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Homework Help: Physics Nuclear Reaction Mechanics Question

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    For some isotopes of some very heavy nuclei, including nuclei of thorium, uranium, and plutonium, the nucleus will fission (split apart) when it absorbs a slow-moving neutron. Plutonium-239, with 94 protons and 145 neutrons, can fission when it absorbs a neutron and becomes Plutonium-240. The two fission fragments can be almost any two nuclei whose charges Q1 and Q2 add up to 94e (where e is the charge on a proton, e = 1.6multiply10-19 coulomb), and whose nucleons add up to 240 protons and neutrons (Pu-240, formed from Pu-239 plus a neutron). One of the possible fission modes involves nearly equal fragments, silver nuclei (Ag) each with electric charge Q1 = Q2 = 47e. The rest masses of the two silver nuclei add up to less than the rest mass of the original nucleus. (In addition to the two main fission fragments there are typically one or more free neutrons in the final state; in your analysis make the simplifying assumption that there are no free neutrons, just two silver nuclei.)

    The rest mass of the Pu-240 nucleus (formed from Pu-239 plus a neutron) is 240.002 u (unified atomic mass units), and the rest mass of each of the two Ag-120 nuclei is 119.893 u, where 1 u = 1.66multiply10-27 kg (approximately the mass of one nucleon). In your calculations, keep at least 6 significant figures, because the calculations involve subtracting large numbers from each other, leaving a small difference. There are three states you should consider in your analysis:

    1) The initial state of the Pu-240 nucleus, before it fissions.
    2) The state just after fission, when the two silver nuclei are close together, and momentarily at rest.
    3) The state when the silver nuclei are very far away from each other, traveling at high speed.
    (a) Calculate the final speed v, when the silver nuclei have moved very far apart due to their mutual electric repulsion. Keep at least 6 significant figures in your calculations. In your analysis it is all right to use the nonrelativistic formulas, but you then must check that the calculated v is indeed small compared to c. (The large kinetic energies of these silver nuclei are eventually dissipated into thermal energy of the surrounding material. In a nuclear reactor this hot material boils water and drives an electric generator.)

    2. Relevant equations

    pf=pi+Fdeltat
    U=q1q2/(couloumb constant)*r
    K=.5*m*v^2

    3. The attempt at a solution
    I tried using the the diameter of an average nucleus as the radius but it doesnt work

    (47*1.6e-9)^2/(9e9)(1.6e-15)=3.9e-10
    3.9e-10=.5*(119.893*1.66e-27)*(v)^2=6.28m/s

    which is wrong
     
  2. jcsd
  3. Oct 6, 2009 #2

    Delphi51

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    Looks like you forgot about the energy of the lost mass.
    There is no mention of any gamma rays or other particles carrying away any of that so you must use E = mc^2 and add this energy to the electrical energy you already found.
     
  4. Oct 6, 2009 #3
    I tried the electical energy + mc^2= .5mv^2 and it was still wrong
     
  5. Oct 6, 2009 #4

    Delphi51

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    What you need is
    q1q2/(couloumb constant)*r + mc^2= .5mv^2
    Both the mass conversion energy and the electric potential energy is converted to kinetic energy at the far away position.
     
  6. Oct 6, 2009 #5

    turin

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    rkd,
    - Whence did you get r=1.6e-15? I would expect it to be a bit larger than that.
    - Kinetic energy is not measured in m/s, but rather in units of energy. You need to be generally more careful about your units.
    - The problem warns you to use 6 sig figs, which you did not do (in the OP).

    Also, I would say that the equation q1q2/(kr) + mc^2 = 0.5mv^2 is not correct. There is at least the obvious flaw that you are using the same symbol for the mass on both sides, but you should be talking about two different masses. And how many kinetic energies are there?
     
  7. Oct 6, 2009 #6

    Delphi51

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    Good point - be sure to use lots of digits in the calculation of the mass that is converted to energy. You'll be subtracting two nearly equal masses to get the amount of mass that is converted.
     
  8. Oct 6, 2009 #7
    I derived another method i just want to see if it makes sense.
    m(pu)c^2-m(ag)c^2=.5m(v)^2
     
  9. Oct 6, 2009 #8

    Delphi51

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    This formula does not include the electric potential energy.
    As for the mass, I think you are better off subtracting before you multiply by c^2 due to the large number of digits of accuracy required.
     
