atom decays.

See - http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli.html

 

[Physicsissuef]

When some nuclei decays, are there alpha, beta and gamma particles at same time?? We are talking here about kinetic energy. Is W the kinetic energy, or it is the inner energy of the electrons?

 

[Physicsissuef]

In my textbook, it says that the when the nuclei decays, it "pull" the electrons with it, so it gives them kinetic energy, but the neutrino is also there which is taking the most kinetic energy. How it is possible when it is electro neutral?

 

[Astronuc]

When some nuclei decays, are there alpha, beta and gamma particles at same time?? We are talking here about kinetic energy. Is W the kinetic energy, or it is the inner energy of the electrons?

Alpha decay and beta decay are mutually exclusive, and while some radionuclides can experience both modes, any given nuclei will do one or the other, but not both simultaneously. Sometimes there is excess energy still remaining and a gamma ray is given off (as in Isomeric Transition).

Particles have energy and momentum, even electroneutral particles like photons (EM) and neutrinos. The neutron and neutral pion, which have mass as well as charge neutrality, have the properties of energy and momentum.

The charged particles, alpha and beta, will interact electrically with the atomic electrons surrounding the nucleus. As alpha and beta particles travel past/through other atoms, they will ionize or excite those atoms.

 

[Physicsissuef]

I can't understand how the neutrinos are "carrying" the most of the kinetic energy when they don't have electro static forces, i.e they are electro neutral. How is this possible?

 

[Physicsissuef]

Astronuc, can you please give me answer to this particular question.

 

[Astronuc]

I can't understand how the neutrinos are "carrying" the most of the kinetic energy when they don't have electro static forces, i.e they are electro neutral. How is this possible?

Why would being electroneutral be an issue. Here is a summary of the neutrino.

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/neutrino.html

 

[Physicsissuef]

Why would being electroneutral be an issue. Here is a summary of the neutrino.

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/neutrino.html

So how will I understand this? Like no possible explanation?

 

[jtbell]

I can't understand how the neutrinos are "carrying" the most of the kinetic energy when they don't have electro static forces, i.e they are electro neutral. How is this possible?

Neutrinos don't interact electromagnetically, but they do interact via the weak nuclear interaction, which is one of the four fundamental interactions ("forces").

 

[Physicsissuef]

And that's why the neutrinos carry the most of the kinetic energy?? When electrons are released with beta decay, how many neutrinos are released?

 

[jtbell]

One neutrino per decay. The basic process is

$$n \rightarrow p + e^{-} + {\overbar \nu}_e$$

A certain fixed total amount of energy is available (for beta decay of a given isotope), and it's divided randomly between the electron, the neutrino and the recoiling nucleus. How much kinetic energy each outgoing particle gets, on the average, depends on their relative masses. When you solve the equations for conservation of momentum and conservation of energy together, you find that in general, the less-massive decay products have more kinetic energy.

This is easier to show if you imagine a heavy particle (at rest) decaying into an almost-as-heavy particle plus a very light one. The two outgoing particles must have a total momentum of zero, so they have equal magnitude momentum in opposite directions. Therefore the lighter particle has a larger velocity, and more kinetic energy.

The recoiling nucleus is very massive, so it carries very little kinetic energy, so little that we usually ignore it. The neutrino has much less mass than the electron, so it carries more energy than the electron, on the average. If you look at the graph at the bottom of this page, you can see that the electrons' kinetic energy distribution is skewed towards the low end, showing that on the average, they carry less than half the total (or maximum) energy, Q.

 

[Physicsissuef]

Ok, thanks. I understand. Just I want to ask you, are there $M_1$ and $M_2$ because of the different kinetic energies of releasing the beta particles (i.e different energy distributions)?

 

[Astronuc]

Ok, thanks. I understand. Just I want to ask you, are there $M_1$ and $M_2$ because of the different kinetic energies of releasing the beta particles (i.e different energy distributions)?

What are M₁ and M₂ to which one is referring? The pairs of nuclear isobars (both parent and daughter of same atomic mass) have different mass relationships (and different binding energies), because each pair has different atomic mass and different Z, so the nuclear properties will be different.

