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'Heavy unstable nuclei are likely to emit helium because it has a high binding energy'. Why is this? If a nucleus has a high binding energy than I thought that meant it has a lower mass than its constituent parts (mass defect) ∴ wouldn't it be better to emit a lower BE nuclei as E=mc^2 so a higher mass nuclei would release more energy when emitted?

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You can think of the neutrons and protons being in energy levels inside the nucleus.
Those nucleons (neutrons and protons) in the highest energy levels do not have enough energy to escape from the nucleus due the attraction of the strong nuclear force.

The prescence of the strong nuclear force results in a a potential barrier ("wall") around the nucleus which the nucleons have to surmount in order to escape.
However if 2 neutrons and 2 protons form a cluster the binding energy of which is very large at about 28 MeV, that cluster has enough extra energy (the 28 MeV) to exist in higher energy levels where the (potential) barrier to them escaping is thinner.
Although the cluster has not enough energy to climb over the barrier, the barrier is thin enough for the cluster to have a relatively high probability of tunnelling (which quantum mechanics allows) through the barrier and escape as an alpha particle.
If the binding energy of 2n & 2p had been lower then not enough energy would have been available for the cluster to have a reasonable probability of tunneling through the barrier lower down where it was thicker.

There is more about the quantum mechanical tunneling process here but remember that to have a reasonably high probability od the alpha (cluster) escaping it must have enough energy to attempt the escape through a relatively thin part of the barrier.
That requires the cluster to have enough energy to rise that fr and that energy comes from the very high binding energy of the cluster (a Helium nucleus).

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wouldn't it be better to emit a lower BE nuclei as $E=mc^2$ so a higher mass nuclei would release more energy when emitted?

This is a confused statement. The relativistic mass $m$ in the formula $E=mc^2$ is completely irrelevant to nuclear models and energies. It depends on the velocity of the object under study, and nucleons in nuclei are at low energy levels , and even in a Bohr type model the velocities would not be relativistic ( i.e. approaching c, the velocity of light).

What is relevant and invariant to Lorenz transformations is the invariant or rest mass of a particle or a system of particles ( that nuclei are). It is the "length of the four vector representing a particle. The binding energy is the difference in MeV of the sum of the invariant masses composing a nucleus and the invariant mass of the nucleus itself.

A tightly bound smaller nucleus within a larger ensemble acts as a "particle" itself and , as the answer by Farcher explains has a probability of tunneling whole , depending on the nucleus and the quantum numbers, which may be larger than that for a beta or gamma decay.

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