  10. Oct 6, 2009 #9
    I am also working on this problem, but cannot get it for the life of me. I've tried multiple ways, but they're all wrong. I basically need to know what "The state just after fission, when the two silver nuclei are close together, and momentarily at rest." is, cause the answer would be equal to something like m(Pu)c^2+ ? = .5*m(Ag)*v^2 where I am trying to find v...I just don't know what the state just after fission would be. And it can't be q1q2/(couloumb constant)*r because it doesn't give out any radii of any of the atoms. Do I have to find the neutron potential energy or something? Please help...
     
  11. Oct 6, 2009 #10
    I'm guessing it would be something like U(pu)+ Neutron = U(ag)initial + K(ag)final
     
  12. Oct 6, 2009 #11

    Delphi51

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    The question isn't clear on how far apart the two silver nucleii are after the fissioning. rkd1312 guessed it would be about 10^-15 m, which he says about the diameter of a nucleus. Sounds reasonable but estimates will vary.

    I offered the clue that the energy immediately after the fission is the
    electric potential energy + energy converted from lost mass
    and said that this energy will be converted to kinetic energy as the nucleii fly apart. You have formulas for all these kinds of energy, so fire up your pencil and do some calcs.
     
  13. Oct 6, 2009 #12
    it doesn't give the radius you have to find it later in the problem which I found out to be 6.41215e-15. But you are able to find v without it.
     
  14. Oct 6, 2009 #13
    I've tried multiple equations but the one thing that would be helpful is what "The state just after fission, when the two silver nuclei are close together, and momentarily at rest" would be represented by.
     
  15. Oct 6, 2009 #14

    Delphi51

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    You could represent them as two charged spheres.
    I would be most interested in how you found their separation distance.
    Surely the energy and the final speed depends on the separation - the closer two charges are pushed together, the greater their potential energy and the faster they will fly apart.
     
  16. Oct 6, 2009 #15
    1) The initial state of the Pu-240 nucleus, before it fissions = m(pu)c^2
    2) The state just after fission, when the two silver nuclei are close together, and momentarily at rest = ?
    3) The state when the silver nuclei are very far away from each other, traveling at high speed = .5*m(ag)*v^2

    the equation will be something like m(pu)c^2 + ? = .5*m(ag)*v^2

    Those are the only state you need worry about, well times almost up, oh well 14/15 suits me
     
  17. Oct 6, 2009 #16
    Anyway the electric potential energy is negligible since it is such a small number and doesn't affect the equation
     
  18. Oct 6, 2009 #17

    Delphi51

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    electric potential energy + energy converted from lost mass = KE
    U=k*q1*q2/(r) + (Pu mass - 2*silver mass)*c^2 = .5*m*v^2
     
  19. Oct 7, 2009 #18

    turin

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    I think that AirChambz raises an important issue. This problem depends critically on the model of the state just after fission. If I calculated correctly, then I determine that at 1 fermi (10-15 m ~ diameter of a single proton) of separation, the electrostatic potential energy between two Ag nuclei should be about 3 times the rest energy of a nucleon, which is much greater than the nuclear energy difference given in the problem. I would expect the separation to be several fermi, based on a naive model that a nucleus is a close-packed arrangement of hard-sphere nucleons. Since the electrostatic potential energy is inversely proportional to the separation, it will be less than the above estimate, but still on the order of the rest energy of a nucleon, and greater than the nuclear energy difference. So, I don't think that the electrostatic potential energy is negligible at all. Maybe I messed up the calculation.

    EDIT:

    I just recalculated in a different unit system, using GeV for energy, fm for length, and e for charge. I determine:

    ke=1.439964415E-3 GeV fm / e2

    (using widely available values for ke and charge of proton=e on wikipedia).

    These units are convenient for nuclear caclulations, because q is then simply the atomic number, and each nucleon is about 1 fm in size and has about 1 GeV of rest energy. This confirms my previous claim that the electrostatic energy is nonnegligible for separations on the order of several nucleon diameters.
     
    Last edited: Oct 7, 2009
  20. Oct 7, 2009 #19

    Delphi51

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    Indeed - I ran it with rkd1312's estimate of 1.6e-15 m separation and got the electric energy about ten times the nuclear. Also odd that the question says the two nucleii are stationary immediately after the fission. The energy converted from mass should appear as kinetic energy. Or more likely be carried off by fast neutrons, neutrinos and gamma rays.
     
