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KE of alpha particle using integer values of nuclear masses

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    (a) Cobalt has only one stable isotope, 59Co. What form of radioactive decay would you expect the isotope 60Co to undergo? Give a reason for your answer.

    (b) The radioactive nuclei 21084Po emit alpha particles of a single energy, the product nuclei being 20682Pb.
    (b) (i) Using the data below, calculate the energy, in MeV, released in each disintegration.
    (b) (ii) Explain why this energy does not all appear as kinetic energy, Eα, of the alpha particle.
    (b) (iii) Calculate Eα, taking integer values of the nuclear masses.

    Nucleus, mass (u): 21084Po, 209.936 730; 20682Pb, 205.929 421; α-particle, 4.001 504.

    (1 atomic mass unit, u = 931 MeV.)

    Answers: (b) (i) 5.40(4) MeV, (iii) 5.30(2) MeV.

    2. The attempt at a solution
    (a) No idea. It can't be α, β, since the difference in the nucleon number is 1: 60Co → 59Co.

    (b) (i) (209.936 730 - 205.929 421 - 4.001 504) * 931 = 5.404455 MeV.

    (b) (ii) No idea. KE like 1 / 2 m v2? Or E = m c2? Maybe because we don't have mass in kg in this situation?

    (b) (iii) Also no idea. Integer values like 209 - 205 - 4? Without digits?

    Any help please?
     
  2. jcsd
  3. Nov 1, 2016 #2

    mfb

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    That is not the decay process that happens. There are nuclei which do that (neutron emission), but those are rare.
    Missing units, but the answer looks good.
    Wrong direction. Assume that the initial nucleus is at rest. What happens after the decay? The alpha particle flies away - what about the remaining nucleus?
    You'll need (ii) to start here.
     
  4. Nov 1, 2016 #3
    So that is the answer for (a)? Form of radioactive decay -- neutron emission?

    Decay makes the nucleus more stable. Aside from that not sure what to add...
     
  5. Nov 1, 2016 #4

    mfb

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    No, you have to find the right decay process. Co-60 does not decay to Co-59.
    Which conservation laws could be relevant in the decay?
     
  6. Nov 2, 2016 #5
    Maybe something like 60Co → 59Co + 10n?

    Conservation of energy?
     
    Last edited: Nov 2, 2016
  7. Nov 2, 2016 #6

    mfb

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    Assuming "C" is a typo (carbon??): No, that is the process where I said already that it does not happen.
    That is relevant, but not the point. Can the alpha particle just start moving in some direction while everything else stays at rest all the time?
     
  8. Nov 2, 2016 #7
    I have no idea.

    I searched for "kinetic energy alpha particle" and got tp this topic, but I don't have the mass and velocity. Found pictures like this and tried to calculate the answer using 209.936 730 - 205 - 4 and that didn't help. Integer values says that the number should have no digits (at least I understood the definition so). But 209 - 205 - 4 = 0 and doesn't help much.

    No idea what to do and how to answer your question : (. What shall read on this?
     
  9. Nov 2, 2016 #8

    mfb

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    What about momentum conservation?

    Your initial nucleus is at rest, if the alpha particle flies away in one direction, the remaining nucleus has to fly away in the opposite direction.

    Don't start with (iii) before you got (ii) right, that won't work.
     
  10. Nov 3, 2016 #9
    I understand the concept of momentum conservation. So p = m v initial and then find the velocities of the alpha particle and the nucleus?
     
  11. Nov 3, 2016 #10

    mfb

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    The initial nucleus is at rest, the total momentum is zero. After the decay the alpha particle moves in one direction, the remaining nucleus moves in the opposite direction. What does that mean for the energy? The alpha particle has some kinetic energy, what about the nucleus?

    For (iii): you know the total energy released, you know the total momentum, and you know the particle masses, that should allow to find the energy of the alpha particle.
     
