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Multiparticle system, KE and PE, distances before/during/after fission

  1. Feb 24, 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. Uranium-235, with 92 protons and 143 neutrons, can fission when it absorbs a neutron and becomes Uranium-236. The two fission fragments can be almost any two nuclei whose charges Q1 and Q2 add up to 92e (where e is the charge on a proton, e = 1.6multiply10-19 coulomb), and whose nucleons add up to 236 protons and neutrons (U-236, formed from U-235 plus a neutron). One of the possible fission modes involves nearly equal fragments, palladium nuclei (Pd) each with electric charge Q1 = Q2 = 46e. The rest masses of the two palladium 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 palladium nuclei.)

    The rest mass of the U-236 nucleus (formed from U-235 plus a neutron) is 235.995 u (unified atomic mass units), and the rest mass of each of the two Pd-118 nuclei is 117.894 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 U-236 nucleus, before it fissions.
    2) The state just after fission, when the two palladium nuclei are close together, and momentarily at rest.
    3) The state when the palladium nuclei are very far away from each other, traveling at high speed.
    (a) Calculate the final speed v, when the palladium 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 palladium 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.)

    (b) Using energy considerations, calculate the distance between centers of the palladium nuclei just after fission, when they are momentarily at rest. Keep at least 6 significant figures in your calculations.

    (c) A proton or neutron has a radius r of roughly 1multiply10-15 m, and a nucleus is a tightly packed collection of nucleons. Therefore the volume of the nucleus, (4/3)piR3, is approximately equal to the volume of one nucleon, (4/3)pir3, times the number N of nucleons in the nucleus: (4/3)piR3 = N(4/3)pir3. So the radius R of a nucleus is about N1/3 times the radius r of one nucleon. More precisely, experiments show that the radius of a nucleus containing N nucleons is (1.3multiply10-15 m)multiplyN1/3. What is the radius of a palladium nucleus?

    (d) You could make a careful scale drawing on paper of the two palladium nuclei in part (b), just after fission, and label the drawing with the distances that you calculated in parts (b) and (c). If the two palladium nuclei are nearly touching, this would be consistent with our model of fission, in which the U-236 nucleus fissions into two pieces that are initially nearly at rest. How big is the gap between the surfaces of the two nuclei? (If you have done the calculations correctly, you will indeed find that the gap is a rather small fraction of the center-to-center distance, which means that our model for the fission process is a pretty good model.)

    2. Relevant equations

    Ue = (9e9)(q1q2/r); K = 1/2 m v^2; ...

    3. The attempt at a solution

    I have absolutely no idea QQ
  2. jcsd
  3. Feb 24, 2009 #2
    yea i was having the same problem. do you know where to start?
  4. Feb 24, 2009 #3
    no idea...I've thought about potential and kinetic energies but then it always ends up with me having two variables...
  5. Feb 24, 2009 #4
    a) 12570700 u = 1/2mv^2
    b)1.57644e-14 u = columbos constant * ((q1q2)/r) solve for r
    c)6.3763e-15 N = 118
  6. Oct 6, 2009 #5
    Is there any way you can explain in detail how you got those answers?

    I would really appreciate it.
  7. Oct 6, 2009 #6
    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
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