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Homework Help: Calculate the momentum of a 249 MeV pion

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    The question is from "Advanced Physics" by Adams and Allday. It is from section 9 "The Physics of Particles", "Practice Exam Questions". It is question 10. The level of this study is between school and University.

    I am uncertain about the answer to part c, ii (3rd meson) and iv.

    The whole part c question is given in case the context is relevant.

    c) The K+ is a meson with strangeness +1. One of its common decay modes is K+ → π+ + π0. Pions are not strange particles.

    i. Name the type of interaction responsible for the K+ → π+ + π0 decay.

    ii. Table 9.4 gives the properties of the relevant quarks. Deduce the possible quark content of the K+, π+ and π0
    • u, charge +2/3, baryon number 1/3, strangeness 0
    • d, charge -1/3, baryon number 1/3, strangeness 0
    • s, charge -1/3, baryon number 1/3, strangeness -1
    iii. The rest mass of a proton is 938 MeV/c2. The K+ rest mass = 0.53 proton masses, and the π+ rest mass = π0 rest mass = 0.15 proton masses. Assuming that the K+ is stationary when it decays, show that the total energy of each pion produced in the decay is 249 MeV.

    iv.Given that E2 = m02c4 + p2c2, calculate the momentum of each pion. Express your answer in units of MeV/c.

    2. Relevant equations
    Given in the question.

    3. The attempt at a solution
    ii. Only unsure about the possible quark content of the π0. Underlining is used below instead of barover to indicate anti-quark.

    The π0 is a meson with charge 0 and strangeness 0.
    Being a meson it is composed of a quark and an antiquark.
    Strangeness 0 so either no s or ss.
    Charge 0 so could be ss, dd. or uu.
    Quark content of π0: ss, dd. or uu.

    References confirmed dd. or uu but not ss. Is ss wrong and, if so, why?

    E2 = m02c4 + p2c2

    Using "natural" units and values given in the question this reduces to

    p = sqrt(249^2 - (0.15 * 938))
    = 249 MeV/c ct3sf.

    Is this correct? I am not familiar with using natural units so have some doubt about whether it's OK just to discard all the "c"s like that. If it is correct, is it just coincidence that the momentum is numerically the same as the energy?


  2. jcsd
  3. Feb 9, 2009 #2

    Tom Mattson

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    It's true that there are no [itex]s[/itex] quarks in the [itex]\pi^0[/itex] wavefunction. But I suppose that if all you know about the pions is their charge then it's not possible to rule them out. What you really need to know about the pions is that they form an isospin triplet and that the only possible quark content for [itex]\pi^+[/itex] is [itex]u\overline{d}[/itex].

    When I do the above calculation I get p=205 MeV/c.
  4. Feb 9, 2009 #3
    Thanks Tom :smile:

    OK with ss not being possible because of spin considerations but can you recheck the calculation? I just pasted sqrt(249^2 - (0.15 * 938)) into gcalctool 5.22.3 and got 248.71730941 again ...


  5. Feb 9, 2009 #4

    Tom Mattson

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    You forgot to square the mass inside the square root sign.
  6. Feb 9, 2009 #5
    Thanks again Tom :smile:

    Oops! I tried to work out whether sqrt(E^2 - m0) is E or whether it is just a coincidence that it gives the same number. Presumably coincidence because the erroneous calculation is dimensionally inconsistent.


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