Register to reply 
Questions from Candymanby theCandyman
Tags: candyman 
Share this thread: 
#1
Jan1105, 11:16 AM

P: 395

I have some questions from reading my Intro to Nuclear engineering text book and did not want to take over Dlockwood's thread.
Why does the atomic number increase when a beta particle is released? Passage from my text book: "The nuclear fission process, as one method of converting mass into energy, is relatively inefficient, since the "burning" of 1 kg of uranium involves the conversion of only 0.87 g of matter into energy. This corresponds to about 7.8 x 10^13 J/kg of the unranium comsumed."* I read this but the J/kg is throwing me off what the author means. Is it that for every 0.87g of uranium, 7.8 x 10^13 J of energy is released? I can not find it in context, but I think I remember MeV being used to represent mass. Is this a valid way to represent small amounts of mass? *Nuclear Energy: An Introduction to the Concepts, Systems, and Applications of Nucler Process. Fifth edition, 2001. Raymond L. Murray. 


#2
Jan1105, 12:06 PM

P: 1,382

"0.87 g of matter into energy" refers to matter literally disappearing wholesale and turning into energy. In contrast, "kg of the uranium consumed" refers to fissioned uranium that has been broken up into smaller particles and thus is no longer fuel. Therefore, "This corresponds to about [itex]7.8 x 10^{13} J/kg[/itex] of the uranium consumed" means that .87g of every kilogram of fissioned uranium vanishes from the universe and is thus turned into energy ([itex]7.8 x 10^{13}[/itex] J).
http://hyperphysics.phyastr.gsu.edu...ne/nucbin.html As you can see, Fe56 has the least energy. It is also the most tightly bound, as the more tightly bound an atom is, the less energy it has relative to the energies its particles would have if they were not bound into an atomic structure at all. The missing energy is missing mass, so Fe56 could be thought of also as the lightest atom relative to the total weight its subatomic particles would have if they happened to be unbound. U235 is on the far right of the binding energy curve, so you can see that if you fissioned it into smaller particles the binding energy of its subatomic particles would increase. Proportionately, the massenergy (which is to say the massweight) of its subatomic particles would decrease and the difference in energy would be released  [itex]7.8 x 10^{13} J/kg[/itex] of fuel fissioned.



#3
Jan1105, 12:52 PM

Sci Advisor
P: 1,160

nucleus has too many neutrons for it to be stable. The extra neutron decays into a proton, an electron, and an antineutrino. The electron and antineutrino escape the nucleus. We detect the electron as "beta" radiation. The antineutrino is too weakly interacting to catch. The proton stays in the nucleus  hence there is one more proton in the nucleus than there used to be  therefore the atomic number increases by 1. Dr. Gregory Greenman Physicist 


#4
Jan1105, 01:04 PM

Admin
P: 21,880

Questions from Candyman
[tex] n \rightarrow p^+\,+\,e^\,+\,\overline{\nu}_e [/tex] where n is the neutron with neutral charge, and the antineutrino is electron associated. 


#5
Jan1205, 10:58 AM

P: 395

Thank you for all of your replies. Why does the neutron decay? Neutons are unstable alone (correct?) so does the atom being an ion cause decay?



#6
Jan1205, 02:48 PM

Admin
P: 21,880

Lone neutrons decay (by beta emssion) with a halflife about 10.23 minutes. Usually decay means that there is an excess of energy in the system, i.e. the system is excited, and the excess energy is then released through a particle emission. The emssion of an electron and antineutrino puts the neutron system into a low energy state, the proton, which is much more stable.
Many atoms are stable. Isotopes with an excess of neutrons may decay  usually by beta decay. However, some heavy isotopes (in elements Ra and up) may decay by alpha emission. The nuclear instability depends on the quantum mechanical aspects of the nucles. Momentarily after beta decay, the nucleus is an ion and it will attract an electron. A succession of atoms will then exchange electrons until the ejected electron is caught by another atom some distance way from where it was emitted. Charge neutrality (balance) is thus maintained. See  http://wwwndc.tokai.jaeri.go.jp/CN03/ for a good reference on radionuclides. 


