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Momentum Formula for Alpha Decay

  1. Apr 27, 2014 #1
    [Mentor's note: this thread does not use the standard homework-help template, because it was started in one of the non-homework forums. It was moved here because it had already gotten some help.]

    Hi everyone. This isn't a homework question, as I am just revising notes for my exams, and after extensive searching both online through search engines, and through browsing this forum, I did not seem to find any resource that can answer my question, so I just hope that somebody here can help me.

    So let's say we have an α-decay where:

    [itex]^{226}_{88}Ra[/itex] → [itex]^{222}_{86}Rn[/itex] + [itex]^{4}_{2}α[/itex] + [itex]Q[/itex]

    and I want to find the kinetic energy of both the daughter Nucleus [itex]^{222}_{86}Rn[/itex], and also of the [itex]^{4}_{2}α[/itex] particle.

    What information I have:


    Equation for momentum p (I am not even sure if THIS is correct, the lecture notes that we have at our University are extremely confusing, and often omits working details and derivations)

    [itex]p=\frac{1}{2M_{Ra}}\sqrt{(M_{Ra}-(M_{Rn}-m_{α})^2)(M_{Ra}-(M_{Rn}+m_{α})^2)}[/itex]

    and I have two equations for Kinetic Energies:

    [itex]Ek_{α}=\sqrt{p^2+m_{α}^2}-m_{α}[/itex]

    [itex]Ek_{Rn}=\sqrt{p^2+M_{Rn}^2}-M_{Rn}[/itex]

    My results:


    [itex]Ek_{α}≈4.8MeV[/itex]
    [itex]Ek_{Rn}≈0.09MeV[/itex]

    Could someone verify or explain me how the momentum p is Actually calculated for these particles? and then how to actually obtain those results (which are supposedly correct) for the Kinetic Energies?
     
    Last edited by a moderator: Apr 27, 2014
  2. jcsd
  3. Apr 27, 2014 #2

    Simon Bridge

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    Ownch: don't blame you.
    You basically go through the normal conservation of momentum and energy approach you learned when you studied collisions.
    In this case there is a possibility that the alpha-particle will be relativistic.
    You know those equations. It is a standard exercise for students to work from first principles - so you should have a go deriving the relations you have.

    In practice it often works out that the mass of the daughter is so much bigger than the mass of the alpha that you can treat the daughter nucleus as stationary in the lab-frame after the decay. Otherwise we may elect to do the computations in a convenient reference frame like "center of mass".
     
  4. Apr 27, 2014 #3
    I get completely wrong results for the momentum equation. To be frank, I just have no idea what value [itex]m_{α}[/itex] really represents and stuff like that.

    What I got is that p should be 189MeV/c, but what I get from my calculations is that p is 100GeV/c, so somewhere something must have gone really wrongly, and I don't have any good learning resources to make sense of this momentum formula...
     
  5. Apr 27, 2014 #4

    Simon Bridge

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    Please show your working and reasoning.

    Note:
    ##\small{m_\alpha = 3.727379240(82)}## GeV/c2 is the rest mass of the alpha-particle.

    ... then do not use that formula. Use conservation of energy and momentum instead.
    What education level are you doing this at?
     
  6. Apr 27, 2014 #5
    Right, so...

    [itex]p=\frac{1}{2\times210.4}\sqrt{(210.4-(206.6-3.72)^2)(210.4-(206.6+3.72)^2)}[/itex]

    ... and the result I get is much too large than what I'm supposed to get.
    Because
    I get 100GeV/c.
    I SHOULD get 189MeV/c.

    We were not taught how to use these methods. Our lecturer just gave us these relationships, and I probably know much less about this than I should do...

    Oh.. this is first year material...
     
    Last edited: Apr 27, 2014
  7. Apr 27, 2014 #6

    Simon Bridge

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    Conservation of energy and momentum are part of the secondary school syllabus in most countries - physics students normally have experience with such calculations by the time they are 17yo.

    1st year at... University?
    And you don't know how to use conservation laws?

    Your equation does not look right to me - did you forget to square some of the terms? Or maybe the squares shouldn't be there? Check you have it right.

    Using your figures I get:
    > sqrt( ( 210.4 - (206.6 - 3.72)^2 ) * ( 210.4 - (206.6 + 3.72)^2 ) ) /(2*210.4)
    ans = 100.90GeV

    Leaving off the squares I get:
    > sqrt( ( 210.4 - (206.6 - 3.72) ) * ( 210.4 - (206.6 + 3.72) ) ) /(2*210.4)
    ans = 0.0018432GeV

    You are probably just using the wrong equation.

    You can look up the radium decay modes:
    http://en.wikipedia.org/wiki/Radium (see table in the right-hand sidebar)
    http://en.wikipedia.org/wiki/Decay_energy (how the calculation is done.)
     
    Last edited: Apr 27, 2014
  8. Apr 27, 2014 #7
    Hmm, I'm not sure how to interpret your response. I understand how conservation laws work, I just do not know how, or why they are applicable to finding the result.

    I need Q and that energy equals the kinetic energy of the two particles added together, I'm aware of that result. However, I have no idea how I can find out the kinetic energy of each of the particles separately, as the alpha kinetic energy IS slightly less than Q, and also how I can calculate momentum p, which is the main focus of the question.

    Also, sorry that my problem statement might have been unclear, nuclear physics is really not my strength.
     
  9. Apr 27, 2014 #8
    That would not surprise me at all, our lecture notes are mind-boggingly confusing, because "worked" examples state only the fundamental equation and the end result, and nothing inbetween. Do you know of any way that I can use to work out the momentum of a particle in a decay process? (Assuming the mother nucleus is at rest)

    This might help: this is what I need to learn to use:
    v7vcs0.png
    (this is taken directly from my lecture notes in an unaltered form)
     
    Last edited: Apr 27, 2014
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