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Uranium 235 Fission Question

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Another mode in which uranium 235 can undergo fission is U + n -> Te + Zr +n. Calculate the energy change when 1g of uranium 235 undergoes fission in this way. The masses are U=235.04u. n=1.0087u. Te=134.92u. Zr=99.92u.


    2. Relevant equations



    3. The attempt at a solution

    The mass of the left side is 236.0487u
    The mass of the right side is 235.8487u

    delta M=-0.2u

    E=mc^2
    =-0.2u x 9x106 J/kg x(1.66x10^-27kg/u)
    =-.2.988x10^-21J/u ? (Please help me with units here!!!)

    E1g=-2.988x10^-21J/u x 1g
    (Units have failed me here, please tell me where I went wrong...)
     
  2. jcsd
  3. Nov 5, 2011 #2

    Delphi51

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    Homework Helper

    E=mc^2
    =-0.2u(1.66x10^-27kg/u)*(3x10^8m/s)²
    = 2.99 x 10^-10 Joules per Uranium nucleus.
    How many Uranium nucleii fission?
     
  4. Nov 5, 2011 #3
    For the speed of light, I used 9 x 10^16 J/kg. Can you show the unit analysis for just squaring 3x10^8 m/s ?

    To answer your question

    1 gram of U-235, the mass of 1 U-235 nucleus is 235.04u given, so 1gram/235.04u=2.5621772x10^21. Then2.99 x 10^-10 x 2.5621772x10^21 = 7.66091009 × 1011 Joules???
     
  5. Nov 6, 2011 #4

    Delphi51

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    Homework Helper

    Pick any energy formula, say E = mgh. So Joule = kg*m/s²*m.
    and Joule/kg = m/s²*m = m²/s²

    In my calc, I have
    E=mc^2
    =-0.2u(1.66x10^-27kg/u)*(3x10^8m/s)²
    The u's cancel out and the answer is in kg*m²/s²
    which is Joules as seen in E = ½mv² or E = mc².
    Your answer looks good now. You just wrote the exponent 16 as 6 in the first calc and there is some confusion about where the decimal place is.
     
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