Calculating Broglie Wavelength for Helium Superfluid

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In summary, helium can become a superfluid at temperatures below 2.18 K, exhibiting no viscosity. This phenomenon can only be explained through quantum physics and occurs when the de Broglie wavelength of a helium atom is comparable to the interatomic spacing of the fluid. The de Broglie wavelength is given by the equation lambda_deBroglie = h/p, where h is Planck's constant and p is the particle's momentum. This concept is covered in modern physics and can be found in the Fundamentals of Physics textbook. Additionally, thermal velocity distributions and Bose-Einstein statistics are related to this topic. The equation for de Broglie wavelength may also be used for multiple choice questions, with m and v being changed.
  • #1
glueball8
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Helium becomes a super°uid at temperatures T <
2:18 K. A super°uid °ows with no viscosity. This behaviour
can only be explained using quantum physics and it can
only happen if the de Broglie wavelength of a helium atom,
of mass m, is comparable to the inter-atomic spacing of the
°uid. What could be an expression for
the de Broglie wavelength?

Can anyone tell me where to learn this?

I'm currently working on Fundamentals of Physics, would I understand this when I finish the book? What level is this?

Thanks in advance
 
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  • #2
The debroglie wavelength is a concept in modern/ quantum physics. If you have an edition of Fundamentals that includes a section on modern physics, then the book should cover the DeBroglie wavelength.

In short, the Debroglie wavelength expresses the wave-particle nature of matter by assigning a wavelength to matter particles. A particle's deBroglie wavelength is given by:

[tex]\lambda_{deBroglie} = \frac{h}{p}[/tex]

where p is the particle's momentum, and h is Planck's constant.

Also, for an introductory summary of the topic:

http://hyperphysics.phy-astr.gsu.edu/Hbase/debrog.html
 
  • #3
I hadn't thought of super-fluids in that way..

The OP may also want to google for thermal velocity distributions (and Bose-Einstein statistics).
 
  • #4
oh I know that equation and stuff but I can't get an answer. Its a mutiple choice and I think the (A) Based on it looks like the mv is just changed {answers shown below on the pic I think}.

Is K a constant?
 
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Related to Calculating Broglie Wavelength for Helium Superfluid

1. What is the Broglie wavelength and how is it related to helium superfluid?

The Broglie wavelength is a quantum mechanical concept that describes the wavelength of a particle in motion. In the context of helium superfluid, it refers to the wavelength of the atoms in the superfluid state. It is related to helium superfluid because the superfluid state is a quantum phenomenon where individual helium atoms behave as a single entity with a collective wave-like behavior.

2. How is the Broglie wavelength of helium superfluid calculated?

The Broglie wavelength of helium superfluid can be calculated using the de Broglie equation, which states that the wavelength (λ) of a particle is equal to Planck's constant (h) divided by its momentum (p). In the case of helium superfluid, the momentum is equal to the mass of the helium atom (m) times its velocity (v). Therefore, the Broglie wavelength can be calculated as λ = h/mv.

3. What is the significance of calculating the Broglie wavelength for helium superfluid?

The calculation of the Broglie wavelength for helium superfluid is significant because it helps to understand the quantum behavior of the superfluid state. It also provides insight into the collective motion of the helium atoms in the superfluid, which is essential for studying and predicting the properties of helium superfluid.

4. How does the Broglie wavelength of helium superfluid differ from that of normal helium?

The Broglie wavelength of helium superfluid is significantly longer than that of normal helium. This is because in the superfluid state, the helium atoms are in a coherent state, and their collective motion results in a larger momentum, hence a longer wavelength. In contrast, normal helium atoms have a shorter wavelength due to their individual and random motion.

5. Can the Broglie wavelength of helium superfluid be experimentally measured?

Yes, the Broglie wavelength of helium superfluid can be experimentally measured using techniques such as neutron scattering or helium atom scattering. These experiments involve measuring the diffraction pattern of the superfluid, which can then be used to calculate the Broglie wavelength. These measurements have been crucial in confirming the quantum nature of helium superfluid and its collective behavior.

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