1
$\begingroup$

Consider two nuclei A and B. Let the binding energy of A is less than the binding energy of B (i.e., A is unstable compared to B) and B is known to be radioactive. Does it mean that A also has to be radioactive? If comparing binding energies is not sufficient criterion to determine whether a nucleus will be radioactive or not, what is?

$\endgroup$
2
  • 1
    $\begingroup$ Set A as helium-4 (with binding energy 28 MeV) and B as uranium-235 (with binding energy 1.8 GeV) so that the binding energy of A is less than the binding energy of B, and B is radioactive; here A is not radioactive. I imagine that you didn't mean your language to include this case, but as you've written the question then this situation needs to be included - so maybe you should think a bit more carefully about what, exactly, you want to ask? $\endgroup$ Jun 7, 2018 at 15:53
  • $\begingroup$ @EmilioPisanty Comparing binding energy is not enough to conclude whether a nucleus will be radioactive or not? If not binding energy, what is/are the criteria for a nucleus to be radioactive? $\endgroup$ Jun 7, 2018 at 16:02

2 Answers 2

4
$\begingroup$

A radioactive process is allowed if the rest masses of the particles in the final states add up to less than the rest mass of the particle in the initial state, as long as the decay obeys some conservation laws. Conservation of energy, momentum, and angular momentum determine the dynamics of the decay. But conservation of baryon number, lepton number, and electric charge determine whether the decay can take place at all.

The binding energy is relevant to this calculation, but isn't much use on its own in determining radioactivity because the proton and neutron masses are different. Consider the decay of the free neutron,

$$ \rm n \to p + e^- + \bar\nu_e $$

The binding energy of the proton in the final state is zero. The binding energy of the neutron in the initial state is also zero, because the neutron is not a molecule that somehow contains the other particles.

For a better counterexample to your proposal that less-tightly-bound nuclei should be radioactive is a more-tightly-bound nucleus is also radioactive, consider the unusual case of tantalum. The isotope $\rm^{180}Ta$ is unstable against $\beta^\pm$ decay to $\rm^{180}Hf$ or $\rm^{180}W$, with a lifetime of only eight hours. But there's a high-spin excited state in $\rm^{180}Ta$ which has, by definition, a smaller binding energy than the ground state, but which has never been observed to undergo either gamma emission or $\beta^\pm$ decay. The decay of $\rm^{180m}Ta$ would be allowed by all the conservation laws I listed above, but because the gamma ray emission would have to shed an angular momentum of $8\hbar$, its lifetime is at least 100,000 times longer than the present age of the universe.

An even better counterexample to your proposal is the beta decay of tritium. There the energy released in the decay is smaller than the neutron-proton mass difference.

In general you'll have better luck figuring out nuclear decays if you use the mass excess instead of the binding energy.

$\endgroup$
4
$\begingroup$

In a lot of ways this is a strange question.

The short answer is no.

To explain, the fact that even radioactive nuclei have positive binding energy tells you that they are not going to spontaneously come completely apart (i.e. go from nucleus to a collection is disparate protons and neutrons).

But radioactive decays doesn't involve the dissolution of nuclei it involves

  • rearrangement of the nuclei (pure gamma decays)
  • conversion of a nucleon to a different type (beta decays, electron capture, ...)
  • or ejection of a subset of the nucleons (alpha decay, neutron emission, proton emission).

So the relationship you should be asking about is the relative energy states of the nucleus and the various possible outcomes of nuclear processes, not between the nucleus and a state of full dispersion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.