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U-238 to Th-234 decay

  1. Jun 27, 2012 #1
    Hello, I need some help understanding this.

    http://atom.kaeri.re.kr/cgi-bin/decay?U-238 A

    it says that 79% of the time U-238 goes to the ground state of Th-234 emitting a 4.198 MeV alpha ray. But the difference between the ground state of U-238 and Th-234 is more than 4.198 MeV. What happened to the rest of the decay energy? Does it have anything to do with uncertainties? Thanks.
     
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  3. Jun 27, 2012 #2

    Bill_K

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    Wikipedia gives the decay energy as 4.270 MeV. Is that any better?
     
  4. Jun 27, 2012 #3
    Thank you for replying Bill_K.
    The website also gives the decay energy as 4.270 MeV. I wanted to know what happened to the remaining 72 keV.
     
  5. Jun 27, 2012 #4

    mathman

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  6. Jun 27, 2012 #5

    Bill_K

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    The missing 72 keV is the recoil energy carried off by the Th-234 daughter nucleus.
     
  7. Jul 2, 2012 #6
    Newton's third law of motion, for every action there is an equal and opposite reaction.
    Momentum must be conserved, so if you take the system, the momentum change from the decay must be 0.
    So, when you split a He off a U238, you have the total energy of the mass difference, which must be divided between the He and the Th234 in such a way that the momentum is still 0. So, the He, being 58.5 times lighter, must have much higher velocity imparted, but the Th must also have some.
     
  8. Jul 2, 2012 #7

    Bill_K

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    ... and the reaction products share the energy in inverse proportion to their masses. Thus the Th-234 gets 4.270 MeV / 58.5 = 73 keV
     
  9. Jul 4, 2012 #8
    Makes sense. Thank you.

    One last thing. For the gamma rays, do the intensities represent the probability of getting that gamma ray energy? They appear to be very low since U-238 does not decay to the ground state of Th-234 ~21% of the time. I would think that Th-234 needs to be in its ground state before it can decay to Pa-234 but I might be wrong.
     
  10. Jul 5, 2012 #9
    If one of the components is excited, then that excitation energy is not kinetic at the reaction time, so the actual energy of the decay is used.

    Later, the excited state will decay to the ground state, and give off a gamma ray in the process. Just like the alpha decay, momentum must be preserved, so the Gamma ray will have most of the energy, but the nucleus must have exactly the same momentum in the opposite direction as the gamma ray.
     
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