How Does Fission Energy Maximize with Equal Division of Charge and Mass?

[stunner5000pt]

Homework Statement

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)?

Homework Equations

The semi empirical mass formula is
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)

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

The Attempt at a Solution

For a fission reaction like

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

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

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

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

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

Then our Q valu becomes

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]

Now i need to show somehow that if A-\alpha = \alpha and Z-\zeta=\zeta, then this value of Q is maximized. So thjis is what I am 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

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

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!

 

[stunner5000pt]

bump

 

[malawi_glenn]

bump

Does this mean that you solved the problem?

 

[stunner5000pt]

Does this mean that you solved the problem?

unfortunately ... no

 

[malawi_glenn]

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.

 

[stunner5000pt]

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.

iwas wondering if there is a 'cleaner' way of doing this problem??

 

[malawi_glenn]

"So thjis is what I am 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.