  12. Nov 3, 2016 #11
    Time 0: p = m v = m * 0 = 0.
    Time 1: pα = mα vα, pnucleus = mnucleus vnucleus.

    They both have some kinetic energy and that is why the energy in (i) does not all appear as kinetic energy Eα?

    Total energy released = 5.404455 MeV.
    Total momentum = 0.
    Particle masses = 205.929 421 u and 4.001 594 u.

    KE = p2 / 2 m? But p = 0.

    Update
    I found a formula here, Eα = E / (1 + mα / mnucleus) = 5.404 455 / (1 + 4.001 504 / 205.929 421) = 5.301 404 MeV. Is this correct? Thought It's not 5.30(2) as in the answer.
     
    Last edited: Nov 3, 2016
  13. Nov 3, 2016 #12

    mfb

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    Correct.

    [quote[Particle masses = 205.929 421 u and 4.001 594 u.[/quote]You are supposed to use 206 and 4 here.
    ##p \neq 0## for both the alpha particle and the produced nucleus.

    While that formula is right, I don't think you are supposed to use it without deriving it. It fits to the given answer, the agreement gets slightly better with the integer values for the masses.
     
  14. Nov 3, 2016 #13
    Yes, completely forgot about the integer values.

    But how to derive the momentum then (velocity is unknown)? Plus in KE = p2 / 2 m I think m only includes one mass of either the alpha particle or the produced nucleus.

    As I understand the original isotope Co-60 has one more neutron than Co-59.

    6027Co → 5927Co + 10n.

    I think it should be neutron emission.
     
    Last edited: Nov 3, 2016
  15. Nov 3, 2016 #14

    mfb

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    You have two unknowns (the two velocities) and two equations - one for conserved momentum, one for the sum of energies of the two particles.

    No, it is not neutron emission. The daughter nucleus is NOT Co-59.
     
  16. Nov 4, 2016 #15
    p = m v = 0
    p = m valpha + m vnucleus = 206 vnucleus + 4 valpha
    0 = 206 vnucleus + 4 valpha
    206 vnucleus = - 4 valpha

    I think this is unnecessary complication. The momentum, the velocities. More chances to do a mistake somewhere.

    We have Co-60 that becomes Co-59. I did that and used neutron emission. I can also use proton emission, but in that case the proton number will decrease by one: 6027Co → 5926Co + 11proton emission.
     
  17. Nov 4, 2016 #16

    mfb

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    It is not unnecessary, it is the only way to derive the equation you found.
    No it does not, and I wonder how often I have to repeat that.

    There is no Cobalt with 26 protons, and proton emission is not the right decay either.
     
  18. Nov 4, 2016 #17
    mD vD = ma va
    KED + KEa = 5.404 MeV

    KED = 1/2 mD vD2
    vD = ma va / mD
    KED = 1/2 mD ma2 va2 / mD2
    KEa = 1/2 ma va2
    KED = 1/2 ma va2 * ma / mD
    KED = KEa * ma / mD

    KEa * ma / mD + KEa = 5.404 MeV
    Take KEa out of the brackets: KEa (ma / mD + 1) = 5.404 MeV
    KEa = 5.404 MeV / (ma / mD + 1)

    I can't find any other type of decay that fits this problem (the nucleon number decreases by one). Maybe not all of them are listed here?
     
  19. Nov 4, 2016 #18

    mfb

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    Looks good with the energy calculation.
    It does not. I don't understand why you got stuck with that idea.

    You can also look up the actual decay mode if you want. It is a very common one that you should be familiar with.
     
  20. Nov 5, 2016 #19
    But it says that. We have Co-60 that decays to Co-59. Or I understand the text wrong?
     
  21. Nov 5, 2016 #20

    mfb

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    Where does it say that?
    The problem statement says that Co-59 does not decay, and it says that Co-60 decays, but not to what. That's up to you to figure out.

    It could also say that oxygen-16 is stable and Co-60 is not, that doesn't mean that Co-60 would decay to oxygen-16.
     
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