SciForums.com > Science > Physics & Math > beta particles PDA View Full Version : beta particles Post ReplyCreate New Thread StMartin03-29-08, 08:54 AMHi! 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.th.jpg (http://img151.imageshack.us/my.php?image=picture001copy1jg2.jpg) http://img329.imageshack.us/img329/3677/picture001copyuc2.th.jpg (http://img329.imageshack.us/my.php?image=picture001copyuc2.jpg) Please help! Thank you. Reiku03-29-08, 03:38 PMDon't know much about beta particles. soz. BenTheMan03-29-08, 05:26 PMThe graph just says that the beta particles come out with different energies from the nucleus. There's nothing particularly special about the fact that tere are two values: it's just a distribution. Compare to something like the distribution of IQ's in society...is there anything particularly special about the fact that 0.1% of the population has an IQ either above 145 or below 55? http://lonelygenius.files.wordpress.com/2007/10/iqnormalcurve.gif StMartin03-29-08, 05:40 PMBecause of the beta particles the kinetic energy of the electrons is not constant? BenTheMan03-29-08, 06:24 PMWell, exactly. We would never expect all of the beta particles to come out with the same energy---there is ALWAYS a distribution of energy in every physical process. temur03-29-08, 06:45 PMBecause of the beta particles the kinetic energy of the electrons is not constant? I think if you consider one particular experiment, beta particle will have certain (constant) energy, but if you consider collection of thousands of experiments (keeping all the parameters you care constant), the energies of the beta particles will be different, but as you increase the number of experiments, the percentage of particles with certain energy tend to something specific. The uncertainty in the energy of beta particle is in part because some factors that you cannot control might be influencing your experiment (some undetected particle production, measurmenet device error, or you just leave some parameters uncontrolled to save money or time etc), in part because of the quantum phenomenon. BenTheMan03-30-08, 02:50 AMArgh. I mean not exactly'' above... as in you weren't exactly correct. Either way, I think temur cleared it up. StMartin03-30-08, 03:27 AMI 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? temur03-30-08, 04:27 AMOne neutron is missing, so you have baryon number conservation. From the fact that neutrino is electro neutral you can deduce it cannot take much energy under electromagnetic interaction, but the main player here is the weak interaction which has little to do with electric charge. What do you mean by "period of half decay possible"? StMartin03-30-08, 04:40 AMWhat is the ratio of the electrons and beta particles (how many beta particles have in ratio with electrons)? Period of half decay T_1_/_2 BenTheMan03-30-08, 12:38 PMCould you explain what you mean? StMartin03-30-08, 02:42 PMare all alpha, beta and gamma particle released from radioactive decay?? What is T_1_/_2 from all this? Walter L. Wagner03-30-08, 06:19 PMBecause the beta particle can come off with a distribution of kinetic energies, in order to conserve momentum and energy, it was hypothesized by Wolfgang Pauli [of Pauli exclusion principle fame] that an unseen, electrically neutral particle was also emitted during beta decay. He called the particle a "neutron". This particle is now known as the neutrino, a name coined by Fermi because Chadwick subsequently used the term "neutron" for the neutrons he discovered that were present in the nucleus. The neutrino is not considered to be present in the nucleus, nor is the electron. Rather, they are considered to be generated during the decay of a neutron into a proton, emitting both the electron and electron-neutrino. Other types of neutrinos are present for other types of similar decays, such as the positron-neutrino; the muon-neutrino, etc. There is a lot of good information on this also from Fermilab, as they've been conducting some excellent neutrino detection experiments. Unlike beta radiation, alpha radiation comes off with a single energy. Nuclear decays that occur with only a single interaction, rather than emission of two particles simultaneously as for beta decay, occur with discrete energies. That is why we have discrete gamma energy spectrums for such decays, allowing us to select for those discrete energies and determine the isotope that decayed by its gamma spectrum. You can check with Wikipedia under Wolfgang Pauli and Neutrino for more information. MetaKron03-30-08, 08:10 PMThe answer to the original question is that such a distribution follows a typical Bell curve. Ben explained it but he didn't use the term for it. A neutrino seems to be a particle that is almost entirely energy but has a very small real mass component