Understanding Beta Particles and Their Role in Radioactivity | Diagram Included

[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.

 

[Physicsissuef]

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.

Thank you very much. I suppose you made excellent posts, but, (maybe it is because of the translation or the difficulty, more of the things I can't understand. Can you explain with simpler words, and maybe give me some simple analogy to understand what you mean?

 

[Physicsissuef]

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.

Why then there are only two dots (M_1 and M_2). If there is unique distribution for every radionuclide, then there will be infinite numbers of distributions, or not?

 

[Physicsissuef]

Astronuc, please answer me, my last question...

 

[malawi_glenn]

Physicsissuef: you can find the forumulas (and how to derive them) for the beta-probability distribution in Kranes nuclear textbook - Introductory Nuclear Physics.

 

[PhilippH]

Beta- emission, may vary from a radionuclide to another but in general the energy distribution graph looks like the same for every radionuclide, apart for the maximum energy.
The spectrum of the Beta emission is continuos due to the random ripartition of momentum and energy between neutrino.

http://r8vogq.blu.livefilestore.com/y1pldgTfHaQFOXoZlF3yYP64dp8coOvIfGhHHg-bWwW5sw7XoC5HzygOSM9EIyG8b_nHVvyr6bx63YEWt9uqwoCuA/Nuovo%20Immagine%20bitmap.JPG"

 

[Physicsissuef]

So there are only 2 kinetic energies for every nuclei?

 

[malawi_glenn]

So there are only 2 kinetic energies for every nuclei?

NO! There exists an energy distribution, which is quite similar for each nucleus. You nay want to look for the derivation of this distribution shape, see for instance the reference i gave you.

 

[Physicsissuef]

Why there are 2 points M_1 and M_2 is my question??

 

[malawi_glenn]

Why there are 2 points M_1 and M_2 is my question??

And as I said, if you want to find out why this probability distribution arises check the reference I gave you..

Perhaps you should google 'probability distribution' to learn more about this concept.

 

[Astronuc]

Why there are 2 points M_1 and M_2 is my question??

Because the function is always positive and it increases from zero to some maximum and then decreases to zero again, so that for each ordinate value, there are two corresponding values on of the abscissa. Note that there is one maximum value.

Draw an inverted parabola, and one would see that for each y there are two values of x, except for the maximum value of which there is one.

There is a continuum of energies (between 0 and E_max), and E_max is an upper limit.

 

[Physicsissuef]

I mean, how is possible for one value of beta particles, to have 2 values for the kinetic energy of the electrons?

 

[malawi_glenn]

now you are just writng things I can't understand.. check your english.

on your y-axis you have the NUMBER of electrons emitted. On the x-axis you have their energy. Please check 'probability distribution function' on google, I think you need this.

Every beta particle has just ONE value of Energy, but what energy - and the fraction of the beta particles who has a perticular energy- is given by the distribution function.

 

[Physicsissuef]

On the x-axis I have number of beta particles, and on the y-axis I have the kinetic energy of electrons...

 

[Astronuc]

On the x-axis I have number of beta particles, and on the y-axis I have the kinetic energy of electrons...

On the very first image posted, the vertical (y) axis (ordinate, dependent variable) is number of particles and the horizontal (x) axis (abscissa, or independent variable) is the energy.

By convention, in Cartesian coordinates y-axis is vertical and x-axis is horizontal when looking.

 

[malawi_glenn]

No. please look at your images again.

y axis goes up, x-axis goes to the right. It is the most used convention.

 

[Physicsissuef]

Yes, sorry it was typo. So how is possible that for one value of beta particles, to have 2 kinetic energies for the electrons?

 

[PhilippH]

It's statistic!

A single electron can't have 2 different energy at the same time, but on a population of milion of electrons emitted by that radionuclide you will know how many ( % ) got a E1 energy and how many got E2 energy and so on from 0 to Emax.

I.E. if you use a distribution of human population weight, you can have that 20% of them that weight "A" Kg and another 20% that weight "B" kg , but that doesn't mean that you have a man that weight both "A" kg and "B" kg at the same time.

