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Degenerate Fermi temperature of 2 fermionic gases |
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| Apr27-09, 07:21 AM | #1 |
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Degenerate Fermi temperature of 2 fermionic gases
1. The problem statement, all variables and given/known data
Consider 2 fermionic gases, each made of 10^6 atoms occupying 10^-3 m^3 volume: tritium, the nucleus containing one proton and two neutrons, and potassium, 19 protons and 21 neutrons. Which gas do you have to cool more in order for a Fermi surface to form? Find the ratio of the temperatures where the 2 gases become degenerate. 2. Relevant equations Fermi Enegery = (h^2/8m)(3N/PiV)^(3/2) Fermi Temperature = Fermi Energy / k = (h^2/8km)(3N/PiV)^(3/2) 3. The attempt at a solution Hmmm...I think I am trying to solve for the Fermi temperature here... since both N and V are the same for both gases...if I just take the ratio of the Fermi temperature of the 2 gases I will only have the ratio of their masses left... Since Potassium is heavier than tritium, does it mean potassium will have to be cooled more to become degenerate? Is the ratio of their degenerate temperature simply the ratio of their masses? Thank you for the help!
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