to go with its relativistic mass, which is also small. That would suggest to me a level of quantization of masses of particles that is much smaller than the mass of an electron, as if the interaction that produces the neutrino can take a "chip" off a proton or neutron. Walter L. Wagner03-30-08, 09:56 PMActually, it is a curve that is not a Bell curve. A Bell curve extends to infinity, but with rapidly decreasing magnitude beyond the first few standard deviations about the center. The betas have a maximum energy, and no value higher than the maximum. It is these maximum values that are typically posted in reference sheets such as the Chart of the Nuclides. That is an excellent source for a background in nuclear physics, and ones of varying degrees of information are readily available. The ones put forth by GE Nuclear remain the best, in my opinion. The neutrino is the carrier of both momentum and energy, to allow conservation of both during the decay. For many decades, it was uncertain whether the neutrino had a rest mass, or whether it was massless like a photon. It is now believed, based on work of just the last few years, to have a small rest mass, allowing it to change from one type of neutrino into another during transit. There's lots of room for more research in this field. StMartin03-31-08, 11:13 AMis alpha, beta, gamma decay same with T_1_/_2 ?? Walter L. Wagner03-31-08, 02:19 PMThey follow the same radioactive decay law, whether it is alpha decay, beta decay, electron capture, etc. Each radioisotope has its own characteristic half-life. StMartin04-04-08, 02:52 PMBecause the beta particle can come off with a distribution of kinetic energies, in order to conserve momentum and energy, it was hypothesized by Wolfgang Pauli [of Pauli exclusion principle fame] that an unseen, electrically neutral particle was also emitted during beta decay. He called the particle a "neutron". This particle is now known as the neutrino, a name coined by Fermi because Chadwick subsequently used the term "neutron" for the neutrons he discovered that were present in the nucleus. The neutrino is not considered to be present in the nucleus, nor is the electron. Rather, they are considered to be generated during the decay of a neutron into a proton, emitting both the electron and electron-neutrino. Other types of neutrinos are present for other types of similar decays, such as the positron-neutrino; the muon-neutrino, etc. There is a lot of good information on this also from Fermilab, as they've been conducting some excellent neutrino detection experiments. Unlike beta radiation, alpha radiation comes off with a single energy. Nuclear decays that occur with only a single interaction, rather than emission of two particles simultaneously as for beta decay, occur with discrete energies. That is why we have discrete gamma energy spectrums for such decays, allowing us to select for those discrete energies and determine the isotope that decayed by its gamma spectrum. You can check with Wikipedia under Wolfgang Pauli and Neutrino for more information. But in that shouldn't there infinite number of M points? Is there different kinetic energy of the electron, for every radionucleide, or for one radionucleide there are only 2 points? Walter L. Wagner04-04-08, 04:59 PMFor any one particular type of radionuclide [e.g. C-14], the kinetic energy of the electron that comes off is always different for each decay, but will have a maximum value. The maximum value is the one reported as the beta energy for that radioisotope, but it is actually a spread in values up to that maximum. The spread in values is represented by the curve you posted. Here's what an old reference text ["Radiation Safety Technician Training Course"] of mine from 1972 reads [page 55-56]: "Unlike alphas, betas are not emitted with discrete energies but show a continuous energy spectrum (see Fig. 3.5 [shows your graph above]). The atom emits electrons of all energies up to some maximum value. The maximum value for the beta spectrum may be found in nuclide tables [e.g. Chart of the Nuclides] and is characteristic of that atom." Hope this answers your question. StMartin04-05-08, 12:58 AMI want to ask you why there are only 2 points for each radionucleid? Is there only 2 kinetic energies of the electrons possible for that radionucleid? Walter L. Wagner04-05-08, 11:37 AMI'm not certain why you're drawing a line across the graph to establish those two points. At those particular percentages, there are two energies, but it doesn't mean that there are only 2 kinetic energies for the electrons. They come off at a wide range of energies, all the way up to their maximum. There is not a particular discrete energy for them. Perhaps you are mis-reading what the graph means. StMartin04-05-08, 11:58 AMFor particular number