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The strong force between two nucleons is attractive. Therefore, if we consider a nucleon on the surface of a nucleus (i.e. at radius $r=R$ where R is the radius of the nucleus), it is pulled only inwards by the strong forces provided by its neighbouring nucleons (all of which are inside the nucleus and pulls inwards). On the other hand, if we consider a nucleon at the interior of the nucleus (i.e. at radius $r<R$), it will be pulled inwards by some neighbouring nucleons as well as outwards by the other neighbouring nucleons.

But if what I described above were correct, the surface nucleons would seem to be more tightly bound than the inner nucleons. But in reality, it's exactly the opposite. But I cannot figure out what is wrong with this way of reasoning.

Solidification
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1 Answers1

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The argument is about the binding energy per nucleon in a nucleus, not about the forces involved.

The binding energy is the energy required to extract the nucleon from the nucleus and place it at infinity. In the semi-empirical "liquid drop" model, this is considerably larger for a nucleon in the centre of the nucleus, sitting deeper in the "potential well" than one that is at the surface, not least because it would have to be moved to the surface first...

Your argument is like suggesting that a piece of rock at the centre of the Earth is less bound than one at the surface. Clearly not.

To expand on this, imagine a potential well in the form of a "U". The force on a particle is given by the gradient of the potential. So this will be zero at the centre of the well, but this also marks the minimum of the potential and hence the biggest binding energy.

ProfRob
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