Particle Density Needed for BEC @ 100 nK, A=100

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SUMMARY

The discussion focuses on calculating the particle density required for Bose-Einstein condensation (BEC) at a temperature of 100 nanokelvin for atoms with a mass number of A = 100. The critical temperature formula used is ##\lambda = \sqrt{ \frac{2\pi \hbar^2}{m k_b T}}##, leading to a calculated thermal de Broglie wavelength of ##\lambda_{T_c} = 3.46 \times 10^{-6} m##. The critical density is derived using the formula ##n_c \approx 2.612 \frac{1}{\lambda_{T_c}^3}##, resulting in a density of ##n_c = 6.3 \times 10^{16}## particles per cubic meter. The calculations and approach are confirmed as correct by forum participants.

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Homework Statement


For the observation of quantum mechanical Bose-Einstein condensation, the interparticle
distance in a gas of noninteracting atoms must be comparable to the de Broglie
wavelength, or less. How high a particle density is needed to achieve these conditions
if the atoms have mass number A = 100 and are at a temperature of 100 nanokelvin?

Homework Equations


I found on the web that at the critical temp. for BEC, ##\lambda = \sqrt{ \frac{2\pi \hbar^2}{m k_b T}} ##

I also found a formula for the critical density for BEC at that temp., ##n_c \approx 2.612 \frac{1}{\lambda_{T_c}^3}##

The Attempt at a Solution


I took m = particle mass ## \approx 100 \times m_p = 1.67 \times 10^{-25} kg##
Mechanically plugging the numbers in I got ##\lambda_{T_c} = 3.46 \times 10^{-6} m ##, and ## n_c = 6.3 \times 10^{16} ## particles per ## m^3##

I'd appreciate if someone could tell me if I am on the right track? The numbers seem OK although I think the ##\lambda## looks high?
 
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The approach is correct. I haven't checked the numbers, but they seem reasonable.
 
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