#7
Jan1205, 11:25 PM

P: 10

Are photons elementary particles of hardons or the make up of them (Z + N) like quarks. When particles fission, the release of energy (Kinetic photons ) are born from the hadrons or they are smaller particles encased inside the protons and neutrons?



#8
Jan1305, 10:28 AM

Sci Advisor
P: 1,160

The lone neutron is unstable and does decay. However, an atom being an ion has no effect on the neutrons. An atom becomes an ion when it loses electrons in orbit about the atom. However, even in an ion; down deep in the nucleus  the neutrons are still surrounded by protons and other neutrons  and that keeps them "happy". Dr. Gregory Greenman Physicist 


#9
Feb505, 03:16 PM

P: 395

I have a few more questions from my text book.
How can an atom have "too few protons or too many neutrons"? I read this in the same book as I quoted above, and I stopped to think about this statement. I thought the element was defined by the number of protons it had, so how is it possible to have too few protons? Would not it always be that there are too many neutrons? What is angular velocity? And how did the units of it become "per second"? 


#10
Feb505, 05:00 PM

Admin
P: 21,880

I find this site useful  Chart of Nuclides 2004 Angular velocity is given by the quotient of an angle divided by the time taken to 'sweep' through the angle. Time is a dimension, angle (orientation) is not. All points on a radius have the same angular velocity, but the linear velocity of a point is proportional to the distance from the origin (rotational axis). Hopefully I am clear about angles. 


#11
Feb705, 10:09 AM

Sci Advisor
P: 1,160

I think I may have been a little loose in my terminology. I should have said too many neutrons  TO BE STABLE! The nucleus of an atom of atomic number 'Z' contains Z protons. Now if Z > 1; those protons repel each other because they are of like charge and like charges repel  and blow the nucleus apart. Each nucleon, like a proton or neutron; brings with it some "nuclear glue". However, the repulsion force due to the like charges repelling goes like Z * (Z1) /2. Each of the Z protons can repel (Z1) other protons [ and you divide by 2 so that you don't double count ]. So the repulsive force goes up quadratically with Z. Neutrons bring in approximately the same amount of nuclear binding force or "glue" that protons do  but without the repulsion. It turns out there is an optimum number of neutrons to have for a given number of protons for the nucleus to be stable. If you don't have the optimum value, or values  you either have an excess of protons or neutrons. If you have too many protons, there is too much electrostatice repulsion for the nucleus to be stable. The nucleus needs to get rid of some of its positive charge. It can do that via "betaplus" decay  one of the protons turns into a neutron, a positron, and a neutrino. The positron and neutrino are ejected and we detect the positron as "betaplus" decay. If you have an excess of neutrons, as I stated before, the neutron can decay into a proton, an electron, and an antineutrino  which gives us "betaminus" decay. Having an excess of protons or neutrons is always with respect to the stable condition. Dr. Gregory Greenman Physicist 


#12
Feb705, 10:17 AM

Sci Advisor
P: 1,160

Are you familiar with the term "RPM" for your car's engine, for example; that's an angular velocity. The units are Revolutions Per Minute, hence the acronym RPM. You could just as well use revolutions per second. Or you could use degrees per second  90 degrees being a quarter turn, and 360 degrees is a revolution. Scientists frequently measure angles in "radians". One radian is the angle by which the length of arc equals the radius. It's about 57 degrees. You can compute the angle in radians by taking the arc length subtended by that angle and dividing by the radius. If you measure both in centimeters  the radian has the units of centimenter per centimeter  so it is dimensionless. Now just as you could express an angular velocity as degrees per second, you could also express it as radians per second. But radians are really dimensionless  so you are left with "per second" as the units of an angular velocity. Dr. Gregory Greenman Physicist 


#13
Feb805, 09:34 PM

P: 395

Again, thank you! I have an exam on Wednesday, so wish me luck.



