Question on Lithium-6 superfluidity

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SUMMARY

Lithium-6 (Li-6) is classified as a fermion gas due to its odd number of fermionic components, totaling nine: three protons, three neutrons, and three electrons. The discussion highlights the superfluid behavior exhibited by Li-6 and poses a question regarding the theoretical superfluidity of Li-6 if the three electrons were removed, resulting in a bosonic nucleus with six fermionic entities (three protons and three neutrons). It is confirmed that experiments cannot produce a fluid from ions due to strong electrostatic repulsion.

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Salman2
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Li-6 atom is a fermion gas because it has an odd number of fermion entities: 3 P, 3 N, 3 e- = 9. It also has been reported that Li-6 shows superfluid behavior (wee wiki for superfluidity).

My question is, would superfluidity behavior be predicted for Li-6 if the three electrons (e-) were removed, making the remaining Li-6 nucleus a boson with even number of fermion entities, e.g., 3 p + 3 N = 6. Has this experiment been conducted ? Thanks.
 
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You won't be able to form a fluid out of ions: the electrostatic repulsion is too strong.
 

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