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What is mass of particle X before the decay?

  1. Mar 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Professor X, a nuclear physicist who works at the MSU FRIB facility, has designed a new particle detector called The da Vinci Decoder. Using this detector, she has discovered a new particle dubbed particle X that violates lepton number conservation. A stationary X is observed to decay spontaneously into an alpha particle (α) plus a proton (p), electron (e), and a neutrino (ν):

    X − −−→ α + p + e + ν .

    The mass of an alpha particle is 4.00260u (this is the rest mass, which accounts for binding energy), the mass of a proton is 1.00727u, and the mass of an electron is 0.000 55 u. Lastly, the mass of a neutrino is less than one billionth of an atomic mass unit – in other words you can neglect its mass.

    (a) After the decay, the alpha, proton, electron, and neutrino, are all mov- ing in different directions, with a total kinetic energy Ktot = 9.819 × 10−13 J = 6.128 MeV. What is the mass of the X particle?



    2. Relevant equations

    k=1/2mv^2
    Erest=mc^2
    p= mv/(sqrt(1-(v/c)^2))


    3. The attempt at a solution

    Add up the velocities of the particles

    6.128MeV=1/2(4.0026)v
    alpha particle v= 3.062m/s

    6.128MeV=1/2(1.00727)v
    proton particle v= 12.168m/s

    6.128MeV=1/2(.00055)v
    electron particle v= 22283.636m/s

    v1+v2+v3 = 22298.866m/s

    Now use k=1/2mv^2 to find mass of particle X

    6.128 MeV = (1/2)m(22298.866m/s)^2

    2.465E-8 kg

    This answer is obviously way off, it should be a little more than the total masses given because of the binding energy in particle X.

    Please help put me on the right track.











     
  2. jcsd
  3. Mar 9, 2015 #2
    The reaction is

    X --> alpha + p + e (+ v, but we neglect) + Kinetic Energy

    If this is the case, then X should have the mass of the alpha+p+e+KE, slightly more than the mass of the constituents.

    For a contrasting situation (where a decay mode is forbidden because it weighs less than the would-be products), see here --> http://hyperphysics.phy-astr.gsu.edu/hbase/particles/deuteron.html
     
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