Quantum tunnelling/alpha decay

[tmv3v]

Im a uni student and a Phd student asked me in an alpha decay,
the helium nuclei is bounded by the strong nuclear force, how do the alpha particle overcome such a strong force and shoot out? And he told me it is because of quantum tunnelling.
To my understanding quantum tunnelling is an effect that due to Heisenberg's Uncertainty principle, a particle can have a distribution of energy, which there is a very low but finite probability that a particle can have enough energy to go over the energy barrier.
But I then don't understand why only certain isotopes of an element is radioactive since for those stable elements, quantum tunnelling will still take place. Someone could explain to me:D? Thanks!

 

[VantagePoint72]

The probability for quantum tunneling to occur is extremely sensitive to the relevant energies (the binding energy of the nucleus, the energy levels occupied by its components, etc.) Technically all elements besides hydrogen are indeed radioactive in that they have a non-zero probability of spontaneously breaking apart. However, the elements we classify as 'stable' have a such a low probability of decaying that their half lives are very long (often many orders of magnitude longer than the age of the universe).

 

[tmv3v]

Ah ok so everything is radioactive and having one more or less neutron will change the probability of quantum tunneling dramatically!
Thank you!

 

[VantagePoint72]

You're welcome. I would also add that the uncertainty principle explanation of tunneling isn't really correct. For one thing, uncertainty in momentum doesn't necessary imply uncertainty in energy. Also, you can have tunneling even when the energy of a system is precisely defined. The wave nature of matter is a better explanation: the wavefunction associated with a particle is a continuous, twice differentiable function (since it has to satisfy the Schrödinger equation) which means it can't abruptly stop if the external potential is well-defined everywhere. So, the particle's wavefunction can't just stop at the barrier, and since it continues through it (though falling off exponentially with distance) there is a non-zero probability of the particle being on the other side.