Binding Energy: Why Does Deuterium Have 1 MeV?

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SUMMARY

The binding energy of a deuterium nucleus is approximately 1 MeV, which is significantly lower than the potential energy minima of 100 MeV observed in nucleon interactions. This discrepancy arises because binding energy, defined as the energy required to separate nucleons, does not need to match the potential energy minima. The binding energy is influenced by the type of nucleons involved and their wave-function solutions derived from the Schrödinger equation. Additionally, the example of the hydrogen atom illustrates that binding energies can vary widely from potential energy minima.

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  • Explore the concept of binding energy in various isotopes, focusing on deuterium
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Students and professionals in nuclear physics, quantum mechanics enthusiasts, and anyone interested in understanding the fundamental principles of binding energy and nucleon interactions.

VishalChauhan
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The graph of potential energy of two nucleons shows a minima at 100 MeV,but the binding energy of a deuterium nucleus is close to just 1 MeV.

Since binding energy is the energy required to rip apart the nucleus ,hypothetically,should the two values not be same?
 
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Can you show the graph?
In general the binding energy of two nucleons depends on the nucleon type (eg nn and pp are not possible)... For having a bound state you need E<0.
 
Please see the attachment.
 

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I guess that this graph wants to describe the fact that the nucleons will prefer existing within the nuclei radius, rather than going too far away or falling on each other
 
VishalChauhan said:
Since binding energy is the energy required to rip apart the nucleus ,hypothetically,should the two values not be same?

No, they don't have to be the same. Look at the hydrogen atom, for instance. The electron's binding energy is 13.6 eV while the minimum energy of the coulomb potential is -∞.
 
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The minimum potential is just a lower bound for the binding energy - and you would need particles of "infinite" mass to approach this value. The real nucleons will form some (3-dimensional) wave-function similar to the electron, and have an energy corresponding to a solution of the Schrödinger equation (ignoring relativistic effects).
 

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