Could Copper's 29th Ionization Energy Be Used to Create Uranium-235?

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Discussion Overview

The discussion revolves around the potential use of copper's 29th ionization energy in relation to uranium-235, exploring the energy calculations and theoretical implications of such a comparison. Participants examine the differences between ionization energy and nuclear energy production, as well as the feasibility of creating a copper ion through nuclear reactions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates copper's 29th ionization energy and compares it to the energy produced by uranium-235 fission, suggesting a theoretical connection.
  • Another participant agrees with the energy calculation but questions the rationale behind comparing an energy-consuming process (ionization) with an energy-producing one (nuclear fission).
  • It is noted that nuclear reactions involve different energy scales compared to electronic ionization, with nuclear energy levels typically in the MeV range.
  • Participants discuss the practical challenges of obtaining copper-29 ions compared to uranium-235, highlighting the rarity of such ions in nature.
  • One participant mentions the possibility of creating bare atoms through advanced techniques like photoionization and collider experiments, though they express uncertainty about the details.

Areas of Agreement / Disagreement

Participants generally agree on the calculations related to ionization energy but disagree on the relevance of comparing ionization processes with nuclear reactions. The discussion remains unresolved regarding the practical implications of these comparisons.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the feasibility of creating copper-29 ions through nuclear reactions and the lack of consensus on the practical applications of the calculations presented.

Go Ved
So I was playing with periodic table and discovered Cu's 29th ionization energy(cu with no electron) is 1116105 KJ/mole .

Then i searched on google to convert to MeV which is 6.9661798e+21 , then I thought that one uranium 235 produce 200mev
so to make that you would need around 3.5e+19 atoms(0.00005812022 mole) which means 0.01366080898g of it(practically more than it)

I'm just wowed !
 
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So the 6.966 E+21 MeV is for one mole=6.02E+23 atoms. The Bohr atom formula has a ## Z^2 ## in the numerator, and with ## Z=1 ## for hydrogen, you get ## E=13.6 \, eV ##. That would make ## E=(29^2)(13.6)=11,440 \, eV ##, and for one mole it would be ## (N_A )(11,440)=(6.02 E+23)(11,440)=6.90 \,E+27 \, eV=6.90 \,E+21 \, MeV ##, so I agree with your calculation. ## \\ ## Welcome to Physics Forums! :) :)
 
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Go Ved said:
then I thought that one uranium 235 produce 200mev
so to make that you would need around 3.5e+19 atoms(0.00005812022 mole) which means 0.01366080898g of it(practically more than it)
I am missing your point here, while your calculation is correct I cannot get your idea why you are comparing the process which consumes energy (to ionize Cu 29th) and the process which produce energy (from 235 U). Nevertheless, the reason there is such big difference is that in the second part you are considering nuclear reaction which only has respect on the nuclear structure and not on the electronic one as is considered in the ionization part. The nuclear energy levels have typical energy spacing in the order of MeV.
Btw it's 200 MeV not meV, I know it's typo but it may cause people not familiar with nuclear reaction to understand that the meV is correct.
 
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blue_leaf77 said:
I am missing your point here, while your calculation is correct I cannot get your idea why you are comparing the process which consumes energy (to ionize Cu 29th) and the process which produce energy (from 235 U). Nevertheless, the reason there is such big difference is that in the second part you are considering nuclear reaction which only has respect on the nuclear structure and not on the electronic one as is considered in the ionization part. The nuclear energy levels have typical energy spacing in the order of MeV.
Btw it's 200 MeV not meV, I know it's typo but it may cause people not familiar with nuclear reaction to understand that the meV is correct.
The electron capture of a ## Cu^{+29} ## ion would produce that much energy. For a first posting, I think the OP came up with something interesting. :)
 
Charles Link said:
The electron capture of a Cu+29 Cu^{+29} ion would produce that much energy.
I see, but it's not like we can mine natural Cu 29th from nature like we do 235 U, despite the low abundance.
 
blue_leaf77 said:
I see, but it's not like we can mine natural Cu 29th from nature like we do 235 U, despite the low abundance.
I think the OP is basically comparing the .0114 MeV per atom from this process, if we could get it to occur, with the 200 MeV for a Uranium atom. And yes, like you stated, the process consumes energy, so there is really no practical use for this=it's simply a calculation that can be done with the Bohr model of the atom for ## Z=29 ##. To produce a ## Cu^{+29} ## ion, the first 28 electrons would need to be removed, and then the 29th one would require this energy to remove it, which is readily calculated.
 
I'm tryin to say that if we give that much energy to Cu by nuclear reaction then we could create that :)
 
Go Ved said:
I'm tryin to say that if we give that much energy to Cu by nuclear reaction then we could create that :)
Theoretically that should be possible but I have no idea whether someone has tried this. Recently however, physicists have been able to create bare atoms using photoionization by free-electron laser such as this one. Also if my memory serves me right, the professor in the class I once took in my college spoke about creating bare atoms in a collider or so. Again, I am certain though if that was what he really said back then.
 
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