Total Angular Momentum and Binding Energy

In summary, according to Hund's third rule, the highest J value would have the lowest energy (in other words, the highest binding energy). However, this does not always hold true because there is an energetic interaction between the spin magnetic moment and the orbital magnetic moment. This interaction can cause the J=1/2 state (spin -1/2) to have a higher binding energy than the J=3/2 (spin 1/2) state.
  • #1
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Why, for states with angular momentum l >0, do states with smaller total angular momentum J have a higher binding energy? For example, why does the 2p1/2 state have a higher binding energy than the 2p3/2 state? If the 2p orbital is filled (2p6), wouldn't Hund's third rule indicate that the highest J value would have the lowest energy (in other words, the highest binding energy)?
 
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  • #2
There is an energetic interaction (typically amounting to a few eV for core levels) between the spin magnetic moment and the orbital magnetic moment. If an orbital is half-filled, the two magnetic moments prefer to line up (have lowest energy). If the orbital is filled (as in the 2p6 configuration) there is only one possible state. Thus in a photoelectron transition:

initial state: 2p6
final states: either 2p5 with spin up, or 2p5 with spin down (non-denegerate).
 
  • #3
gadong said:
There is an energetic interaction (typically amounting to a few eV for core levels) between the spin magnetic moment and the orbital magnetic moment. If an orbital is half-filled, the two magnetic moments prefer to line up (have lowest energy). If the orbital is filled (as in the 2p6 configuration) there is only one possible state. Thus in a photoelectron transition:

initial state: 2p6
final states: either 2p5 with spin up, or 2p5 with spin down (non-denegerate).

So why would the j=1/2 state (spin -1/2) have a higher binding energy than the j=3/2 (spin 1/2)?
 
  • #4
Sorry, my mistake - lowest energy [final] state is when the spin and orbital moments are in opposite directions (j = 1/2).
 
  • #5
gadong said:
Sorry, my mistake - lowest energy [final] state is when the spin and orbital moments are in opposite directions (j = 1/2).

But from Hund's 3rd rule (http://en.wikipedia.org/wiki/Hund's_rules), would the higher J (3/2) have lower energy (higher binding energy)?
 
  • #6

What is total angular momentum?

Total angular momentum is a physical quantity that describes the rotational motion of a system. It is the sum of the individual angular momenta of all the particles within the system.

How is total angular momentum calculated?

Total angular momentum is calculated by multiplying the moment of inertia (a measure of an object's resistance to rotational motion) by the angular velocity (the rate at which an object rotates). It is represented by the symbol L.

What is binding energy?

Binding energy is the amount of energy required to separate a system of particles into individual particles. In the context of total angular momentum, binding energy refers to the energy required to break apart a rotating system into its individual components.

How is binding energy related to total angular momentum?

Binding energy and total angular momentum are related through the conservation of angular momentum. As the total angular momentum of a system decreases, its binding energy increases. This means that a system with a large binding energy will have a lower total angular momentum, and vice versa.

What is the importance of studying total angular momentum and binding energy?

Understanding total angular momentum and binding energy is crucial in many fields of physics, such as astrophysics, nuclear physics, and quantum mechanics. These concepts help us understand the behavior of rotating systems and the stability of bound states, and they have practical applications in industries such as energy production and materials science.

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