 

[Physicsissuef]

But in that case, there will be infinite numbers of M points, right?

 

[Astronuc]

But in that case, there will be infinite numbers of M points, right?

I would repeat, to what masses (M) is one referring - the atomic mass of the nuclear isobars, or the mass of the electron and neutrino. Remember that rest mass is more or less fixed (in this case with respect to the decaying radionuclide), whereas mass varies according to velocity in a particular inertial frame.

The energy spectrum of the beta particle and neutrino in the beta decay of a particular nuclide represents a continuum, and that is consequence of the continuum of angles at which the beta particle and neutrino could be ejected from the nucleus, or more precisely from intermediate W vector boson.

http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html#c5

 

[sanman]

Hey, what about the process of double-beta decay? Is that doubly-energetic?

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

Since we were discussing radiovoltaics in that nanomaterial thread in the nuclear engineering forum, I was googling around to find out more, and came across that.

Is beta-decay capable of high power output, comparable to fission?

I'd read somewhere else about how buckyballs have been used to accelerate the beta-decay rate. How is that possible? Would the acceleration be even more pronounced in a double-beta decay situation?

 

[sanman]

http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/decay_rates.html

 

[Astronuc]

Is beta-decay capable of high power output, comparable to fission?

No. A single fission produces about 200 MeV of energy per fissioned nucleus, and beta are low MeV, and mostly < 2 MeV.

 

[sanman]

Okay, but half-life is also a consideration in power output. Since half-lives of various nuclides can vary widely, then are there any shorter-lived species whose decay chains might radiate a lot of betas within a practically usable timeframe (eg. days, weeks), so as to provide meaningfully high power output?

 

[sanman]

Regarding this interesting page:

http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/decay_rates.html

Alpha decay and spontaneous fission might also be affected by changes in the electron density near the nucleus, for a different reason. These processes occur as a result of penetration of the "Coulomb barrier" that inhibits emission of charged particles from the nucleus, and their rate is very sensitive to the height of the barrier. Changes in the electron density could, in principle, affect the barrier by some tiny amount. However, the magnitude of the effect is very small, according to theoretical calculations; for a few alpha emitters, the change has been estimated to be of the order of 1 part in 107 (!) or less, which would be unmeasurable in view of the fact that the alpha emitters' half lives aren't known to that degree of accuracy to begin with.

So an example of "breaching the Coulomb barrier" is Rubbia's proton-beam in an accelerator-driven system. But of course it's not yet efficient enough to achieve breakeven.

What about bombarding with deuterons? Would a little extra inertial mass help to overcome the Coulomb barrier more easily? Would it reduce the probability of absorption and increase the probability of alpha emission?

 

[Physicsissuef]

I would repeat, to what masses (M) is one referring - the atomic mass of the nuclear isobars, or the mass of the electron and neutrino. Remember that rest mass is more or less fixed (in this case with respect to the decaying radionuclide), whereas mass varies according to velocity in a particular inertial frame.

The energy spectrum of the beta particle and neutrino in the beta decay of a particular nuclide represents a continuum, and that is consequence of the continuum of angles at which the beta particle and neutrino could be ejected from the nucleus, or more precisely from intermediate W vector boson.

http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html#c5

M_1 and M_2 are just points like S_1 and S_2. They don't refer mass, and also W is referring the kinetic energy of the electrons. So I probably think the kinetic energy of the electrons is discontinius because of the different "speed of ejection". Is this correct?

 

[Astronuc]

M_1 and M_2 are just points like S_1 and S_2. They don't refer mass, and also W is referring the kinetic energy of the electrons. So I probably think the kinetic energy of the electrons is discontinius because of the different "speed of ejection". Is this correct?

Ah, my apologies, I was thinking of mass, rather than one's OP.

The population distribution is continuous, more or less. It is based on the measurement of a large population in which each decay is a discrete event with a unique beta energy. Taken together, with some large N, e.g. 10²⁰ (arbitrary example), one observes that population distribution when one plots the number of particles of energy between E and some ∆E. When ∆E gets very small the distribution looks like a continuum.

Each radionulide has a unique distribution, with a unique mean and max value, but the shapes are much the same, because although the energies are different, the same weak process applies.