Question

Problem 7.67. In the first achievement of Bose-Einstein condensation with atomic hydrogen, a gas of approximately $2\times {10}^{10}$atoms was trapped and cooled until its peak density was$1.8\times {10}^{14}atoms/c{m}^3$. Calculate the condensation temperature for this system, and compare to the measured value of$50\mu K$.

Short answer

The condensation temperature for the atomic hydrogen system is$50.96\mu K$

Step-by-step solution

Step 1. Given information

Condensation temperature for the substance,

$T_c=0.527\left(\frac{h^2}{2\pi mk}\right){\left(\frac{N}{V}\right)}^{\frac{2}{3}}$

${\text{Planck's constant,h=6.63×10}}^{-34}$

${\text{mass of the particle,m=1.67×10}}^{-27}\text{kg}$

$N\text{=number of theparticles}$

$V\text{=volume of the substance.}$

Density of atomic hydrogen$=$$\frac{N}{V}=1.8\times {10}^{14}\text{atoms}/{\mathrm{cm}}^3$$=1.8\times {10}^{20}atoms/m^3$

Step 2.  Substituting the values of h, N/V, m, k  in the equation  Tc=0.527h22πmkNV23

we get,

$T_c=(0.527)\frac{{\left(6.63\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)}^2}{\left(2\pi \left(1.67\times {10}^{-27}\mathrm{kg}\right)\right)\left(1.381\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)}{\left(1.8\times {10}^{20}\text{atoms}/{\mathrm{m}}^3\right)}^{\frac{2}{3}}$

$=50.96\times {10}^{-6}\mathrm{K}$

$=50.96\times {10}^{-6}\mathrm{K}\left(\frac{1\mu \mathrm{K}}{1\times {10}^{-6}\mathrm{K}}\right)$

$=50.96\mu \mathrm{K}$

The value of condensate temperature for the atomic hydrogen system is $50.96\mu K$,

which is approximately equal to the measured value of Bose-Einstein condensation with atomic hydrogen$\left(50\mu K\right)$

width="220">=50.96×10-6K1μK1×10-6K

$=50.96\mu \mathrm{K}$