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Homework Help: Semi empirical mass formula

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data
    For a nucleus described by the semi-empricial mass formula show that neglecting the pairing and the sassymmetry terms the energy released in fission into two ffragments is a macximum for equal divison of charge and mass. For what value of Z^2/A does this division become favourable energetically? How might the inclusion of the asymmetry term change these results?? (discuss qualitatively)?


    2. Relevant equations
    The semi empirical mass formula is
    [tex] E_{B} = a_{V} A - a_{S} A^{2/3} - a_{C} \frac{Z(Z-1)}{A^{1/3}} - a_{A} \frac{(A - 2Z)^{2}}{A} + \delta(A,Z)[/tex]

    since we are ignoring the pairing and symmtery terms, we ned not care about the last two terms

    3. The attempt at a solution
    For a fission reaction like

    [tex] ^A_{Z} X \rightarrow ^{A-\alpha}_{Z-\zeta} Y_{1} + ^{\alpha}_{\zeta}Y_{2} [/tex]

    the eenrgy released in this reaction Q is given by the differenc ein teh binding energies of the reactant and the product

    [tex] Q = B_{0} -B_{1}-B_{2} [/tex]
    [tex] B_{0} = a_{V} A - a_{S} A^{2/3} - a_{C} \frac{Z(Z-1)}{A^{1/3}} [/tex]

    [tex] B_{1} = a_{V} (A-\alpha} - a_{S} (A-\alpha)^{2/3} - a_{C} \frac{(Z-\zeta)(Z-\zeta-1)}{(A-\alpha)^{1/3}} [/tex]

    [tex] B_{2} = a_{V} \alpha - a_{S} \alpha^{2/3} - a_{C} \frac{\zeta(\zeta-1)}{\alpha^{1/3}} [/tex]

    Then our Q valu becomes

    [tex] Q = B_{0} -B_{1}-B_{2} = -a_{S} \left[ A^{2/3} - (A-\alpha)^{2/3} -\alpha^{2/3}\right] - a_{C} \left[ \frac{Z(Z-1)}{A^{1/3}} - \frac{(Z-\zeta)(Z-\zeta-1)}{(A-\alpha)^{1/3}} - \frac{\zeta(\zeta -1)}{\alpha^{1/3}} \right][/tex]

    Now i need to show somehow that if [itex] A-\alpha = \alpha [/itex] and [itex] Z-\zeta=\zeta [/itex], then this value of Q is maximized. So thjis is what im wondering about... should i differentiate Q wrt zeta or alpha?? Neither of them are constant and in fact one depends on the other...


    TO show the rato of Z^2/A to make the divsion energetically favourable


    welll if this was to be enrgetically favourable hten the Q value would have to be >0. And for an equidivision of both charge, Z and mass, A, and Q = 0

    [tex] -a_{S}A^{2/3} + a_{C} \frac{Z(Z-1)}{A^{1/3}} =0[/tex]
    [tex] \frac{a_{S}}{a_{C}} = \frac{Z(Z-1)}{A}[/tex]

    for an approxiamtion lie.. are we supposed ot assume that Z>>1?? After all the SEMF only works well for large value of A and Z...

    Inclusion of the asymmetry term

    Now the ratio of Z^2/A would be obviously skewed by the fact taht the Neutron number is playing a role. For N=Z the asymmtery term would not play a role. However N > Z the value of Q becomes more favourable?? Ill add the reasoning for this guess later on

    thanks for any help!!
     
  2. jcsd
  3. Apr 10, 2007 #2
  4. Apr 11, 2007 #3

    malawi_glenn

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    Does this mean that you solved the problem? :smile:
     
  5. Apr 11, 2007 #4
    unfortunately ... no :confused:
     
  6. Apr 11, 2007 #5

    malawi_glenn

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    Well a hint: A = N + Z, that makes the differentiation of Q easier, if you now want to proove that max Q is obtained when fission is done into equal masses and charges.
     
  7. Apr 11, 2007 #6
    iwas wondering if there is a 'cleaner' way of doing this problem??
     
  8. Apr 11, 2007 #7

    malawi_glenn

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    "So thjis is what im wondering about... should i differentiate Q wrt zeta or alpha?? Neither of them are constant and in fact one depends on the other...
    "

    Well i would do the same thing you came up with, then express A in terms of N and Z, then differentiate.
     
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