Degenerate Fermi temperature of 2 fermionic gases

In summary, we calculated the Fermi temperature for tritium and potassium and found that potassium needs to be cooled more in order for a Fermi surface to form. We also found that the ratio of the temperatures at which the two gases become degenerate is approximately 1.10.
  • #1
Odyssey
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Homework Statement



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.

Homework Equations



Fermi Enegery = (h^2/8m)(3N/PiV)^(3/2)
Fermi Temperature = Fermi Energy / k = (h^2/8km)(3N/PiV)^(3/2)

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! :smile:
 
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  • #2


Hello, thank you for your question! First, let's clarify some concepts. The Fermi temperature is the temperature at which the Fermi-Dirac distribution function becomes valid and the gas becomes degenerate. This is not the same as the Fermi energy, which is the energy of the highest occupied state at absolute zero temperature.

Now, to answer your question, we need to calculate the Fermi temperature for both gases. Using the Fermi energy equation, we get:

For tritium:
Fermi Energy = (6.626 x 10^-34 J s)^2 / (8 x 3 x 9.109 x 10^-31 kg) x (3 x 10^6 / 3.14 x 10^-3 m^3)^(3/2) = 1.06 x 10^-20 J
Fermi Temperature = 1.06 x 10^-20 J / (1.38 x 10^-23 J/K) = 7.68 x 10^2 K

For potassium:
Fermi Energy = (6.626 x 10^-34 J s)^2 / (8 x 3 x 3.274 x 10^-25 kg) x (3 x 10^6 / 3.14 x 10^-3 m^3)^(3/2) = 9.58 x 10^-20 J
Fermi Temperature = 9.58 x 10^-20 J / (1.38 x 10^-23 J/K) = 6.95 x 10^2 K

As we can see, the Fermi temperature for potassium is lower than that of tritium. This means that potassium needs to be cooled more in order for a Fermi surface to form.

To find the ratio of the temperatures at which the two gases become degenerate, we simply take the ratio of their Fermi temperatures:

Ratio = 7.68 x 10^2 K / 6.95 x 10^2 K = 1.10

Therefore, the temperature at which potassium becomes degenerate is about 10% lower than that of tritium.

I hope this helps! Let me know if you have any other questions.
 
  • #3




Yes, you are correct in your approach. The ratio of the degenerate temperatures of the two gases will simply be the ratio of their masses. This can be seen from the equation for Fermi temperature, which is inversely proportional to the mass. Since potassium has a higher mass than tritium, it will need to be cooled more in order for a Fermi surface to form. The exact ratio of the degenerate temperatures can be calculated by dividing the Fermi temperatures of the two gases, which would be (19/3)^(3/2) = 24.5. This means that the degenerate temperature of potassium will be approximately 24.5 times higher than that of tritium.
 

1. What is the degenerate Fermi temperature of 2 fermionic gases?

The degenerate Fermi temperature is the temperature at which fermionic gases exhibit quantum degeneracy, meaning that most of the particles are in their lowest possible energy state. It is a characteristic temperature for a system of fermions, such as electrons in a metal.

2. How is the degenerate Fermi temperature calculated?

The degenerate Fermi temperature can be calculated using the formula kBTF = (h^2/2m)(3N/8πV)2/3, where kB is the Boltzmann constant, h is the Planck's constant, m is the mass of the fermion, N is the number of particles, and V is the volume of the system.

3. What is the significance of the degenerate Fermi temperature?

The degenerate Fermi temperature is an important characteristic of fermionic systems because it marks the transition from classical behavior to quantum behavior. Below this temperature, the fermions start to exhibit wave-like properties and behave differently from classical particles.

4. How does the degenerate Fermi temperature affect the behavior of fermionic gases?

Below the degenerate Fermi temperature, fermionic gases exhibit properties such as zero-point energy, Bose-Einstein condensation, and superfluidity. These properties are a result of the particles being in their lowest possible energy state and behaving collectively as a wave instead of individual particles.

5. Can the degenerate Fermi temperature be reached in real-life systems?

Yes, the degenerate Fermi temperature has been observed in various physical systems such as ultracold atomic gases and white dwarf stars. It is also relevant in fields such as condensed matter physics and astrophysics.

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