As the atomic number $Z$ increases, is the Hartree-Fock approximation getting better and better, or worse and worse?

kokomocutie88r1
2022-07-21
Answered

As the atomic number $Z$ increases, is the Hartree-Fock approximation getting better and better, or worse and worse?

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Killaninl2

Answered 2022-07-22
Author has **20** answers

Hartree-Fock is less accurate for heavier atoms. There are more configurations and these are closer in energy to each other, so multiconfigurational Hartree-Fock (MCHF) is required. Relativistic effects become more important so the Dirac equation must be solved, in case you are referring to Schrödinger based Hartree-Fock.

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Nuclear Mass and stability. plotting the $Ag$$-109$ Mass parabola putting the Atomic number on the $x-$axis and the Mass of the nuclei on the $y-$axis (calculated as $M=Z{m}_{p}+(A-Z){m}_{n}-B$).

Noticed that even if the $Ag-109(Z=47)$ is the stable element the minimum of the parabola is on $Z=48$. Is that possible?

Noticed that even if the $Ag-109(Z=47)$ is the stable element the minimum of the parabola is on $Z=48$. Is that possible?

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Calculating energy released in nuclear fission

Consider the neutron induced fission $\text{U-235}+n\to \cdots \to \text{La-139}+\text{Mo-95}+2n$, where $\dots $ denotes intermediate decay steps.

I want to calculate the released energy from this fission. One way would be to calculate the difference of the binding-energies ($B$):

$\mathrm{\Delta}E=B(139,57)+B(95,42)-B(235,92)\approx 202,3\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

(Btw. I didn't use the binding energies from a semi-empirical binding energy formula but calculated them directly via mass defect).

Another way is:

$\mathrm{\Delta}E=(m(\text{U-235})+{m}_{\text{Neutron}}-m(\text{Mo-95})-m(\text{La-139})-2{m}_{\text{Neutron}}){c}^{2}\approx 211,3\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

Which one gives the correct result? Why?

You notice that $57+42\ne 92$, if that was the case, it would be equal, but I don't clearly see where the difference physically comes from and what to add or subtract (and why) from to the first or from the second term to get the other result. How to make this clear?

A slightly other point of view: What different questions do both calculations answer?

Consider the neutron induced fission $\text{U-235}+n\to \cdots \to \text{La-139}+\text{Mo-95}+2n$, where $\dots $ denotes intermediate decay steps.

I want to calculate the released energy from this fission. One way would be to calculate the difference of the binding-energies ($B$):

$\mathrm{\Delta}E=B(139,57)+B(95,42)-B(235,92)\approx 202,3\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

(Btw. I didn't use the binding energies from a semi-empirical binding energy formula but calculated them directly via mass defect).

Another way is:

$\mathrm{\Delta}E=(m(\text{U-235})+{m}_{\text{Neutron}}-m(\text{Mo-95})-m(\text{La-139})-2{m}_{\text{Neutron}}){c}^{2}\approx 211,3\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

Which one gives the correct result? Why?

You notice that $57+42\ne 92$, if that was the case, it would be equal, but I don't clearly see where the difference physically comes from and what to add or subtract (and why) from to the first or from the second term to get the other result. How to make this clear?

A slightly other point of view: What different questions do both calculations answer?

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