of beta particles, why there are only 2 kinetic energies for the electrons? (I said why there are only 2 points for particular number of beta particles). I am not misunderstanding the graphic. StMartin04-06-08, 05:52 AMWalter? StMartin04-07-08, 03:01 PMHelp? Billy T04-07-08, 04:19 PMHi Walter: You have made many 100% correct and quite informative posts in this thread. Makes me think you may know answer to a question about book I have (I was once told it is "collector's item."): It is titled: Nuclear Physics with rather long subtitle (In format of cover): A Course Given by ENRICO FERMI at the University of Chicago. Notes compiled by Jay Orear, A.H. Rosenfeld, and R. A. Schluter Then below on faded blue cover is: Revised Edition Next the UOC press seal With at bottom of cover: THE UNIVERSITY OF CHIAGO PRESS Guy who told me this years ago offered to buy it, but as I intended to read it, I refused. It is a "paper back" in pretty good shape, no notes written in etc. but the as was in humid area for several years, the staples have rusted some. Know anything about it? In Brazil it will just end up in the trash when I am gone, so I would like to see it go to some good hands, if it really is a collector’s item. I am not a collector and do not know how to evaluate this book. (Except as a very-intersting, historically-important physics text.) If you (or anyone reading do) even that information would be helpful. My copy is from the "fifth impression" made in 1955. The fount appears to be a type-writter. Book was probably made by some photo-offset printing process in small editions as demand occurred. Fermi gave the course in the first six month of 1949. Walter L. Wagner04-07-08, 10:09 PMBillyT: I'm flattered. It does indeed sound like a collector's item. Probably quite rare, and of some historical interest. Perhaps Fermilab would be interested in it. I'm sure they must have a museum of some sort from Fermi's work. I wonder what happened to the slide rule he used on the day that his first pile went critical? I would imagine the text itself contains some good physics, but there's been so much more discovered since then, that it's likely not the best source. However, I frequently find that young physicists [you'll note that to be the case here on this forum] have some misperceptions about laws and facts of nuclear physics established many decades ago while they try to unravel some of the more topical or timely mysteries. It could prove an eye-opener for them. For my own part, I too have some relatively older texts, such as that Radiation Safety Tecnician Training Program, which itself was a compilation of documents of physics from the 1950s. It's nice to maintain an historical perspective. If Fermilab is not interested [or if they already have it], email me and I'll give you my address where you can mail it. Regards, Walter ------------ Billy T04-08-08, 12:37 PM...I wonder what happened to the slide rule he used on the day that his first pile went critical?...Likewise the bucket that held the boron water (Probably a strong commercial "borax solution") that he was ready to dump, as his last Earthly act, over the graphite moderating bricks if the reaction did not leveloff after "going critical." (As you no doubt know, but some reading will not, Boron has a huge neutron capture cross section, NCS.) - I am not sure, but think it was the first measures of Boron´s thermal NCS that caused the NCS to be measured in "Barns" to this day. I.e. when the NCS of boron was learned, someone said: "My God!; It is as big as a barn." I should note that the graduate students building the first nuclear reactor, carbon birck, by carbon brick, would periodically stop for the neutron flux (or someting indicating the sub-critical reaction rate) to be measured. It was alway found to be within the measurement error of Fermi´s predictions. In view of this, I do not think Fermi was scared as critcality was achieved, but he did stand above the pile with that bucket. Physicists like Fermi are few and far between. kevinalm04-08-08, 04:55 PMI thought it was a cadmium solution in the buckets? At least most accounts say so. Walter L. Wagner04-08-08, 11:04 PMThere were three cadmium control rods. One could be gravitationally dropped into place with the swing of an axe cutting a rope. Another was a backup. The third actually controlled the reactor. In addition, "a 'liquid control squad' composed of Harold Lichtenberger, W.Nyer, and A.C. Graves, stood on a platform above the pile. They were prepared to flood the pile with cadmium-salt solution in case of mechanical failure of the control rods." "The First Reactor", US AEC. Billy T04-09-08, 10:48 AMI thought it was a cadmium solution in the buckets? At least most accounts say so.Probably so. - I post from memory and almost never search