# Binding Energy of a Nucleus

Mark Zhu
Homework Statement:
See Attachment
Relevant Equations:
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This is an example from my textbook that I am having trouble understanding. So the binding energy of Beryllium-8 is positive 56.6 MeV, so it means the nuclide is stable, right? My textbook seems to use the reference of positive binding energy as being stable. And so that means alpha decay for Beryllium-8 is unfavorable, because that binding energy is negative. Then why does the textbook say, "From the standpoint of energy, there is no reason why a 8Be nucleus will not decay into two alpha particles" and that Beryllium-8 is unstable, yet it has a positive binding energy? Thank you.

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• Capture.PNG
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I don't think they should really be using the term 'binding energy' to refer to the difference in energy of the products and reactions of the decay [they ought to use mass/energy defect, or similar], since binding energy refers specifically to how much energy you need to supply to separate the system into its individual parts.

Anyway, what they're saying is that the two helium nuclei produced by the decay have a lower combined rest mass than the original nucleus, i.e. it's an energetically favourable process for the decay to proceed in the forward direction [i.e. "unstable w.r.t. ##\text{He}##"]. The 'lost' energy becomes kinetic in the decay products.

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Okay, so given that definition of binding energy, higher binding energy = more stable nucleus.

Yes, a nucleus with a higher binding energy per nucleon is generally more energetically stable with respect to its constituent protons and neutrons.

If so, then wouldn't a loss in the net binding energy in the decay mean the product (two alpha particles) has lower binding energy than 8Be and is thus less stable than the 8Be?

You're confusing what binding energy means, and that's probably the fault of the author's bad wording. If you have a nucleus of some variety, the binding energy is the amount of energy you need to supply to separate it into its constituent protons and neutrons, at infinity.

The two helium nuclei produced have a lower combined mass, and thus a lower combined energy, but that means they have a higher binding energy [i.e. you need to put in more energy to separate the nucleons!]. That means, they're more stable.

• Mark Zhu
Mark Zhu
If we are talking in terms of mass in mass-energy, then that makes sense. Because later there is another equation later introduced in the attachment below, and in that equation, the last term delta is positive for even-even nuclei and negative for odd-odd nuclei. As even-even nuclei are more stable, then higher binding energy = more stable.

Edit: sorry forgot to add the attachment.

#### Attachments

• etotheipi
If we are talking in terms of mass in mass-energy, then that makes sense. Because later there is another equation later introduced in the attachment below, and in that equation, the last term delta is positive for even-even nuclei and negative for odd-odd nuclei. As even-even nuclei are more stable, then higher binding energy = more stable.

Edit: sorry forgot to add the attachment.

Yes, the attachment is the so-called 'liquid drop model (or if you want to be fancy, 'semi-empirical mass formula') approximation.

Mark Zhu
Yes. The semi-empirical liquid drop model. Thank you for your help. I understand it now. I confused change in energy with binding energy.

• etotheipi