# If greater the binding energy per nucleon, more the stability, then why is Uranium (7.7MeV/nucleon) unstable?

My book says that if the binding energy per nucleon is high, then the nucleus is stable, which is why Fe (iron), with 8.8 MeV, is most stable. Why then is Uranium unstable and subject to nuclear decay?

• Because splitting it up yields more binding energy per nucleon... Mar 5 at 19:01

## 2 Answers

The $${^{280}}{U}{}$$ is far from the line of stability and can decay get closer to the line of stability. Consider the nucleus $${^{280}}{U}$$, the alpha decay reaction is given by $${^{280}}{U}\rightarrow {^{226}}{\text{Th}}+\alpha$$ If you go about finding $$Q$$-value for the reaction then you will find that $$Q=[M({^{280}}{U})-M({^{226}}{\text{Th}})-M({^{4}}{\text{He}})]c^2=6.0\ \text{MeV}$$ So that alpha decay is allowed, because $$Q>0$$.

These heavy nuclei have increasingly stronger Coulomb repulsion as protons are added. The expulsion of two protons (along with two neutrons) in the form of an alpha particle may decrease this column energy and make the resulting nucleus more stable.

• Oh yeah, I forgot about the stability belt. Thanks! Mar 6 at 2:54

the nuclear binding energy curve results from a tug-of-war between the strong nuclear force, which makes protons and neutrons stick together, and the electrostatic repulsion between protons, which makes them want badly to fly apart. Because the strong force is extremely short-ranged, the nucleons need to be nearest neighbors for that attractive force to act. On the other hand, the electrostatic repulsion is long-ranged, so that every proton in the assemblage is actively repelling every other proton present.

As you make nuclei bigger than iron, the repulsion effect grows progressively stronger, the resulting nucleus progressively less stable, and thereby more susceptible to either fissioning spontaneously or undergoing some other type of decay process.

• This was a great explanation! Thanks for taking the time. Mar 6 at 3:44