Polarizability of Liquid Helium

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The polarizability of liquid helium is calculated using the formula alpha = (3(permittivity of free space))/N) / ((X-1)/(X+2)), where N is determined from the density and molecular volume. Given a density of 145 kg/m^3 and a relative permittivity of 1.0566, N is calculated to be approximately 2.18 x 10^28 molecules/m^3. Substituting these values into the equation yields a polarizability of about 2.26 x 10^-41. The accuracy of this result is uncertain due to the lack of prior experience in similar calculations. Further verification or guidance is sought to confirm the correctness of the result.
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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??
 
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well that's great, thanks for all the help :/
 
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