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A basic conservation of energy question

  1. Apr 5, 2016 #1
    • Member advised to use the formatting template for questions posted to the homework forums
    The question asks: Uranium nuclei are unstable and decay by emitting a helium nucleus, which contains two protons and two neutrons. Suppose the original nucleus has 92 protons and 146 neutrons before it emits the helium nucleus, and that the helium nucleus has an knknown speed as it departs from the surface of the remaining nucleus, which is spherical with a radius of about 7.4 fm. What is the helium nucleus's minimum kinetic energy far from the emitting nucleus and why is this the lower limit?

    My attempt at a solution:

    First, note that m_p = 1.673 * 10^-27kg, m_n = 1.675 * 10^-27kg.
    also, e = 1.602 * 10^-19C

    Initially the energy of the system is 0. We can ignore gravitational effects because they are very small compared to the potential energy from the electrical attraction. Therefore, KE_f = (kq_1q_2/r) + (1/2)m_1v_1^2 + (1/2)m_2v_2^2.

    To calculate q_1 and q_2 I took q_2 = 2 * e, and q_1 = -q_2.
    By conservation of momentum, m_1v_1 = m_2v_2 so that v_1 = (m_2/m_1)*v_2.
    If you let r = 7.4 fm, and plug all of this in, and set KE_f = 0, you will find that (1/2)m_2v_2^2 = .12 pJ.

    The answer in the book is 5.6 pJ. I don't see how they get this, and I'm confused about the question asking why the energy is minimal. Based on the equations, the minimal kinetic energy occurs when r --> infinity, in which case the kinetic energy goes to 0.
     
  2. jcsd
  3. Apr 5, 2016 #2

    gneill

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    I think that they are assuming that the uranium remnant remains fixed in place (not a terrible assumption if the atom is held in a lattice of other atoms).

    For the second part, consider the concept of escape velocity.
     
  4. Apr 5, 2016 #3
    Even when I follow escape velocity derivation where the potential is an electric potential, I still get the same answer I have written above. The only thing I can imagine is that I'm making a charge calculation error, but I don't believe it.
     
  5. Apr 5, 2016 #4

    gneill

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    Can you show details of your calculation for a fixed nucleus? I'm seeing the book's result.
     
  6. Apr 5, 2016 #5
    The equation is -4kq_p^2/r + (1/2)*m2*(v2^2) = 0 right? where q_p is the charge of a proton, and m2 is 2(m_p + m_n). It's just plug and chug from there to solve for the kinetic energy which is the second term on the left hand side. I get .12pj
     
  7. Apr 5, 2016 #6

    gneill

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    The charge of the nucleus and the charge of the alpha particle both contribute to the electrical potential energy. Your first term would give the electric potential (Volts), not the potential energy (Joules).
     
  8. Apr 5, 2016 #7
    Look closer, it's a squared term. Right? They have equal but opposite charges?
     
  9. Apr 5, 2016 #8

    gneill

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    No they don't. The nucleus has +90 protons, the alpha +2.
     
  10. Apr 5, 2016 #9
    the potential energy equation is kq1q2/r. Positive terms for both q1 and q2 would result in a positive term and would result in nonsense. v would have to be 0 in that case.
     
  11. Apr 5, 2016 #10

    gneill

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    ##PE_o + KE_o = PE_f + KE_f##

    Assuming the initial KE is zero (alpha particle just detaches without initial velocity), and assuming that the PE at infinity is zero (definition) then

    ##k \frac{q_1 q_2}{r^2} + 0 = 0 + \frac{1}{2} m v^2##

    ##k \frac{q_1 q_2}{r^2} = \frac{1}{2} m v^2##
     
  12. Apr 5, 2016 #11
    I literally just figured that out as you wrote it. It all makes sense now. Thanks a lot for the discussion!
     
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