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Questions from Candyman

  1. Jan 11, 2005 #1
    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. jcsd
  3. Jan 11, 2005 #2
    "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).

    No. .87g of any kind of matter equals [itex]7.8 x 10^{13}[/itex] J of energy. As a kilogram of uranium fissions into smaller atoms - since the subatomic particles of those smaller atoms are more tightly bound and thus have less total mass than they did when they were bound together as uranium atoms - .87 grams of matter disappear and are replaced by energy. Here is the [curve of binding energy (hopefully it will make it clearer):

    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 mass-energy (which is to say the mass-weight) 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.

    It is on page 9 of your textbook:

    • For one unite of atomic mass, [itex]1.66 x 10^{27}[/itex] kg, which is close to the mass of a hydrogen atom, the corresponding energy is 931 MeV.

      Thus we see that matter and energy are equivalent, with the factor [itex]c^2[/itex] relating the amounts of each.

    I imagine it should be a valid way to represent any amount of mass. However, it is important to distinguish between rest energy of a particle and the energy of that particle at a given velocity. Even though a perticle will have energy at rest - since mass converts directly to energy - if it is moving, it will have a total energy above that of its rest mass. Thus, you will hear the phrase "high energy neutrons." This simply means the neutrons are moving fast.
    Last edited: Jan 11, 2005
  4. Jan 11, 2005 #3


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    You get "beta" decay - or specifically here "beta minus" decay, when the
    nucleus has too many neutrons for it to be stable.

    The extra neutron decays into a proton, an electron, and an anti-neutrino.
    The electron and anti-neutrino escape the nucleus. We detect the
    electron as "beta" radiation. The anti-neutrino 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.

    The rest energy of the proton [ ~ 1 amu ] is 938 MeV.

    Dr. Gregory Greenman
  5. Jan 11, 2005 #4


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    Very simply, in beta decay, a neutron decays into a proton + electron (beta particle and antineutrino. Thus while the atomic mass (integer) remains unchanged, the Z increases.

    [tex] n \rightarrow p^+\,+\,e^-\,+\,\overline{\nu}_e [/tex]

    where n is the neutron with neutral charge, and the anti-neutrino is electron associated.
  6. Jan 12, 2005 #5
    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?
  7. Jan 12, 2005 #6


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    Lone neutrons decay (by beta emssion) with a half-life 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 anti-neutrino 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/ [Broken] for a good reference on radionuclides.
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  8. Jan 12, 2005 #7
    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?
  9. Jan 13, 2005 #8


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    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
  10. Feb 5, 2005 #9
    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"?
  11. Feb 5, 2005 #10


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    That is an awkward statement. I think the author means 'too few protons for a given number of neutrons', but it is more accurate to say 'too many neutrons' or refer to 'neutron excess'. Basically, if an atom is 'stable', i.e. does not 'decay' or transform into another element, then it is considered to have the 'right' amount of neutrons.

    I find this site useful - http://wwwndc.tokai.jaeri.go.jp/CN04/index.html [Broken]

    You are correct. The number of protons determines the element, by definition. The atomic nucleus of each element has unique number of protons (and electrons, when neutral, i.e. not ionized). The neutrons determine the isotope - same Z, but different atomic mass. There can be too many neutrons, and the nucleus will emit a particle such as a beta (electron) particle or alpha (He nucleus) particle. If too 'few' neutrons, then positron (+ electron) emission or electron capture are possible. Refer to the 'Chart of Nuclides'.

    Angle refers to orientation and it is dimensionless. An arc length of a circular arc is given by the product of the radius (distance from axis of rotation to the arc) and the angle subtended by the arc.

    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.
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  12. Feb 7, 2005 #11


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    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 * (Z-1) /2.

    Each of the Z protons can repel (Z-1) 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 electro-statice repulsion for the nucleus to be stable. The
    nucleus needs to get rid of some of its positive charge.

    It can do that via "beta-plus" 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 "beta-plus" decay.

    If you have an excess of neutrons, as I stated before, the neutron can
    decay into a proton, an electron, and an anti-neutrino - which gives us
    "beta-minus" decay.

    Having an excess of protons or neutrons is always with respect to the
    stable condition.

    Dr. Gregory Greenman
  13. Feb 7, 2005 #12


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    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
  14. Feb 8, 2005 #13
    Again, thank you! I have an exam on Wednesday, so wish me luck.
  15. Feb 16, 2005 #14
    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.
  16. Feb 16, 2005 #15


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    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) = Io 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 10B) is the appropriate absorption cross-section.

    Now remember that it is 10B that reacts, and that is only 20% of natural boron. So the N(B) has to be 5 times that of 10B.
  17. Feb 17, 2005 #16
    .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.
  18. Feb 17, 2005 #17


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    The current changes because the neutrons interact with the B-10 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)/Io = 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 cross-section ([itex]\sigma[/itex]) is often given in barns, but 1 barn = 1 E-24 cm2.

    The macroscopic cross-section, [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 B-10 and 80.1% is B-11. The reaction of interest is with B-10, so one needs 5 times the amount of boron (B-10 + B-11) to get the correct amount of B-10. On the other hand, one could adjust the microscopic cross-section according to the isotopic content.

    Also, I mention natural boron because companies which process boron, e.g. http://www.eaglepicher.com/EaglePicherInternet/Technologies/Boron/ [Broken] can enrich boron in B-10. That way it takes less 'boron' to get a certain amount of B-10.

    Hopefully all your questions have been answered to your satisfaction.
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