Understanding Nuclear Stability: The Role of Binding Energy

In summary, the conversation discusses the concept of binding energy and its relation to nuclear stability. The speaker questions the use of the term "binding energy" and suggests using "mass/energy defect" instead. It is clarified that a higher binding energy per nucleon indicates a more stable nucleus. The attachment provided introduces the semi-empirical liquid drop model as an approximation for calculating nuclear stability.
  • #1
Mark Zhu
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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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  • #2
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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  • #3
Mark Zhu said:
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.

Mark Zhu said:
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.
 
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  • #4
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.
 

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  • #5
Mark Zhu said:
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.
 
  • #6
Yes. The semi-empirical liquid drop model. Thank you for your help. I understand it now. I confused change in energy with binding energy.
 
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Related to Understanding Nuclear Stability: The Role of Binding Energy

What is the binding energy of a nucleus?

The binding energy of a nucleus is the amount of energy required to completely separate all the nucleons (protons and neutrons) in a nucleus.

Why is the binding energy of a nucleus important?

The binding energy of a nucleus is important because it determines the stability of an atom. A higher binding energy means that the nucleus is more tightly bound and therefore more stable.

How is the binding energy of a nucleus calculated?

The binding energy of a nucleus is calculated using the famous equation E=mc^2, where E is the energy, m is the mass defect (difference between the mass of the nucleus and the sum of the masses of its individual nucleons), and c is the speed of light.

What factors affect the binding energy of a nucleus?

The binding energy of a nucleus is affected by the number of nucleons, the arrangement of nucleons (whether they are arranged in shells or not), and the strong nuclear force that holds the nucleus together.

How does the binding energy of a nucleus relate to nuclear reactions?

The binding energy of a nucleus plays a crucial role in nuclear reactions. When a nucleus undergoes a nuclear reaction, the difference in binding energy between the initial and final states is released as either energy or mass. This is the basis for nuclear power and nuclear weapons.

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