Q. 7.67 - An Introduction to Thermal Physics

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 $50\mu K$.

Step-by-Step Solution

Answer

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

Step 1. Given information

Condensation temperature for the substance, 

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

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

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

$N\text{ =number of the particles}$

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

  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 T c = 0 . 527 h 2 2 π m k N V 2 3

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" style="max-width: none; vertical-align: -19px;" $=50.96\times {10}^{-6} \mathrm{K}\left(\frac{1\mu \mathrm{K}}{1\times {10}^{-6} \mathrm{K}}\right)$

$=50.96\mu \mathrm{K}$