(adsbygoogle = window.adsbygoogle || []).push({}); 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!!

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# Homework Help: Semi empirical mass formula

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