If during a nuclear fission U(235) decomposes into Xe(140), Sr(94) and n(1), how is it possible that the original U(235) has bigger mass than the three resulting nuclei together?
Should it not be the case that the U(235) has a deeper bonding energy so the lower mass; this way its binding energy is released during the fission and the resulting elements have smaller mass defect?
[Physics] Nuclear fission problem
nuclear-physics
Related Solutions
To understand binding energy and mass defects in nuclei, it helps to understand where the mass of the proton comes from.
The news about the recent Higgs discovery emphasizes that the Higgs mechanism gives mass to elementary particles. This is true for electrons and for quarks which are elementary particles (as far as we now know), but it is not true for protons or neutrons or for nuclei. For example, a proton has a mass of approximately $938 \frac{\mathrm{MeV}}{c^2}$, of which the rest mass of its three valence quarks only contributes about $11\frac{\mathrm{MeV}}{c^2}$; much of the remainder can be attributed to the gluons' quantum chromodynamics binding energy. (The gluons themselves have zero rest mass.) So most of the "energy" from the rest mass energy of the universe is actually binding energy of the quarks inside nucleons.
When nucleons bind together to create nuclei it is the "leakage" of this quark/gluon binding energy between the nucleons that determines the overall binding energy of the nucleus. As you state, the electrical repulsion between the protons will tend to decrease this binding energy.
So, I don't think that it is possible to come up with a simple geometrical model to explain the binding energy of nuclei the way you are attempting with your $\left(1\right)$ through $\left(15\right)$ rules. For example, your rules do not account for the varying ratios of neutrons to protons in atomic nuclei. It is possible to have the same total number of nucleons as $\sideset{^{56}}{}{\text{Fe}}$ and the binding energies will be quite different the further you move away from $\sideset{^{56}}{}{\text{Fe}}$ and the more unstable the isotope will be.
To really understand the binding energy of nuclei it would be necessary to fully solve the many body quantum mechanical nucleus problem. This cannot be done exactly but it can be approached through many approximate and numerical calculations. In the 1930's, Bohr did come up with the Liquid Drop model that can give approximations to the binding energy of nuclei, but it does fail to account for the binding energies at the magic numbers where quantum mechanical filled shells make a significant difference. However, the simple model you are talking about will be incapable of making meaningful predictions.
EDIT: The original poster clarified that the sign of the binding energy seems to be confusing. Hopefully this picture will help:
$\hspace{75px}$
.
This graph shows how the potential energy of the neutron and proton that makes up a deuterium nucleus varies as the distance between the neutron and proton changes. The zero value on the vertical axis represents the potential energy when the neutron and proton are far from each other. So when the neutron and proton are bound in a deuteron, the average potential energy will be negative which is why the binding energy per nucleon is a negative number - that is we can get fusion energy by taking the separate neutron and proton and combining them into a deuteron. Note that the binding energy per nucleon of deuterium is $-1.1 \, \mathrm{MeV}$ and how that fits comfortably in the dip of this potential energy curve.
The statement that $\sideset{^{56}}{}{\text{Fe}}$ has the highest binding energy per nucleon means that lighter nuclei fusing towards $\text{Fe}$ will generate energy and heavier elements fissioning towards $\text{Fe}$ will generate energy because the $\text{Fe}$ ground state has the most negative binding energy per nucleon. Hope that makes it clear(er).
By the way, this image is from a very helpful article which should also be helpful for understanding this issue.
They don't. Here's a figure from Wikipedia:
Typically there's daughter with mass around 95, a daughter with mass around 140, and two or three extra free neutrons. In discussion of environmental contamination after nuclear accidents, you hear a lot about iodine-133 and strontium-90, because they are relatively long-lived and biologically active. Iodine-133 lives for about a week and accumulates in the thyroid; strontium-90 lives for about 30 years and can replace calcium in bones.
There are several heavy isotopes which can spontaneously fission; the big ones are uranium, plutonium, and californium.
Best Answer
You can see that U(235) will have more mass than the particulate elements that are blown off if you explicitly write the equations for energy conservation.
If you consider that Xe(140) Sr(94) and n(1) have kinetic energy when they are blown off then you can write the following equation for the initial and final states:
$$M_\text Uc^2 =\frac{m_\text{Xe}c^2}{\sqrt{1-\frac{v_\text{Xe}^2}{c^2}}} +\frac{m_\text{Sr}c^2}{\sqrt{1-\frac{v_\text{Sr}^2}{c^2}}} +\frac{m_\text{n}c^2}{\sqrt{1-\frac{v_\text{n}^2}{c^2}}}$$
where the M's correspond to the rest mass of each object. Note that (In case you don't know) $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$ is the total energy in relativistic dynamics. (I haven't written it this way for uranium since it is assumed that its stationary and $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=1$ for an object with zero velocity.)
So if you analyse the equation I wrote above, you'll notice that it is necessary that $M_U>m_{Xe}+m_{n}+m_{Sr}$ since the $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is always greater than one.