# Energy Release through Fission and SEMG

1. Nov 1, 2009

### TFM

1. The problem statement, all variables and given/known data

On neutron-capture induced fission, $$^{235}_{92}U$$ typically splits into two new “fission product" nuclei with masses in the ratio 1:1.4. These are born with the same proton to neutron ratio as the original uranium, so they have too many neutrons to be stable at their mass number and are highly radioactive. Energy is released in two stages:

First an intermediate or prompt release leading to radioactive fission products in their ground state;
And then a much slower release via the beta and gamma decays of the fission product nuclei, which continue until they become stable.

Use the semi-empirical mass equation to estimate the magnitudes of the energy release per fission in each of the two stages. You may take the final Z/A ratios from appropriate known stable nuclei.

2. Relevant equations

SEMG: $$M(Z,A) = ZM_p +(A-Z)M_n - a_vA + a_sA^{2/3} + a_C\frac{Z^2}{A^{1/3}} + a_A\frac{(A-2Z)^2}{A} + \left(\frac{-1}{\frac{0}{1}}\right)\frac{a_P}{A^{1/2}}$$

Energy-Mass Relationship: $$E = mc^2$$

3. The attempt at a solution

Okay I have worked out that the Fission goes like so:

Firstly:

$$^{235}_{92}U + ^1_0n_1 \rightarrow ^{236}_{92}U \rightarrow ^{98}_{36}Kr_{62} + ^{138}_{50}Sn_{88}$$

And then through a series of beta decays:

$$^{98}_{36}Kr_{62} \rightarrow ^{98}_{42}Mo_{57}$$

$$^{138}_{50}Kr_{88} \rightarrow ^{138}_{58}Mo_{80}$$

However I am slightly unsure how to work out the energy using the SEMG.

I think I have to put in E = mc^2 into SEMG, and rearrange for E:

$$E = c^2ZM_p + c^2(A-Z)M_n - c^2a_vA + c^2a_sA^{2/3} + c^2a_C\frac{Z^2}{A^{1/3}} + c^2a_A\frac{(A-2Z)^2}{A} + c^2\left(\frac{-1}{\frac{0}{1}}\right)\frac{a_P}{A^{1/2}}$$

Then work out the energy for each part and then work out the difference, ie
Work out the energy in U, Mo, Kr

Then work out the Energy relaesed from U compared to Kr and Sn, and then Kr to Mo and Sn to Ce?

Does this make Sense?