Beta particles

Main Question or Discussion Point

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 [Broken]

http://img329.imageshack.us/img329/3677/picture001copyuc2.jpg [Broken]

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

Astronuc
Staff Emeritus
Try

http://img151.imageshack.us/img151/7273/picture001copy1jg2.jpg [Broken]

http://img329.imageshack.us/img329/3677/picture001copyuc2.jpg [Broken]

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

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

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.

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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
Staff Emeritus
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.

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.

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
Staff Emeritus
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.

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?

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Astronuc
Staff Emeritus
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
Mentor
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.

so because of the weak ineraction with the nucleus, they carry the kinetic energy of it, right?
By period of half-decay, I meant $T_1_/_2[/tex] Astronuc Staff Emeritus Science Advisor 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 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? 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 Staff Emeritus Science Advisor 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. 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? Astronuc, can you please give me answer to this particular question. Astronuc Staff Emeritus Science Advisor jtbell Mentor 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"). 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 Mentor 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. Last edited: Ok, thanks. I understand. Just I want to ask you, are there [itex]M_1$ and $M_2$ because of the different kinetic energies of releasing the beta particles (i.e different energy distributions)?