#14
Feb1605, 05:40 PM

P: 395

I am completely stumped on the homework I have assigned. Here is the question, if anyone could push me in the right direction on where to begin, I would greatly appreciate it.
"Exercise 10.1 Find the number density of molecules of [tex]BF_3[/tex] in a detector of 2.54 cm diameter o be sure that 90% of the thermal neutrons incident along a diameter are caught ([tex]\sigma_a[/tex] for natural boron is 760 barns)." This is part A of the question and the only equation that I can remember that has number density in it is [tex]R = nN\sigma_a[/tex] so that means [tex]n = R/N\sigma_a[/tex], still I feel that I cannot do much with this equation. 


#15
Feb1605, 07:45 PM

Admin
P: 21,880

Another way of looking at this problem is that the intensity of the neutron flux/current must be 0.1 of the initial flux/current.
Then apply, I(L) = I_{o} exp([itex]\Sigma[/itex] L) where L is the length, in this case 2.54 cm. [itex]\Sigma[/itex] = N [itex]\sigma_a[/itex] where N is the atomic density and [itex]\sigma_a[/itex] (assume for ^{10}B) is the appropriate absorption crosssection. Now remember that it is ^{10}B that reacts, and that is only 20% of natural boron. So the N(B) has to be 5 times that of ^{10}B. 


#16
Feb1705, 11:20 AM

P: 395

.1 is from 10%? Why does the current change from the inital?
Does "exp()" means exponent? So you are raising the inital current to the power of [tex]N\sigma_aL[/tex]? How do I go about finding the inital current, or do I solve this problem in variables? Is atomic density the number of atoms for my volume (I mean, is N just the total number of atoms of [tex]BF_3[/tex])? Barns are the unit for absorbition cross section? What does your last statement mean? What is N(B) and how do you say that boron is 20% of natural boron? Is some of it made by humans? Sorry for all the questions, I was not expecting I had so many either. This problem is the first one so I expect it is easier than the rest.I cannot help but suspect that this Intro is class is for upper classmen. 


#17
Feb1705, 12:28 PM

Admin
P: 21,880

The current changes because the neutrons interact with the B10 nucleus in an (n,[itex]\alpha[/itex]) reaction. If 90% of neutrons are absorbed along L, then 10% survive, and the equation gives the number surviving. This problem is effectively one of beam attenuation.
So one would want to find I(L)/I_{o} = 0.10. This comes from the first order differential equation dN(x)/dx = [itex]\Sigma[/itex] N(x), although some texts use [itex]\mu[/itex] as attenuation coefficient, so the equation would be dN(x)/dx = [itex]\mu[/itex] N(x). exp (x) = [itex]e^x[/itex]. Atomic density, N, is the number of atoms per unit volume. All units need to be consistent. The microscopic crosssection ([itex]\sigma[/itex]) is often given in barns, but 1 barn = 1 E24 cm^{2}. The macroscopic crosssection, [itex]\Sigma[/itex] = N [itex]\sigma[/itex]. Units of [itex]\Sigma[/itex] are cm^{1}, so that when multiplying length, the argument of the exponential function is dimensionless. N(B) is the atomic density of B. Natural boron, i.e. the boron mined or extracted from the ground contains 2 isotopes  19.9% of boron is B10 and 80.1% is B11. The reaction of interest is with B10, so one needs 5 times the amount of boron (B10 + B11) to get the correct amount of B10. On the other hand, one could adjust the microscopic crosssection according to the isotopic content. Also, I mention natural boron because companies which process boron, e.g. EaglePicher can enrich boron in B10. That way it takes less 'boron' to get a certain amount of B10. Hopefully all your questions have been answered to your satisfaction. 


Register to reply 
Related Discussions  
A few questions about QM and QFT.  Quantum Physics  11  
Questions about GR  Special & General Relativity  4  
Two questions: Please Help  Introductory Physics Homework  3  
Greetings, and a few questions.  Astronomy & Astrophysics  6  
Three little questions  Electrical Engineering  1 