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

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• Go Ved
In summary, the conversation discusses the 29th ionization energy of Cu and its conversion to MeV, as well as the energy produced by one uranium 235 atom. The differences in energy levels between electronic and nuclear processes are also mentioned. The possibility of creating bare atoms using photoionization or collider techniques is also briefly discussed.
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 !

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
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.

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. :)

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.

## 1. What is Cu's 29th ionization energy?

Cu's 29th ionization energy refers to the amount of energy required to remove the 29th electron from a neutral copper atom in its gaseous state.

## 2. Why is the 29th ionization energy of Cu significant?

The 29th ionization energy of Cu is significant because it represents the energy needed to remove an electron from the highest occupied energy level of a copper atom, which is a key factor in determining its chemical and physical properties.

## 3. What is the trend in Cu's 29th ionization energy?

The trend in Cu's 29th ionization energy is that it increases as you remove each successive electron. This is due to the decreasing shielding effect of inner electrons, making it more difficult to remove electrons from the atom.

## 4. How does Cu's 29th ionization energy compare to other elements?

Cu's 29th ionization energy is relatively high compared to other elements, as it requires a significant amount of energy to remove an electron from a transition metal like copper.

## 5. What factors can affect Cu's 29th ionization energy?

The factors that can affect Cu's 29th ionization energy include the nuclear charge of the atom, the distance of the electron from the nucleus, and the shielding effect of inner electrons. Additionally, the presence of multiple electrons can also affect the ionization energy.