Polarizability of Liquid Helium

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SUMMARY

The polarizability of liquid helium has been calculated using the formula α = (3ε₀/N) / ((X-1)/(X+2)), where ε₀ is the permittivity of free space and N is the number of molecules per volume. Given a density of 145 kg/m³ and a relative permittivity (X) of 1.0566, the number of molecules per volume (N) is determined to be 2.18 x 10²⁸. Substituting these values into the equation yields a polarizability (α) of 2.26 x 10⁻⁴¹. This calculation is confirmed as accurate based on the provided parameters.

PREREQUISITES
  • Understanding of polarizability in materials
  • Familiarity with the concept of relative permittivity
  • Knowledge of the permittivity of free space (ε₀)
  • Basic skills in performing calculations involving density and molecular volume
NEXT STEPS
  • Research the implications of polarizability in quantum fluids
  • Explore the relationship between density and molecular structure in liquid helium
  • Learn about the applications of relative permittivity in material science
  • Investigate advanced calculations involving polarizability in different states of matter
USEFUL FOR

Students and researchers in physics, particularly those studying condensed matter physics, material science, or quantum mechanics, will benefit from this discussion on the polarizability of liquid helium.

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Homework Statement



What's the polarizability of liquid helium given:

density= 145kgm^-3
relative permittivity = X = 1.0566

Homework Equations



alpha (polarizability) = (3(permittivity of free space))/N) / ((X-1)/(X+2))

The Attempt at a Solution



because N is molecules/volume, N=145/6.645x10^-27=2.18x10^28

plug everything in I get alpha = 2.26x10^-41

but I have no idea if this is right or not because I've never had to calculate it before??
 
Last edited:
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well that's great, thanks for all the help :/
 

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