  21. Oct 7, 2009 #20

    turin

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    Indeed. At least the problem states that assumption clearly, though. I wonder if there is missing information in the problem statement. Maybe somewhere the assumed separation of the Ag nuclei just after fission is stated, and the OP just neglected this tidbit.

    BTW, I editted my previous post for more detail and affirmation.
     
  22. Oct 7, 2009 #21
    By the way the answer turned out to be 1.27e+07 m/s for the velocity. No idea how to calculate it just wanted to let everyone know.
     
  23. Oct 7, 2009 #22

    Delphi51

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    Thanks, Air. We got quite interested in your problem. With the separation distance you used (assumed?) the speed works out to about 5 x 10^7 m/s, so your estimate was a wee bit low. The velocity is very sensitive to that initial distance.

    I was a student like you about 40 years ago, and if an old guy may offer a tip to a young one, it would be to never let go of a problem without making a real effort to find out how to do it. In this case, there appears to be a definite answer to the question so there must have been a definite initial separation distance. Why didn't you have it? Either you didn't see it or the person who made up the question didn't provide it. Either way, you need to know how you missed it and if it wasn't your fault you should make your marker aware of the fact that the question was incomplete and impossible to answer precisely. In the latter case, you ought to get at least some respect and likely credit for the problem if you can talk intelligently about how it is solved. It appears your generation is suffering from acute answeritis - only the answer matters for your marks, but in fact the process is far, far more important. You'll have to fight that to get a good education.
    Over and out . . . and good luck in your education!
     
  24. Oct 7, 2009 #23

    turin

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    Hmm... That is extremely close to (less than 1% off from) Δ=U, where Δ is the nuclear energy difference between the Pu and 2Ag states, and U is the electrostatic potential energy between the 2 Ag's just after fission. I have no idea why that should be true...
     
  25. Oct 8, 2009 #24

    Delphi51

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    Oh! No wonder there is no KE immediately after the fission - all the mass-conversion energy has gone into the electric potential energy. So you can CALCULATE the separation of the nucleii!

    Airchambz's prof is cleverly teaching us nuclear physics. And you can't get the problems unless you go to class and listen!
     
  26. Oct 8, 2009 #25

    turin

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    This does not seem correct (but then again, I am very naive to nuclear physics). The protons are already there, and I pressume closely packed together as a Pu nulceus, so there is already electrostatic potential energy. Are you suggesting that the decay and resulting nuclear energy difference causes the protons to get pulled closer together? I was always under the impression that the nuclear energy difference tends to convert into kinetic energy of the decay products. Why does the nuclear energy difference prefer to pull the nuclei closer together rather than giving the nuclei some kinetic energy? Do you understand this, and can you explain it to us?

    EDIT:

    Actually, I think that this issue of whether the nuclear energy is converted into electrical or kinetic energy is irrelevant, and just dodges the real issue. There is already electrostatic potential energy present due to the fact that there are 94 protons close together in the Pu nucleas. So, assuming that all of the nuclear energy is converted into electrostatic potential energy:

    ΔUe = Δmc2 = Uef - Uei

    However, in order to determine the final speed, we need:

    2KEAg = Uef = Uei + Δmc2

    So we still need to know Uei, and the suggestion simply trades one ambiguous Ue for another. We would get the same answer, I think, if we just assumed that the nuclear energy is converted into kinetic energy in the first place, and simply add that KE to the KE that results from the initial electrostatic potential energy. The answer for the velocity suggests, in fact, that:

    Uei = Δmc2

    (within some small error), and this is what requires explanation, regardless of what kind of energy Δmc2 is directly converted into.

    DOUBLE EDIT:

    No, dangit, I messed up. This is even more complicated, because the electrostatic "self-energies" of the Ag nuclei need to be subtracted. Self-energy should not be converted into kinetic energy. So:

    2KEAg = Uef - 2UeAg(self) = UePu(self) - 2UeAg(self) + Δmc2

    The notation is getting a bit cumbersome, sorry. So, now, the question has been refined once again. Why is it true that:

    UePu(self) - 2UeAg(self) = Δmc2

    In other words, why should the entirely electrical L.H.S. be equal to the (entirely?) nuclear R.H.S.?
     
    Last edited: Oct 8, 2009
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