anything. Now that Walter simulates my memory, I recall reading about the axe. Perhaps that was what Fermi had in his hands. Fortunately none was needed. I am almost sure the reason is that not all of the fissions, which produce neutrons do so immedaitely*. I forget the percent, but it is on the order of 1% are delayed for on the order of a second, I seem to recall. This makes time for the control systems, even today in power plants, to respond mechanically with "control rods." Few know about this small, but I think, extremely important fraction of "late neutrons." -If that is not true, please correct my belief that it is. -------------- *I don't know, but assume some of the fission fragments are in some excited nuclear state that decays with emission of a neutron. P.S. I have always wondered a little about this and the closely related question as to why not lots of different "heavy isotope"? In my simple POV neutrons are adding to the nuclear binding that over comes the Coulumb forces. I know the strong forces can not even "reach" across a High atomic number nucleus**, but do not understand why there seems to be an upper limit on them. E.e. we have C12, C14 perhaps a few more, but why not C34, etc? **As the Coulumb foces can, this explains why the neutron to proton, N/P, ratio is much greater than unity in high A nuclei. Why is there any upper bound on the N/P ratio?*** Why can we not have barion of only N? For examle N145 - Why only Nzillion, otherwise known as a "neutron star"? ***By edit, after pondering this again a few minutes: Perhaps the neutron should be considered to be a proton containing a tightly "orbiting" electron inside it. This makes a dynamic dipole of extremly high frequency. Perhaps it even radiates outside the "composite neutron" but that radiation escapes only a short range before the field radiated reverses (half an orbit later) Then as the wavelength is so short, some combination of system state quantization aqnd the uncertainity principle make the very high frequency "far field" zero so no energy is lost, but inside the far field region the radiation pressure would overcome the strong force if too many of these "composite neutrons" tried to assemble to make my C34, etc. - Just a wild, passing thought, by one quite ignorant of nuclear physics. Walter L. Wagner04-09-08, 12:12 PMBilly: Way too many questions to answer in a short post. Generally, there is a "sea of stability" wherein lie the stable and radioactive nuclei. This is the chart of the nuclides. It is bounded on one side by the "neutron-drip" line in which extra neutrons would immediately 'drip' away. Generally, the farther from the center line, the shorter the half-life. It is bounded on the other side with a line in which the excess positive charge would immediately force either a positron emission, or even proton. The Coulombic repulsion overcomes the strong-nuclear binding force. A C-34 could not exist because, as you mentioned, the strong-nuclear force is a short-range force, and the excess neutrons would be sufficiently spaced far enough away that the binding force is not sufficient to keep the excess neutrons, and they 'drip' away. There has been a lot of experimentation to exactly delineate wherein lie those two boundary lines, obtaining many radioisotopes with extremely short half-lifes. You should Google for 'Chart of the Nuclides' and take a look at any of the charts that come up. Billy T04-09-08, 02:53 PMBilly: Way too many questions to answer in a short post. Generally, there is a "sea of stability" wherein lie the stable and radioactive nuclei. This is the chart of the nuclides. It is bounded on one side by the "neutron-drip" line in which extra neutrons would immediately 'drip' away. ... A C-34 could not exist because, as you mentioned, the strong-nuclear force is a short-range force, and the excess neutrons would be sufficiently spaced far enough away that the binding force is not sufficient to keep the excess neutrons, and they 'drip' away.I am aware of the "chart of the nuclides" - in fact there is Fermi's 1949 version as a fold-out in the book I mentioned earlier in post 25. I even know about the "magic numbers", "doubly magic" numbers, favored stabilities of alpha grouping numbers in the chart. (Most of the well separated "out lyers" are of these types.) I am afraid "neutron drip" does not tell anything mechanistic - just restates the observed facts. I am almost sure the strong force is insentive to the charge. I.e. N-P P-P or N-N all have the same force at same separation if only the strong force is considered. Thus, I see no reason (assuming it did exist) why C33 could not capture another quite "thermal" neutron that happen to hit the "surface" of C33 to become C34. The strong force is very short range but surely reaches out more than