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Conservation of momentum uranium atom problem

  1. May 27, 2014 #1
    1. The problem statement, all variables and given/known data
    A uranium atom of mass 238 u decays by emitting an alpha particle (the nucleus of a helium atom) of mass 4 u at a speed of 2 x 107 m/s. (Note that "u" is the symbol for atomic mass units and 1 u = 1.67 x 10-27 kg.) What is the recoil speed of the resulting nucleus immediately after the decay?


    2. Relevant equations
    m1v1i + m2v2i = m1v1f + m2v2f


    3. The attempt at a solution
    The first thing I did was realize that the total momentum before the uranium emittes the alpha particle is zero so the momentum after the alpha particle is released is zero. So m1v1 + m2v2 = 0. I plugged in the masses of the two particles and the speed of the smaller particle and solved for the velocity of the larger particle to get the recoil speed.
     
  2. jcsd
  3. May 27, 2014 #2
    That is correct (in the inertial rest frame of the nucleus).
    It would be a tiny bit better to have said it this way:
    "Since (linear) momentum is conserved, the total momentum before the decay is exactly the same as after it."
    This avoids the problem of which rest frame you are referring to. Nothing much wrong with claiming momentum (or velocity) is zero, aside from the fact that it is only zero for one particular set of coordinates. What you can calculate is the difference in velocity of the two particles after the decay, compared to before it. That is
    MuVu = MthVth+MaVa (where V are the velocity vectors, and u=uranium, th= thorium and a= alpha particle). Choosing Vu to be zero greatly simplifies the math. Picking the right coordinates is a very important consideration in solving Physics problems.
    Oops!
    2E7 m/s ?? 2E7 m/c ÷ 3E8 m/s ~ 0.067c...you *might* want to add the relativistic correction to the momentum of the alpha particle - meaning its momentum is actually more than just ma*va. The Th nucleus will have a velocity substantially lower, so you don't need to worry about it's relativistic momentum. √(1-(v/c)²) ~
    √(1-0.067²) ~ 0.998; meaning given velocity with one place precision, no relativistic correction is necessary even for the alpha particle.
     
    Last edited: May 27, 2014
  4. May 27, 2014 #3
    Where are you getting Thorium?
     
  5. May 27, 2014 #4

    SammyS

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    That's element 90 .
     
  6. May 27, 2014 #5
    Is it the same thing as the alpha particle?
     
  7. May 27, 2014 #6

    SammyS

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    No.

    What is an alpha particle's composition?
     
  8. May 27, 2014 #7
    I don't know I haven't learned anything about particles on an atomic level.
     
  9. May 27, 2014 #8
    I haven't learned anything about particles of that size.
     
  10. May 27, 2014 #9

    SammyS

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    What size ?


    From your OP:

    ... emitting an alpha particle (the nucleus of a helium atom) of mass 4 u ...
     
  11. May 28, 2014 #10
    Ok so the only particles involved are the alpha particle that has a mass 4u and the uranium particle with mass 238? If so what did I do wrong in my original calculations?
     
  12. May 28, 2014 #11

    SammyS

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    You didn't really answer this. You could easily look it up, but here's the answer.

    The alpha particle is composed of 2 protons & 2 neutrons.

    If a Uranium (atomic number, 92) nucleus loses 2 protons (and 2 neutrons), what will be the resulting element ?




    .
     
  13. May 29, 2014 #12
    OK it is Thorium. So what do I do after I realize that it is Thorium?
     
  14. May 29, 2014 #13

    haruspex

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    The fact that the other resultant particle is Thorium was an interesting discussion, but not relevant to solving the question. You know its atomic mass, and that's all that matters.
    In the OP, you didn't indicate what answer you got or whether it was known to be wrong. Are you now saying your answer was wrong? If so, either you made a calculation error or you need to take into account the relativistic momentum of the alpha particle, as abitslow suggested. Please post your answer and the one given as correct.
     
  15. May 29, 2014 #14
    OK here is what I did to calculate my answer
    238u(v) = -(4u)(2 x 107 m/s)
    v = -3.36 x 105 m/s
    the correct value for v should be 3.42 x 105 m/s
     
  16. May 29, 2014 #15
    Sorry for the confusion
     
  17. May 29, 2014 #16

    tms

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    With that small a discrepancy, I'd suggest just using the mass of thorium instead of uranium. Since your initial data have only one significant figure, you can probably ignore the binding energy difference and any relativistic corrections due to the speed of the α, and also just take the mass of thorium to be ##238 - 4##. As for the sign, I'd guess they just chose a different positive direction.
     
  18. May 30, 2014 #17

    haruspex

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    BrainMan, in case it's not clear, you should have used 234 in the first place. The m1 and m2 of your equation here map to the 234u and 4u, as though two separate objects from the start.
    The OP asks for speed, not velocity, so an unsigned answer is required.
     
  19. May 30, 2014 #18

    tms

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    So it does.
     
  20. May 30, 2014 #19
    Ok I get it now. Thanks!
     
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