 

[malawi_glenn]

Yes, sorry it was typo. So how is possible that for one value of beta particles, to have 2 kinetic energies for the electrons?

What is 'value of beta particles'??

beta particle = electron, same thing, different names..


And it is a PROBABILITY DISTRIBUTION. The figures you have is a distrubtion cure where one has plotted the number of electrons with repsect to their energy. To obtain this figure, one has recorded several milions of beta-particles.

So each electron has ONE value of kinetic energy, and several electrons can have the same kinetic energy.

 

[PhilippH]

Ah yes I'm sorry, statistic is another thing...sorry again for my italian interpretaion of english words.

 

[Physicsissuef]

Hm... I still can't understand why two points...

 

[malawi_glenn]

Hm... I still can't understand why two points...

Ok, let's take this thing to the very beginning..

Do you know how a histogram works?

You measure a lots of object of same kind, and record their property. Then you keep track on how many of those objects who has a certain property.

Lets take cars or instance.

You stand at the road, and look at the colors of the cars passing by. For each color you have a column, and when a car with color green passes by you increase the number of green cars by 1, and when a red car passes by you increase the number of red cars with 1, etc.

Then when you don't want to recornd anymore; you plot the result - you will get a histogram (http://en.wikipedia.org/wiki/Histogram ), the number of cars with respect to their color.

Do the same thing with a huge amount of radioactive nuclei, and record the number of electrons (cars) of a certain energy (color), here energy is a CONTINOUS variabe, wheras color is a discrete, so you will obtain a probability distribution (a continuous historgraf if you like) - with the number of recorded electrons with respect to energy.

The reason why you have two ponits is just that you record the same number of electrons with energy M_1 as electrons having energy M_2; nothing strange - it is the same thing you would get if you recorded 20 red cars and 20 black cars, nothing strange at all!

 

[Physicsissuef]

So you want to say that for different radioactive nuclei I will have different kinetic energies of the electrons?

 

[Astronuc]

So you want to say that for different radioactive nuclei I will have different kinetic energies of the electrons?

For two different radionuclei, which decay by beta-emission, the range of beta particle energies may overlap.

The two points (same number) on the energy distribution curve (number of particles as a function of energy) simply means that the probability of measuring two particles with those two energies is the same. It's a bit like rolling a pair of dice, and having two faces with the same value occur. Most of the time, the dice will have different values.

 

[malawi_glenn]

So you want to say that for different radioactive nuclei I will have different kinetic energies of the electrons?

yes.

and for each kind of isotopes you will get different SHAPES of this distribution. i.e C-17's shape (histogram) will not look 100% as the distribution from Cl-41. But they will be similar due to theory of beta-decay (see the reference I gave you for instance).

You can compare this with the color of the cars at the road which I told you of before. changing isotopic kind means in this analogy that you change the place (rood) where you study the color of the cars. Maybe in the city of New York you will get a larger number of yellow cars (Taxi cars) than you would do in a small city in greece..

 

[Physicsissuef]

yes.

and for each kind of isotopes you will get different SHAPES of this distribution. i.e C-17's shape (histogram) will not look 100% as the distribution from Cl-41. But they will be similar due to theory of beta-decay (see the reference I gave you for instance).

You can compare this with the color of the cars at the road which I told you of before. changing isotopic kind means in this analogy that you change the place (rood) where you study the color of the cars. Maybe in the city of New York you will get a larger number of yellow cars (Taxi cars) than you would do in a small city in greece..

so for one beta particle there may have infinite number of M points, right?

For two different radionuclei, which decay by beta-emission, the range of beta particle energies may overlap.

The two points (same number) on the energy distribution curve (number of particles as a function of energy) simply means that the probability of measuring two particles with those two energies is the same. It's a bit like rolling a pair of dice, and having two faces with the same value occur. Most of the time, the dice will have different values.

The two points are for same number of paricles... So it means that the probability of measuring two particles, the particles are same, right?