a barion (N or P) diameter to buind it making C34. The only reason it would not stick is that the stong forces does not simply decrease with range but actually turns repulsive* for some greater range. I think that (A REPULSIVE REGION EXIST) is indeed believed to be true, but it sure seems like circular reasoning to me if the only reason for thinking this is the observed "neutron drip." I.e. with no other reason for believing that the strong force does turn negative, this is again just renaming the observation in a more mechanistic-sounding way, IMHO. If the there is really a repulsive region then neutron on neutron scattering should show that as a funtion of impact parameter. Impact parameters slightly LESS than the + to - transition region range of the strong force should have scattering angles LESS due to the weak attrication than the scatterning angles associated with impact parameters equal or slightly greater than the + to - transition separation. Normally, the smaller the impact parameter becomes, the GREATER the scattering angle. I do not know if one can do N on N scattering as function of impact parameter or not.** But it would take something additional, like this, for me to say postulating an outer repulsive region for the strong forces was not just a more sophisticated, but circular, form of saying "NEUTRON DRIP." ---------------- *Then most of the 33-6 neutrons in my C33 could be together pushing the approaching thermal neutron away before it entered the attractive range of the few on the side it was approaching. **N on P should show the same dip in scattering angle even without making a Coulumb correction as the electric force is monotonic function of the impact parameter. Surely one can shoot neutrons thru thin layer*** of hydrogen gas and observe their beams impact patterns on film or something. At least a neutron beam crossing a proton beam should be a feasible experiment. *** Formed, for example, by fine "needle nozzel" allowing low pressure stream of H2 to enter large high vaccum chamber in "puffs" with the vacuum pumps continuously "sucking " away. Walter L. Wagner04-09-08, 03:17 PMBillyT: I suspect there is not presently known a good theoretical explanation for the behavior. This is what is observed, and we only have empirical knowledge. I have seen attempts to explain the attraction, and how adding additional neutrons with increasing Z is required to maintain nuclear stability [so that the ratio of N/Z deviates from 1, to a larger value, after the first ten or so nuclides of Z = 1-10]. However, I do not believe there is any complete theoretical explanation. There are attempts to use 'shell structure' for the neutrons/protons, including 'shell structure' for them expressed as quarks. We know there is some internal 'shell structure' because we can have an 'excited state' that will decay to a 'ground state' via an isomeric transition, such as Tc-99m decaying to Tc-99 with a 6.02 hour T-1/2. However, I certainly do not have a complete understanding of what that shell structure is, and I suspect no one else does either [though I don't doubt we might have some pundits weigh in with their opinions]. Certainly that is some of the purpose of engaging in hadronic collisions in colliders, to obtain further insight. Billy T04-09-08, 03:26 PMBillyT: I suspect there is not presently known a good theoretical explanation for the behavior. This is what is observed, and we only have empirical knowledge. I have seen attempts to explain the attraction, and how adding additional neutrons with increasing Z is required to maintain nuclear stability [so that the ratio of N/Z deviates from 1, to a larger value, after the first ten or so nuclides of Z = 1-10]. However, I do not believe there is any complete theoretical explanation. There are attempts to use 'shell structure' for the neutrons/protons, including 'shell structure' for them expressed as quarks. We know there is some internal 'shell structure' because we can have an 'excited state' that will decay to a 'ground state' via an isomeric transition, such as Tc-99m decaying to Tc-99 with a 6.02 hour T-1/2. However, I certainly do not have a complete understanding of what that shell structure is, and I suspect no one else does either [though I don't doubt we might have some pundits weigh in with their opinions]. Certainly that is some of the purpose of engaging in hadronic collisions in colliders, to obtain further insight.I hope you too have only 30+ year old knowledge of this field. It would be sad to think that not more than this is now known - I knew what is in your post about nuclear shells, excited states, resonate scattering etc. 40+ years ago! (Perhaps not about quarks until 30+ years ago.) Really sad if there has been nothing more understood. :bawl: Post ReplyCreate New Thread